Subject: Optional Mathematics

The collection of numbers in which numbers are arranged in a certain rule is called a sequence, when the numbers of the sequence are connected by signs '+' or '-' then it is called series.

A Sequence is a list of things (usually numbers) that are in order. let us consider the following of numbers.

(i) 1, 4, 7, 10,.....

(ii) 20, 18, 16, 14,....

(iii) 1, 3, 9, 27, 81,....

(iv) 1, 2, 3, 4, ...

We observe that each term after the first term

(i) is formed by adding 3 to the preceding term;

(ii) is formed by subtracting 2 from the preceding term;

(iii) is formed by multiplying the preceding term by 3; each term in

(iv) is formed by squaring the natural numbers 1, 2, 3, 4,.....

In all the above case, we see that set of number follow a certain rule and we can easily say what number will come next to given number. thus, the numbers come in succession in accordance with a certain rule or low. A succession of numbers formed and arranged in a definite order according to a certain definite rule is called a sequence. the successive number in a sequence are called its terms.

A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.

eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.

The successive numbers forming the series are called the terms of the series and the successive terms are denoted by t_{1}, t_{2}, t_{3},....., t_{n },which denotes the 1^{st}, 2^{nd}, 3^{rd}, ...... n^{th} term respectively. The n^{th} term, t_{n}, of a series, is called its general term. Thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.

A sequence of number is said to be a progression if the difference or ratio between its two successive terms is constant throughout the whole sequence. An example of progression is as follows.

(i) 1, 3, 5, 7,..... (ii) 1, 3, 9, 27,.....

In (i), the difference between two successive terms is equal to 2.

In (ii), the ratio of two successive terms is equal to 3.

Progression is divided into following two types.

(i) Arithmetic progression

(ii) Geometric progression

A sequence is called an arithmetic progression if the difference between its two successive terms is constant throughout the whole sequence. An arithmetic progression can be denoted by A.P. The constant number obtained by subtracting succeeding term from its preceding term is called the common difference.

**For example:-**

(i) 1, 3, 5, 7, 9,....

(ii) 15, 12, 9, 6,......

From (i), we find that

second term - first term = 3 -1 = 2,

third term - second term = 5 - 3 = 2,

fourth term - third term = 7 - 5 = 2 and so on.

From (ii), we find that

second term - first term = 12-15 = -3,

third term - second term = 9 - 12 = -3,

fourth term - third term = 6 - 9 = -3 and so on.

**Hence,** the common difference 'd' is calculated by

d = succeeding term - proceeding term = t_{n} - t_{n-1}

Here, we find that the difference between two successive terms, in both sequences, are same or constant. So, such sequence is called arithmetic progressions. The C.D. of the two progressions are 2 and -3 respectively. Thus, arithmetic progressions is a series in which the successive terms increase or decrease by the common difference.

To find the n^{th }term of an A.P.

Let, t_{1} be the first term, n be the number of terms and 'd' the common difference of an A.P. respectively. Then,

t_{1 }= a = a + (1-1)d

t_{2 }= a + d = a + (2-1)d

t_{3 }= a + 2d = a + (3-1)d

t_{4 }= a + 3d = a + (4-1)d

In general, t_{n }= a + (n-1)d

**Formula:** If t_{n }denotes the n^{th} term, of the arithmetic progression whose first term, common term and number of terms are a, d and n respectively.

With this term, arithmetic sequence and series can be written as:

**Arithmetic sequence:** a, a+d, a+2d, a+3d, ............

**Arithmetic series:** a+ (a+d) + (a+2d) + (a+3d), ..........

The terms between the arithmetic progression are known as arithmetic mean. Such as the three numbers 2, 4, 6 are in arithmetic progression with the common difference d = 2, then 4 is the arithmetic mean between 2 and 6.

For example:

Let a, b,c are in arithmetic progression

b-a = c-b

or, b+b = a+c

or, 2b = a+c

or, b = \(\frac{a+c}{2}\)

Hence the arithmetic mean between a and c is (\(\frac{a+c}{2}\))

**n Arithmetic Means between two numbers a and b**

Let m_{1}, m_{2}, m_{3}, .........m_{n} be the arithmetic means between the given term a and b. Then, a, m_{1}, m_{2}, m_{3}, .........m_{n}, b are in A.P.

Here, numbers of arithmetic means = n

So, numbers of terms of A.P. = n+2

It means,

b = (n+2)^{th} term of AP

or, b = a + (n+2-1)d, where d is common difference

or, b =a + (n+1)d

or, (n+1)d = b-a

∴ d= \(\frac{b-a}{n+1}\)

Now, m_{1} = a+d = a + \(\frac{b-a}{n+1}\)

m_{2} = a + 2d = a + \(\frac{2(b-a)}{n+1}\)

m_{3} = a + \(\frac{3(b-a)}{n+1}\)

.............................................

m_{n} = a + \(\frac{n(b-a)}{n+1}\)

Let us consider an arithmetic series

a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l

Here, the first term = a,

first term = a,

common difference = d,

number of terms= n,

last term (t_{n}) = l

the term before last term = l-d

if the sum of n terms is denoted by S_{n}, then

S_{n} = a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l .... (i)

Writing term in the reverse order,

S_{n} = l + (l-d) + (l-2d) + ...... + (a+3d) + (a+2d) + (a+d) + a .... (ii)

Adding the corresponding terms of (i) and (ii)

\(\frac{S_n \;= \;a \;+\; (a+d)\; +\; (a+2d)\; +\; (a+3d)\; + \;......\; + \;(l-2d) \;+\; (l-d)\; + \;l\\S_n\;=\;l\; +\;(l-d)\;+\;(l-2d)\;+\;......\;+\;(a+3d)\;+\;(a+2d)\;+\;(a+d)\;+a\:}{2S_n\;= \;(a+l) \;+ \;(a+l)\; + \;(a+l)\; +\; ............ \;+\; (a+l)\; + \;(a+l)\; +\; (a+l)}\)

= n times (a+l)

= n (a+l)

= \(\frac{n}{2}\)(a+l)

But, the last term l = a + (n-1)d

So, S_{n} = \(\frac{n}{2}\)(a+l) = \(\frac{n}{2}\)[a+a+(n-1)d] = \(\frac{n}{2}\)[2a+(n-1)d]

∴ S_{n} = \(\frac{n}{2}\)[2a+(n-1)d]

Thus, if d is unknown, S_{n} = \(\frac{n}{2}\)(a+l)

And, if l is unknown, S_{n} = \(\frac{n}{2}\)[2a+(n-1)d]

**1. Sum of first n natural numbers**

the numbers 1, 2, 3, 4, ......, n are called the first n natural numbers.

Here, first term (a) = 1

Common difference (d) = 2-1 = 1

Number of terms (n) = n

If S_{n} denotes the sum of these first n natural numbers, then

S_{n} = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.1+(n-1).1] = \(\frac{n}{2}\)[2+n-1] = \(\frac{n}{2}\)(n+1)

**2. Sum of first n odd numbers**

1, 3, 5, 7, ......., (2n-1) are the first n odd numbers.

Here, first term (a) = 1

Common difference (d) = 3-1 = 2

Number of terms (n) = n

If S_{n} denotes the sum of these first n odd numbers, then

S_{n} = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.1+(n-1).2] = \(\frac{n}{2}\)(2+2n-2) = \(\frac{n}{2}\) × 2n = n^{2}

**3. Sum of first n even numbers**

2, 4, 6, 8, ......., 2n are the first n even numbers.

Here, first term (a) = 2

Common difference (d) = 4-2 = 2

Number of terms (n) = n

If S_{n} denotes the sum of these first n even numbers, then

S_{n} = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.2+(n-1).2] = \(\frac{n}{2}\)(4+2n-2) = \(\frac{n}{2}\)(2n-2) = n(n+1)

In mathematics, a **geometric progression**, also known as a **geometric sequence**, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common* ratio*. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly,10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.

Examples of a geometric sequence are powers *r*^{k} of a fixed number *r*, such as 2^{k }and 3^{k}. The general form of a geometric sequence is

a, ar, ar^{2}, ar^{3}, ............ar^{n}

where *r* ≠ 0 is the common ratio and *a* is a scale factor equal to the sequence's start value.

**General Term or n ^{th} term of G.P.**

We use the following notations for terms and expression involved in a geometrical progression:

The first term = a the n^{th} term = toor b

The number of terms = n Common ratio = r.

The expression **ar ^{n-1}** gives us the n

**∴t _{n} = ar^{n-1}**

With the help of this general term, geometric sequence and series can be written in the following ways:

Geometric Sequence: a, ar, ar^{2}, ar\(^3\), .....

Geometric series: a + ar + ar^{2} + ar\(^3\) + ........

If the three numbers are in G.P., then the middle term is called the geometric mean of the other two terms. In other words, the geometric mean of two non-zero numbers is defined as the square root of their product.

Let a, G, b be three numbers in G. P., then the common ratio is the same i.e.

\(\frac{G}{a}\) =\(\frac{b}{G}\)

or, G^{2} = ab

or, G =\(\sqrt{a}{b}\)

Hence, the geometric mean of two numbers a and b is the square root of their product i.e. \(\sqrt{a}{b}\).

So, the geometric mean between two number 2 and 8 is G =\(\sqrt ab\) = \(\sqrt2*8\) = \(\sqrt16\) = 4.When

When any number of quantities are in G. P., all the terms in between the first and last terms are called the geometric means between these two quantities.

Here, G_{n} = ar^{n} = a \(\begin{pmatrix}b\\a\\ \end{pmatrix}\)^{\(\frac{n}{n + 1}\)}

*"Arithmetic mean (A. M) is always greater than Geometric mean (G. M.) between two position real unequal numbers".*

Let us consider two numbers 2 and 8

Here, AM between 2 and 8 =\(\frac{2 + 8}{2}\) = 5

GM between 2 and 8 = \(\sqrt 2 * 8\)) = 4

∴ AM > GM.

**The sum of n terms of a series in G. P.**

Let us consider geometric series a + ar + ar^{2} + ar\(^3\) + .......+ ar^{n -3}+ ar^{ni2}+ ar^{n-1}

Here, first = a common ratio = r number of terms = n last term (l) = ar^{n-1}

∴ S_{n} = \(\frac{lr - a}{r - 1}\)

If the number of terms is odd, we take the middle term as **a**and the common ratio as** r**. If the number of terms is even, we take \(\frac{a}{r}\) and **ar** as the middle terms and** r ^{2}** as the common ratio.

A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.

eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.

The successive numbers forming the series are called the terms of the series and the successive terms are denoted by$$ t_1, t_2, t_3,....., t_n,$$ where$$ t_1, t-2, t-3, ........ tn$$ denote the$$ 1^st, 2^nd, 3^rd ,...... .n^{nt}$$ term respectively. The n^th term, t_n, of a series, is called its general term. thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.

The terms between the first term and last term of an A.P are called arithmetic mean.,

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Write \(\sum_{n=2}^5 (n^2- 3)\) in an expanded form and also find its value.

\(\sum_{n=2}^5 (n^2- 3)\)

Putting n = 2, 3, 4, 5

= (2^{2} - 3) + (3^{2} - 3) + (4^{2} - 3) + (5^{2} - 3)

= (4 - 3) + (9 - 3) + (16 - 3) + (25 - 3)

= 1 + 6 +13 +22

= 42 _{Ans}

u_{1}, u_{2}, u_{3} ...................... are the terms of a sequence where u_{n + 1} = 1 - \(\frac {1}{u_n}\). If u_{1} = 3, find the values of u_{2} and u_{3}.

u_{n + 1} = 1 - \(\frac {1}{u_n}\) and u_{1} = 3, u_{2} and u_{3} = ?

If n = 1,u_{n + 1} = 1 - \(\frac {1}{u_n}\)

u_{1 + 1} =1 - \(\frac {1}{u_1}\)

or, u_{2} = 1 - \(\frac 13\) = \(\frac {3 - 1}{3}\) = \(\frac 23\)

If n = 2, u_{n + 1} =1 - \(\frac {1}{u_n}\)

u2 + 1 =1 - \(\frac {1}{u_2}\)

or, u_{3} = 1 - \(\cfrac{1}{\cfrac{2}{3}}\) = 1 - \(\frac 32\) = \(\frac {2 - 3}{2}\) = -\(\frac 12\)

∴ u_{2} = \(\frac 23\) and u_{3} = -\(\frac 12\) _{Ans}

What are the sequences and series? Illustrate each of them with an example.

Sequence: A sequence is a set of numbers or quantities, which are formed according to some governed laws.

i.e. 1, 4, 16 ............................ and 2, 3, 4, 5...................

Series: The sum of the term of a sequence is called a series.

i.e. 1 + 4 + 16 + ....................... and 2 + 3 + 4 + 5 + ......................

Write \(\sum_{n=4}^7 (3n- 2)\) in an expanded form and also find its value.

Here,

\(\sum_{n=4}^7 (3n- 2)\)

where: n = 4, 5, 6, 7

\(\sum_{n=4}^7 (3n- 2)\)

= (3×4-2) + (3×5-2) +(3×6-2) +(3×7-2)

= (12-2) + (15-2) + (18-2) + (21-2)

= 10 + 13 + 16 + 19

= 58 _{Ans}

Write \(\sum_{k=3}^7 (k^2 + 1)\) in an expanded form and also find its value.

\(\sum_{k=3}^7 (k^2 + 1)\)

= (3^{2} + 1) +(4^{2} + 1) +(5^{2} + 1) +(6^{2} + 1) +(7^{2} + 1)

= (9 + 1) +(16 + 1) +(25 + 1) +(36 + 1) +(49 + 1)

= 10 + 17 + 26 + 37 + 50

= 140 _{Ans}

Find the sum of 1 + 4 + 7 + ..................... + 34

Here,

First term (a) = 1

Second term (b) = 4

Common Difference (d) = b - a = 4 - 1 = 3

Last term (l) = 34

We know,

last term (l) = a + (n - 1)d

or, 34 = 1 + (n - 1) 3

or, 34 = 1 + 3n - 3

or, 3n - 2 = 34

or, 3n = 34 + 2

or, n = \(\frac {36}{3}\)

∴ n = 12

Again,

S_{n} = \(\frac n2\) [a + l]

S_{n} = \(\frac{12}{2}\) [1 + 34] = 6× 35 = 210

∴ S_{n} = 210 _{Ans}

Find the sum of 2 + 7 + 12 + ........ upto 16 terms.

Here,

first term (a) = 2

second term (b) = 7

common difference (d) = b - a = 7 - 2 = 5

number of items (n) = 16

sum (S_{16}) = ?

We know that,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

S_{16} = \(\frac {16}{2}\) [2 × 2 + (16 - 1) 5] = 8 [4 + 15× 5] = 8 [4 + 75] = 8 × 79 = 632 _{Ans}

Which term of the series 3\(\frac12\) + 1 - 1\(\frac 12\) - 4 ............. is -21\(\frac 12\) ?

Here,

first term (a) =3\(\frac12\) = \(\frac 72\)

second terrn (b) = 1

common difference (d) = b - a = 1 - \(\frac 72\) = \(\frac {2 - 7}{2}\) = -\(\frac 52\)

no. of items (n) = ?

last term (l) =-21\(\frac 12\) = -\(\frac {43}{2}\)

We know that,

l = a + (n - 1) d

or,-\(\frac {43}{2}\) = \(\frac 72\) + (n - 1) (-\(\frac 52\))

or, - 43 = 7 - 5n + 5

or, 5n = 12 + 43

or, n = \(\frac {55}{5}\) = 11

∴-21\(\frac 12\) is the 11^{th} term. _{Ans}

Find the 7^{th} term of an Arithmetic sequence in which the first term is 20 and common difference is -3.

Here,

first term (a) = 20

common difference (d) = -3

numbers of term (n) = 7

seventh term (t_{7}) = ?

We know that,

t_{n}= a + (n - 1) d

t_{7} = 20 + (7 - 1) (-3) = 20 + 6 (-3) = 20 - 18 = 2 _{Ans}

Find the sum of the series 2 + 4 + 6 + to 20 term.

Here,

first term (a) = 2

second term (b) = 4

common difference (d) = b - a = 4 - 2 = 2

no. of terms (n) = 20

sum (S_{20}) = ?

We know that,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

S_{20} = \(\frac {20}{2}\) [2× 2 + (20 - 1)× 2] = 10 [4 + 38] = 10× 42 = 420 _{Ans}

Find the 20^{th} term of 1, 3, 5, 7 ................

Here,

first term (a) = 1

second term (b) = 3

common difference (d) = b - a = 3 - 1 = 2

number of items (n) = 20

twenty term (t_{20}) = ?

We know,

t_{n}= a + (n - 1) d

t_{20} = 1 + (20 - 1) 2

t_{20} = 1 + 19× 2 = 1 + 38 = 39 _{Ans}

How many terms are there in sequence 25, \(\frac {45}{2}\), 20, ........................, - 15?

Here,

first term (a) = 25

second term (b) = \(\frac {45}{2}\)

common difference (d) = b - a = \(\frac {45}{2}\) - 25 = \(\frac {45 - 50}{2}\) = -\(\frac 52\)

last term (l) = - 15

number of terms (n) = ?

We know that,

l = a + (n - 1) d

or, - 15 = 25 + (n - 1) - \(\frac 52\)

or, - 15 - 25 = -\(\frac 52\)n + \(\frac 52\)

or, - 40 - \(\frac 52\) = -\(\frac 52\)n

or - \(\frac {80 - 5}{2}\) = -\(\frac {5n}{2}\)

or, -\(\frac {85}{2}\)× -\(\frac {2}{5}\) = n

∴ n = 17

∴ The number of term = 17. _{Ans}

If 3, x, y, -9 are the terms of an A.S., find the value of x and y.

Here,

first term (a) = 3

last term (l) = -9

number of terms (n) = 4 [including a and b]

common difference (d) = \(\frac {b - a}{n - 1}\) = \(\frac {-9 - 3}{4 - 1}\) = -\(\frac {12}{3}\) = - 4

x = a + d = 3 - 4 = - 1

y = a + 2d = 3 + 2× -4 = 3 - 8 = - 5

∴ x = -1 and y = -5 _{Ans}

Find the sum of the 2 - 9 - 20 - ................. 130.

Here,

first term (a) = 2

second term (b) = - 9

common difference (d) = b - a = -9 - 2 = -11

last term (l) = -130

number of terms (n) = ?

We know,

l = a + (n - 1) d

or, -130 = 2 + (n - 1) (-11)

or, -130 - 2 = - 11n + 11

or, -132 - 11 = -11n

or, -11n = -143

or, n = \(\frac {-143}{-11}\)

∴ n = 13

Again, Sn = \(\frac n2\) (a + 1) = \(\frac {13}{2}\) (2 - 130) = 6.5× -128

∴ Sn = -832 _{Ans}

Find the first term of an arithmetic series whose sum of its first seven terms is 0 and the common difference is -3.

Here,

common difference (d) = -3

number of terms (n) = 7

sum (S_{7}) = 0

first term (a) = ?

We know,

Sn = \(\frac n2\) [2a + (n - 1) d]

or, 0 = \(\frac 72\) [2a + (7 - 1) (-3)]

or, 0× \(\frac 27\) = 2a + 6× (-3)

or, 0 = 2a - 18

or, 2a = 18

or, a = \(\frac {18}{2}\) = 9

∴ first term (a) = 9 _{Ans}

Find the number of terms \(\frac 14\), \(\frac 12\), 1, 2 ....................... 128.

Here,

first term (a) = \(\frac 14\)

second term (b) = \(\frac 12\)

common ratio (r) = \(\frac ba\) =\(\cfrac{\frac{1}{2}}{\cfrac{1}{4}}\) = \(\frac 12\)× \(\frac 41\) = 2

last term (l) = 128

number of terms (n) = ?

We know that,

l = ar^{n-1}

or, 128 = \(\frac 14\) (2)n-1

or, 2^{7}× 2^{2} = 2^{n-1}

or, 2^{n-1} = 2^{9}

or, n -1 = 9

or, n = 9 + 1

∴ n = 10

∴ The number of term = 10 _{Ans}

Find the geometric mean between 16 and 36.

Here,

first term (a) = 16

last term (l) = 36

We know that,

geometric mean (G.M.) =\(\sqrt {ab}\)

G.M. =\(\sqrt{({16}×{36})}\)

= \(\sqrt {({4^2}×{6^2})}\)

= \(\sqrt {({4×6})^2}\)

= \(\sqrt {({24})^2}\)

∴ Geometric mean = 24 _{Ans}

Find the eighth term of a geometric sequence \(\frac 13\), 1, 3, ....................

Here,

first term (a) = \(\frac 13\)

second term (b) = 1

common ratio (r) = \(\frac ba\) = \(\cfrac {1}{\cfrac 13}\) = 3number of terms (n) = 8

number of terms (n) = 8

eighth term (t_{8}) = ?

Using formula,

t_{n} = ar^{n-1}

t_{8} = (\(\frac 13\)) (3)^{8-1} = \(\frac 13\)× 3× 3^{6} = 729 _{Ans}

Find the number of terms of the 3, -1, \(\frac 13\), ....................., -\(\frac {1}{81}\).

Here,

first term (a) = 3

second term (b) = 1

common ratio (r) = \(\frac ba\) = -\(\frac 13\)

last term (l) = -\(\frac {1}{81}\)

number of terms (n) = ?

We know that,

l = ar^{n-1}

or, -\(\frac {1}{81}\) = 3 (-\(\frac 13 \))^{n-1}

or, - \(\frac {1}{3^4× 3}\) = (-\(\frac 13\))^{n-1}

or, (-\(\frac 13\))^{5} =(-\(\frac 13\))^{n-1}

or, n - 1 = 5

or, n = 5 + 1

∴ n = 6

∴ The number of term = 6 _{Ans}

Find the 5^{th} term of geometric series \(\frac 23\) - \(\frac 16\) + \(\frac {1}{24}\).

Here,

first term (a) = \(\frac 23\)

second term (b) = -\(\frac 16\)

common ratio (r) =\(\frac ba\) = \(\cfrac{\frac{-1}{6}}{\cfrac{2}{3}}\)= -\(\frac 16\) × \(\frac 32\) = -\(\frac 14\)

number of items (n) = 5

5^{th} term (t_{5}) = ?

We know that,

t_{n}= ar^{n-1}

or, t_{5} = \(\frac 23\)(-\(\frac 14\))^{5-1}

or, t_{5} = \(\frac 23\)×(-\(\frac 14\))^{4}

or, t_{5} = \(\frac 23\)× \(\frac {1}{16}\)×\(\frac {1}{16}\) = \(\frac {1}{384}\)

∴ fifth term = \(\frac {1}{384}\) _{Ans}

\(\frac 23\), x, \(\frac 32\), y are in G.S. find the value of x and y

Here,

first term (a) = \(\frac 23\)

second term (b) = x

third term (c) = \(\frac 32\)

We know that,

G.M. = \(\sqrt {ab}\)

x = \(\sqrt {(\frac 23) × (\frac 32)}\)

x = 1

r = \(\frac ba\) =\(\cfrac{1}{\cfrac{2}{3}}\) = \(\frac 32\)

y = ar = \(\frac 32\)× \(\frac 32\) = \(\frac 94\)

∴ x = 1 and y = \(\frac 94\) _{Ans}

Find the sum of 2 + 4 + 8 ........................... 10 term.

Here,

first term (a) = 12

second term (b) = 4 common ratio (r) = \(\frac ba\) = \(\frac 42\) = 2

number of terms (n) = 10

sum of the ten terms (S_{10}) = ?

We know that,

S_{n} = \(\frac {a(r^n - 1)}{r - 1}\) = \(\frac {2(2^10 - 1)}{2 - 1}\) =\(\frac {2(1024 - 1)}{1}\) = 2× 1023 = 2046

∴ The sum of tenth terms = 2046 _{Ans}

Find the sum of first five terms of 81 + 27 + 9 .......................

Here,

first term (a) = 81

second term (b) = 27

common ratio (r) = \(\frac ba\) = \(\frac {27}{81}\) = \(\frac 13\)

number of terms (n) = 5

sum of five terms (S_{5}) = ?

We know,

S_{n} = \(\frac {a(r^n - 1)}{r - 1}\)

S_{5} = \(\frac {81[(\frac 13)^5 - 1]}{\frac 13 - 1}\)

=\(\frac {81[(\frac {1}{243}) - 1]}{\frac {1 - 3}{3}}\)

=\(\frac {81 × (\frac {1 - 243}{243})}{\frac {-2}{3}}\)

= 81× -\(\frac {242}{243}\)× -\(\frac 32\)

= 121

∴ The sum of five terms = 121 _{Ans}

If the sum of first 6 terms of a G.P. is 9 times the sum of the first 3 terms find the common ratio.

Let,

first term = a

common ratio = r

From equation,

S_{6} = 9 S_{3}

or,\(\frac {a(r^6 - 1)}{r - 1}\) = 9 [\(\frac {a(r^3 - 1)}{r - 1}\)] [\(\because\)S_{n} =\(\frac {a(r^n - 1)}{r - 1}\)]

or, \(\frac {a[(r^3)^2 - (1)^2]}{r - 1}\)×\(\frac {r - 1}{a(r^3 - 1)}\) = 9

or, \(\frac{(r^3 + 1)(r^3 - 1)}{r^3 - 1}\) = 9

or, r^{3} = 9 - 1

or, r^{3} = 8

or, r^{3} = 2^{3}

∴ r = 2

∴ The common ratio = 2 _{Ans}

If the product of first 3 term in G.P. is 729, find the second term.

Let: the three terms of an G.P. are \(\frac ar\), a and ar

From Question,

\(\frac ar\)× a× r = 729

or, a^{3} = 9^{3}

∴ a = 9

∴ The second term = 9 _{Ans}

The AM of two numbers 6 and x is 30 find G.M.

Here,

first term (a) = 6

third term (b) = x

A.M. = 30

A.M. = \(\frac {a + b}{2}\)

or, 30 = \(\frac {6 + x}{2}\)

or, 60 - 6 = x

∴ x = 54

G.M. = \(\sqrt {ab}\)

G.M. = \(\sqrt {6 × 54}\)

G.M. = \(\sqrt {2^3 × 3^3 ×3^3}\)

G.M. = 18

∴ The G.M. = 18 _{Ans}

Insert 3 geometric means between \(\frac 19\) and 9.

Here,

first term (a) = \(\frac 19\)

last term (l) = 9

number of Geometric mean (n) = 3

total number of term (n) = 5

3 Geometric mean (m_{1}), (m_{2}), (m_{3}) = ?

l = ar^{n-1}

or, 9 = \(\frac 19\)r^{5-1}

or, 81 = r^{4}

or, 3^{4} = r^{4}

∴ r = 3

m_{1} = ar = \(\frac 19\)× 3 = \(\frac 13\)

m_{2} = ar^{2} = \(\frac 19\)× 32 = \(\frac 19\)× 9 = 1

m_{3} = ar^{3} = \(\frac 19\)× 33 = \(\frac 19\)× 27 = 3

∴ m_{1} = \(\frac 13\), m_{2} = 1, m_{3} = 3 _{Ans}

Prove that A.M. is greater than G.M.

If M and G be the arithmetic mean and geometric mean between two positive numbers a and b, then

m = \(\frac {a + b}{2}\) .................... (1)

G = \(\sqrt {ab}\) ................................(2)

Equation (1) - (2) we get

M - G = \(\frac {a + b}{2}\) - \(\sqrt {ab}\)

= \(\frac {a + b - 2\sqrt{ab}}{2}\)

= \(\frac {(\sqrt a)^2 - 2\sqrt a ⋅ \sqrt b + (\sqrt b)^2}{2}\)

= \(\frac 12\) (\(\sqrt a\) - \(\sqrt b\))^{2}

\(\sqrt a\) - \(\sqrt b\) gives positive value

∴ M - G = \(\frac {(\sqrt a - \sqrt b)^2}{2}\)≥ 0

M - G≥ 0 or, M ≥G

∴ A.M.≥ G.M. _{Ans}

Find the sum of the first n odd natural number.

Let:

S_{n} = 1, 3, 5, ..................... n

First term (a) = 1

Second term (b) = 3

Common difference (d) = b - a = 3 - 1 =2

number of items = n

We know,

S_{n} = \(\frac n2\)[2a + (n - 1) d]

or, S_{n} = \(\frac n2\)[ 2× 1 + (n - 1) 2]

or, S_{n} = \(\frac n2\)[2 + 2n - 2]

or, S_{n} = \(\frac n2\)× 2n

∴ S_{n} = n^{2}_{Ans}

How many numbers of terms are in G.P. whose first term is 1, common ratio 2 and sum 255?

Here,

first term (a) = 1

common ratio (r) = 2

sum (S_{n}) = 255

We know,S

S_{n} = \(\frac{a(r^n - 1)}{r - 1}\)

or, S_{n} =\(\frac{1(2^n - 1)}{2 - 1}\)

or,\(\frac{2^n - 1}{1}\) = 255

or, 2^{n} = 255 + 1

or, 2^{n} = 256

or, 2^{n} = 2^{8}

∴ n = 8

∴ The no. of terms = 8 _{Ans}

Insert 3 geometric means between 3 and 243.

Let:

first term (a) = 3

last term (b) = 243

no. of geometric means (n) = 3

common ratio (r) = ?

3 G.M are X_{1}, X_{2}, X_{3} = ?

r = (\(\frac ba\))^{\(\frac {1}{n+1}\)}= (\(\frac {243}{3}\))^{\(\frac {1}{3+1}\)}= (81)^{\(\frac 14\)}= (3)^{\(\frac {4×1}{4}\)} = 3

X_{1} = ar = 3× 3 = 9

X_{2} = ar^{2} =3× (3)^{2} = 27

x_{3} = ar^{3} =3× (3)^{3} = 81

∴ 9, 27, 81 _{Ans}

If 27, p, q, r, s, \(\frac {32}{9}\) are in G.P., find the value of p, q, r, and s.

Let:

first term (a) = 27

last term (l) = \(\frac {32}{9}\)

number of terms (n) = 4

common ratio (r) = (\(\frac ba\))^{\(\frac {1}{n+1}\)}

= (\(\cfrac{\frac{32}{9}}{27}\))^{\(\frac {1}{4+1}\)}

=( \(\frac {32}{9×27}\))^{\(\frac 15\)}

= (\(\frac 23\))^{\(\frac {5×1}{5}\)}

=\(\frac 23\)

p = ar = 27×\(\frac 23\) = 18

q = ar^{2}=27×\(\frac 23\)×\(\frac 23\) = 12

r = ar^{3} =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = 8

s = ar^{4} =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = \(\frac {16}{3}\)

∴ p = 18, q = 12, r = 8 and s = \(\frac {16}{3}\) _{Ans}

Find the sum of 2 - 6 + 18 - .................... to 6 terms.

Here,

first term (a) = 2

second term (b) = -6

common ratio (r) = \(\frac ba\) = \(\frac {-6}{2}\) = -3

number of terms (n) = 6

We know that,

S_{n} = \(\frac {a(1 - r^n)}{1 - r}\)

S_{n} = \(\frac {2(1 - (-3)^6)}{1 - (-3)}\)

S_{n} = \(\frac {2(1 -729)}{1 + 3}\)

S_{n} = \(\frac {2× (-729))}{4}\)

∴S_{n}= -364.5_{Ans}

Find the sum of the series: 3 + 6 + 12 + ............... + 1535.

Here,

first term (a) = 3

second term (b) = 6

common ratio (r) = \(\frac ba\) =\(\frac 63\) = 2

last term (l) = 1535

sum of the series (S_{n}) = ?

Using formula,

S_{n} = \(\frac{lr - a}{r - 1}\) = \(\frac {1535 × 2 - 3}{2 - 1}\) = \(\frac {3070 - 3}{1}\) = 3067

∴ Sum (S_{n}) = 3067 _{Ans}

The sum of first 4 terms of an A.P. is 26 and the sum of first 8 terms is 100. Find the sum of first 8 terms is 100. Find the sum of first 28 terms.∴

Let:

first term = a

common difference = d

number of terms (n) = 4

sum of 4 terms (S_{4}) = 26

We know that,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or, S_{4}= \(\frac 42\) [2a + (4 - 1) d]

or, 26 = 2[2a + 3d]

or, 2a + 3d = 13 ........................(1)

Again,

number of terms (n) = 8

sum of 8 terms (S_{8}) = 100

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or, S_{8}= \(\frac 82\) [2a + (8 - 1) d]

or, 100 = 4[2a + 7d]

or, 2a + 7d = 25 ........................(2)

Subtracting the equation (1) from equation (2)

2a | + | 7d | = | 25 |

2a | + | 3d | = | 13 |

- | - | - | ||

4d | = | 12 |

or, d = \(\frac {12}{4}\)

∴ d = 3

Substituting the value of d in equation (2)

2a + 7d = 25

or, 2a + 7× 3 = 25

or, 2a = 25 - 21

or, a = \(\frac 42\)

∴ a = 2

Sum of 28 terms (S_{28}) = ?

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or,S28 = \(\frac {28}{2}\) [2(2) + (28 - 1) d]

or,S28 = 14[4 + 27× 3]

or,S28 = 14 (4 + 81)

or,S28 = 14× 85

∴S28 = 1190 _{Ans}

How many terms of the series 27 + 24 + 21 + ................... must be taken in order that the sum may be 132? Explain the double answer.

Here,

first term (a) = 27

second term (b) = 24

common difference (d) = b - a = 24 - 27 = -3

Sum of the series (S_{n}) = 132

We know that,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or, 132 = \(\frac n2\) [2×27 + (n - 1) (-3)]

or, 264 = n[52 - 3n + 3]

or, 264 = n(57 - 3n)

or, 264 = 57n - 3n2

or, 3n^{2} - 57n + 264 = 0

or, 3(n^{2} - 19n + 88) = 0

or, n^{2} - 11n - 8n + 88 = 0

or, n(n - 11) -8(n - 11) = 0

or, (n - 11) (n - 8) = 0

Either, n - 11 = 0 ∴ n = 11

Or, n - 8 = 0 ∴ n = 8

∴n = 11 or 8

∴ The double answer of n show that there are two series in A.S. having the same sum.

The series to 8 terms 27 + 24 + 21 + 18 + 15 + 12 + 9 + 6 and the series to 11 terms is27 + 24 + 21 + 18 + 15 + 12 + 9 + 6 + 3 + 0 - 3

∴ S_{8} = S_{11} =132 _{Ans}

The 3^{rd} and 13^{th} terms of an arithmetic sequence are 40 and 0 respectively. Find the 25^{th} term.

Here,

third term (t_{3)} = 40

number of terms (n) = 3

Using formula,

t_{n}= a + (n - 1)d

t_{3} = a + (3 - 1)d

40 = a + 2d ............................(1)

Again,

thirteen term (t_{13}) = 0

number of terms (n) = 13

Using formula,

t_{n}= a + (n - 1)d

t_{13} = a + (13 - 1)d

0 = a + 12d ...........................(2)

Subtracting equation (2) from (1)

a | + | 2d | = | 40 |

a | + | 12d | = | 0 |

- | - | - | ||

- | 10d | 40 |

∴ d = -4

Substituting the value of d in equation (2)

a + 12d = 0

a + 12× (-4) = 0

a - 48 = 0

∴ a = 48

Now,

a = 48

d = -4

n = 25

t_{25} = ?

We know,

t_{n}= a + (n - 1)d

∴ t_{25} = 48 + (25 - 1)(-4) = 48 + 24 (-4) = 48 - 96 = - 48 _{Ans}

The sum of first three terms of an arithmetic series is 21. If the sum of first two terms is subtracted from the third term then it would be 9, find the three terms of the series.

Let:

first term = a - d

second term = a

third term = a + d

common difference = d

Let: the threeterms of A.S. are a-d, a, a+d

From first condition.

a - d + a + a+ d = 21........................(i)

or, 3a = 21

or, a = \(\frac {21}{3}\)

∴ a = 7

From second condition,

a + d - (a + a - d) = 9

or, a + d - 2a + d = 9

or, 2d - a = 9.......................... (ii)

Putting the value of a in equation (ii)

2d - 7 = 9

or, 2d = 9 + 7

or, d = \(\frac {16}{2}\) = 8

∴ first term = a - d = 7 - 8 = -1

second term = a = 7

third term = a + d = 7 + 8 = 15

∴Required termsare -1, 7, 15 _{Ans}

If the fourth term of an A.P. is 1 and the sum of its first eight terms is 18, find the tenth term of the series.

Here,

fourth term (t_{4}) = 1

no. of terms (n) = 4

t_{n}= a + (n -1)d

t_{4} = a + (4 - 1)d

1 = a + 3d

a = 1 - 3d ..................(1)

sum of eight terms (S_{8}) = 18

no. of terms (n) = 8

S_{n} = \(\frac n2\) [2a + (n - 1) d]

Putting the value of a from equation (1),

or, 18 = \(\frac 82\) [2(1 - 3d) + (8 - 1) d]

or, 18 = 4 (2 - 6d + 7d)

or, 18 = 4 (2 + d)

or, 18 = 8 + 4d

or, 4d = 18 - 8

or, d = \(\frac {10}{4}\)

∴ d = \(\frac 52\)

Putting the value of d in equation (1)

a = 1 - 3d = 1 - 3× \(\frac 52\) = 1 - \(\frac {15}{2}\) = \(\frac {2 - 15}{2}\) = -\(\frac {13}{2}\)

Again,

a = -\(\frac {13}{2}\)

d = \(\frac 52\)

n = 10

t_{10} = ?

We know,

t_{n} = a + (n - 1)d

t_{10} = -\(\frac {13}{2}\) + (10 - 1)× \(\frac 52\)

= \(\frac {-13}{2}\) + \(\frac {45}{2}\)

= \(\frac {-13 + 45}{2}\)

= \(\frac {32}{2}\)

= 16

∴ 10^{th} term = 16 _{Ans}

If 5^{th} and 10^{th} term of an arithmetic series are 17 and 42 respectively, find its 20^{th} term.

Let:

first term = a

fifth term (t_{5}) = 17

number of terms (n) = 5

common difference = d

We know,

t_{n}= a + (n - 1)d

or, t_{5} = a + (5 - 1)d

or, 17 = a + 4d ....................(1)

Again,

10^{th} term (t_{10}) = 42

number of terms (n) = 10

We know,

t_{n}= a + (n - 1)d

or, t_{10} = a + (10 - 1)d

or, 42 = a + 9d .......................(2)

Subtracting equation (1) and (2)

a | + | 4d | = | 17 |

a | + | 9d | = | 42 |

- | - | - | ||

-5d | = | -25 |

or, d = \(\frac {-25}{-5}\)

∴ d = 5

Putting the value of d in equation (1)

a + 4d = 17

or, a + 4×5 = 17

or, a = 17 - 20

∴ a = -3

Again,

a = -3

d = 5

n = 20

t_{20} = ?

We know,

t_{n}= a + (n - 1)d

t_{20} = -3 + (20 - 1)5 = -3 + 19× 5 = -3 + 95 = 92

∴ 20th term (t_{20}) = 92 _{Ans}

In an arithmetic series, the sum of first ten terms is 520. If its seventh term is double of its third term, calculate the first term and the common difference of the series.

Let;

first term (a) =

common difference (d) = ?

number of terms (n) = 10

sum of the 10 terms (S_{10}) = 520

We know that,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or, S_{10}= \(\frac {10}{2}\) [2a + (10 - 1) d]

or, 520 = 5[2a + 9d]

or, 2a + 9d = \(\frac {520}{5}\)

or, 2a + 9d = 104 ..........................(1)

Again,

t_{7} = 2 t_{3}

In L.H.S. n = 7 and R.H.S. n = 3

We know, t_{n}= a + (n - 1)d

a + (n - 1)d = 2 [a + (n - 1)d]

or, a + (7 - 1)d = 2 [a + (3 - 1)d]

or, a + 6d = 2(a + 2d)

or, a + 6d = 2a + 4d

or, 2a - a = 6d - 4d

or, a = 2d ...............................(2)

Putting the value of a in equation (1)

2× 2d + 9d= 104

or, 4d + 9d = 104

or, 13d = 104

or, d = \(\frac {104}{13}\)

∴ d = 8

Putting the value of d in equation (2)

a = 2d = 2× 8 = 16

∴ first term = 16

∴ common difference = 8 _{Ans}

The age of students in a certain class are in A.P. with the difference of 4 months. If sum of the ages of all the students are 200 years. Find the number of students in the class where the age of youngest one is 10 years old.

Here,

first term (a) = 10 years

common difference (d) = 4 months

\(\frac {4}{12}\) years = \(\frac 13\) years

sum of the series = 200 years

number of students (n) = ?

We know,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

or, 200 = \(\frac n2\) [2×10 + (n - 1)×\(\frac 13\)]

or, 400 = n[\(\frac {20 × 3 + n - 1}{3}\)]

or, 1200 = 60n + n^{2} - n

or, 1200 =n^{2} + 59n

or, n^{2} + 59n - 1200 = 0

or, n^{2} + 75n - 16n - 1200 = 0

or, n(n + 75) - 16(n + 75) = 0

or, (n + 75) (n - 16) = 0

Either: n + 75 = 0 ∴n = -75 (Impossible)

Or: n - 16 = 0 ∴n = 16

∴The number of students = 16 _{Ans}

If the third term of an A.P. is 1 and its fifth term is 7, find the sum of first ten terms of the series.

Here,

third term (t_{3}) = 1

no. of terms (n) = 3

first term (a) = ?

common difference (d) = ?

We know,

t_{n} = a + (n - 1)d

t_{3} = a + (3 - 1)d

1 = a + 2d

a = 1 - 2d ....................(1)

Again,

fifth term (t_{5}) = 7

number of terms (n) = 5

We know,

t_{n} = a + (n - 1)d

t_{5} = a + (5 - 1)d

7 = a + 4d.....................(2)

Putting the value of a in equation (ii)

7 = 1 - 2d + 4d

or, 7 - 1 = 2d

or, d = \(\frac 62\)

∴ d = 3

Putting the value of d in equation (i)

a = 1 - 2× 3 = 1 - 6 = - 5

Again,

a = -5

d = 3

n = 10

S_{10} = ?

We know,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

S_{10} = \(\frac {10}{2}\) [2×(-5) + (10 - 1) 3] = 5[- 10 + 27] = 5×17 = 85 _{Ans}

Find the sum of first 12 terms of an A.P. with fifth term 17 and tenth term 42.

Here,

fifth term (t_{5}) = 17

number of term (n) = 5

tenth term (t_{10)} = 42

number of terms (n) = 10

We know,

t_{n}= a + (n -1)d

t_{5} = a + (5 - 1)d

17 = a + 4d ................(1)

t_{n}= a + (n -1)d

t_{10}= a + (10 -1)d

42 = a + 9d .................(2)

Subtracting eq^{n}(1) from eq^{n}(2)

a | + | 9d | = | 42 |

a | + | 4d | = | 17 |

- | - | |||

5d | = | 25 |

d = \(\frac {25}{5}\)

∴ d = 5

putting the value of d in eq^{n} (1)

a + 4d = 17

or, a + 4× 5 = 17

or, a = 17 - 20

∴ a = -3

Again,

a = -3

d = 5

n = 12

S_{12} = ?

S_{n} = \(\frac n2\) [2a + (n - 1) d]

S_{12} = \(\frac {12}{2}\) [2(-3) + (12 - 1) 5] = 6[-6 + 55] = 6× 49 = 294 _{Ans}

If the 5^{th} term of an A.P. is 19 and 8^{th} term is 31, which term is 67?

Here,

fifth term (t_{5}) = 19

number of terms (n) = 5

t_{n}= a + (n - 1)d

t_{5} = a + (5 - 1)d

19 = a + 4d ......................(1)

Again,

eighth term (t_{8}) = 31

number of terms (n) = 8

t_{n}= a + (n - 1)d

t_{8} = a + (8 - 1)d

31 = a + 7d .........................(2)

Subtracting equation (2) from equation (1)

a | + | 4d | = | 19 |

a | + | 7d | = | 31 |

- | - | - | ||

-3d | = | -12 |

∴ d = \(\frac {-12}{-3}\) = 4

Putting the value of d in eq^{n}(1)

a + 4d = 19

or, a + 4× 4 = 19

or, a + 16 = 19

or, a = 19 - 16

∴ a = 3

last term (l) = 67

first term (a) = 3

common difference (d) = 4

We know,

l = a + (n - 1) d

or, 67 = 3 + (n - 1) 4

or, 67 - 3 = 4n -4

or, 4n - 4 = 64

or, 4n = 64 + 4

or, 4n = 68

or, n = \(\frac {68}{4}\)

∴ n = 17

∴ The 67 number be 17^{th} term. _{Ans}

If the third term and ninth term of an A.P. are 20 and 5 respectively, find the 19^{th} term of the series.

Here,

third term (t_{3}) = 20

number of terms (n) = 3

t_{n}=a + (n - 1)d

t_{3} = a + (3 - 1)d

20 = a + 2d .......................(1)

9th term (t_{9}) = 5

number of terms (n) = 9

t_{n}=a + (n - 1)d

t_{9} = a + (9 - 1)d

5 = a + 8d ..........................(2)

subtracting eq^{n}(1) from eq^{n}(2)

a | + | 8d | = | 5 |

a | + | 2d | = | 20 |

- | - | - | ||

6d | = | -15 |

or, d = -\(\frac {15}{6}\)

∴ d = -\(\frac 52\)

Putting the value of d in eq^{n} (1)

a + 2d = 20

or, a + 2× \(\frac {-5}{2}\) = 20

or, a - 5 = 20

∴ a = 25

∴ d = \(\frac {-5}{2}\)

19^{th} term (t_{19}) = ?

number of terms (n) = 19

t_{n}= a + (n - 1)d

or, t_{19} = 25 + (19 - 1) × -\(\frac {5}{2}\)

or, t_{19} = 25 + 18 × -\(\frac {5}{2}\)

or, t_{19} = 25 - 14

∴ t_{19} = - 20

∴ 19^{th} term be - 20. _{Ans}

Find the 4 arithmetic mean between 140 and -60.

Here,

first term (a) = 140

last term (l) = -60

number of arithmetic mean (n) = 4

common difference (d) = \(\frac {b - a}{n + 1}\) = \(\frac {-60 - 140}{4 + 1}\) = \(\frac {-200}{5}\) = -40

Let: m_{1}, m_{2}, m_{3} and m_{4} are the A.M. then,

m_{1} = a + d = 140 - 40 = 100

m_{2} = a + 2d = 140 - 2 × 40 =140 - 80 = 60

m_{3} = a + 3d = 140 - 3 × 40 =140 - 120 = 20

m_{4} = a + 4d = 140 - 4 × 40 =140 - 160 = -20

∴ Required 4 A.M. are 100, 60, 20 and - 20 _{Ans}

There are n A.Ms between 12 and 33. If the fourth mean is 24. Find the value of n.

Here,

first term (a) = 12

last term (b) = 33

fourth term (m_{4}) = 24

number of mean (n) = ?

common difference (d) = \(\frac {b - a}{n + 1}\) = \(\frac {33 - 12}{n + 1}\) = \(\frac {21}{n + 1}\)

fourth mean (m_{4}) = a + 4d

or, 24 = 12 + 4× \(\frac {21}{n + 1}\)

or, 24 - 12 = \(\frac {84}{n + 1}\)

or, 12 = \(\frac {84}{n + 1}\)

or, n + 1 = \(\frac {84}{12}\)

or, n = 7 - 1

∴ n = 6

∴The number of means = 6 _{Ans}

There are n A.Ms between 3 and 39. Find n so that the third mean : the last mean = 3 : 7.

Here,

first mean (a) = 3

last term (b) = 39

number of terms (n) = ?

\(\frac {third mean}{last mean}\) = \(\frac {m_3}{m_n}\) = \(\frac 37\)

common difference (d) = \(\frac {b - a}{n + 1}\)

d = \(\frac {39 - 3}{n + 1}\)

∴ d = \(\frac {36}{n + 1}\)

From Question,

\(\frac {m_3}{m_n}\) = \(\frac {a + 3d}{a + 4n}\) [\(\because\) m_{n} = a + nd]

or, \(\frac 37\) =\(\frac {3 + 3 × (\frac {36}{n + 1})}{3 + n × (\frac {36}{n + 1})}\) = \(\cfrac{\frac{3n + 3 + 108}{n + 1}}{\cfrac{3n + 3 + 36n}{n + 1}}\)

or, \(\frac 37\) = \(\frac {3n + 111}{39n + 3}\)

pr, 117n + 9 = 21n + 777

or, 117n - 21n = 777 - 9

or, 96n = 768

or, n = \(\frac {768}{96}\)

∴ n = 8

∴The number of mean = 8 _{Ans}

The last term of an arithmetic series of 20 terms is 195 and the common difference 5. Calculate the sum of the series.

Here,

number of terms in A.S (n) = 20

last term (l) = 195

common difference (d) = 5

first term (a) = ?

sum of 20^{th} term (S_{20}) = ?

We know that,

l = a + (n - 1)d

or, 195 = a + (20 - 1) 5

or, 195 = a + 19× 5

or, a = 195 - 95

∴ a = 100

S_{n} = \(\frac n2\) [a + l]

S_{20} = \(\frac {20}{2}\) [100 + 195] = 10× 295 = 2950 _{Ans}

The first and last terms of an arithmetic series are -24 and 72 respectively. If the sum of all terms of the series is 600, find the number of terms and the common difference of the series.

Here,

first term in A.S (a) = - 24

last term (l) = 72

sum of the series (S_{n}) = 600

number of terms (n) = ?

common difference (d) = ?

We have,

S_{n} = \(\frac n2\) (a + l)

or, n = \(\frac {1200}{48}\) = 25

∴ n = 25

Again,

l = a + (n - 1)d

or, 72 = -24 + (25 - 1)d

or, 72 + 24 = 24d

or, d = \(\frac {96}{24}\) = 4

∴ d = 4

∴ number of terms (n) = 25

and common difference (d) = 4 _{Ans}

The fourth and eighth terms of an arithmetic progression are 4 and 32 respectively. Find the sum of first ten terms.

Let:

4^{th} term (t_{4}) = 4

no. of items (n) = 4

first term (a) = ?

common difference (d) = ?

Using formula,

t_{n}= a + (n - 1)d

t_{4} = a + (4 - 1)d

4 = a + 3d ............(1)

Again,

8^{th} term (t_{8}) = 32

no. of terms (n) = 8

first term (a) = ?

common difference (d) = ?

Using formula,

t_{n}= a + (n - 1)d

t_{8} = a + (8 - 1)d

32 = a + 7d ..........(2)

Equation (2) - (1) we get:

a | + | 7d | = | 32 |

a | + | 3d | = | 4 |

- | - | - | ||

4d | = | 28 |

or, d = \(\frac {28}{4}\)

∴ d = 7

Putting the value of d in equation (1)

a + 3d = 4

a + 3× 7 = 4

a = 4 - 21

∴ a = -17

Again,

sum of 10^{th} term (S_{10}) = ?

first term (a) = - 17

number of term (n) = 10

common difference (d) = 7

Using formula,

S_{n} = \(\frac n2\) [2a + (n - 1) d]

S_{10} = \(\frac {10}{2}\) [2× -17 + (10 - 1)^7] = 5[-34 + 63] = 5× 29 = 145 _{Ans}

Find the 5 arithmetic means between 10 and 40.

Here,

Let: m_{1}, m_{2}, m_{3}, m_{4} and m_{5} are the five A.Ms between 10 and 40.

first term (a) = 10

last term (b) = 5

total number of terms (n) = 5 + 2 = 7

common difference = d

Using formula,

t_{n}= a + (n - 1)d

40 = 10 + (7 - 1) d

or, 40 - 10 = 6d

or, d = \(\frac {30}{6}\)

∴ d = 5

m_{1} = a + d = 10 + 5 = 15

m_{2} = a + 2d = 10 + 2× 5 = 10 + 10 = 20

m_{3} = a + 3d = 10 + 3 × 5 = 10 + 15 = 25

m_{4} = a + 4d = 10 + 4 × 5 = 10 + 20 = 30

m_{5} = a + 5d = 10 + 5 × 5 = 10 + 25 = 35

Hence, required A.Ms are 15, 20, 25, 30 and 35. _{Ans}

5 A.Ms are inserted between two numbers a and b. The first and fifth means are 18 and 46 respectively. Find the other means and find the values of a and b.

Here,

first men (m_{1}) = 18

fifth mean (m_{5}) = 46

Let: 18, m_{2}, m_{3}, m_{4}, 46 be 5 A.Ms between a and b then a, 18, m_{2}, m_{3}, m_{4}, 46, b are in Arithmetic sequence.

We know that,

t_{n} = a + (n - 1) d

m_{1}= a + (2 - 1)d

18 = a + d ............(1)

m_{5}= a + (6 - 1)d

46 = a + 5d ..........(2)

Subtracting equation (1) from (2)

a | + | 5d | = | 46 |

a | + | d | = | 18 |

- | - | - | ||

4d | = | 28 |

or, d = \(\frac {28}{7}\)

∴ d = 7

Substituting the value of d in equation (1)

a + d = 18

a + 7 = 18

a = 18 - 7

∴ a = 11

m_{2} = a + 2d = 11 + 2× 7 = 11 + 14 = 25

m_{3} = a + 3d = 11 + 3 × 7 = 11 + 21 = 32

m_{4} = a + 4d = 11 + 4 × 7 = 11 + 28 = 39

b = a + 6d = 11 + 6 × 7 = 11 + 42 = 53

∴ a = 11, m_{2} = 25, m_{3} = 32, m_{4} = 39 and b = 53 _{Ans}

The sum of 3 numbers in A.P. be 15 and the sum of their squares is 83. Find the A.P.

Let: the three numbers are in A.P. are a - d, a and a + d.

By question,

Condition 1^{st}:

a - d + a + a + d =15

or, 3a = 15

or, a = \(\frac {15}{3}\)

∴ a = 5

Condition 2^{nd}:

(a - d)^{2} + a^{2} + (a - d)^{2} = 83

or, (5 - d)^{2} + 5^{2} + (5 + d)^{2} = 83

or, 25 - 10d + d^{2} + 25 + 25 + 10d + d^{2} = 83

or, 2d^{2}+ 75 = 83

or, 2d^{2} = 83 - 75

or, d^{2} = \(\frac 82\)

or, d^{2} = 4

∴ d = ±2

Taking +ve sign:

a - d = 5 - 2 = 3

a = 5

a + d = 5 + 2 = 7

Taking -ve sign:

a - d = 5 + 2 = 7

a = 5

a + d = 5 - 2 =3

Hence, the A.P. is 3, 5, 7 or, 7, 5, 3. _{Ans}

There are some geometric means between 5 and 80. If the second mean is 20. Find the number of means between the two numbers. Also find the last mean.

Here,

first term (a) = 5

last term (l) = 80

third term (t_{3}) = 20

common ratio = r

We know that,

t_{2} = ar

t_{3} = ar^{2}

or, 20 = 5× r^{2}

or, r^{2} = \(\frac {20}{5}\)

or, r^{2} = 4

∴ r = 2

All geometric means between 5 and 80.

m_{1} = ar = 5× 2 = 10

m_{2} = ar^{2}= 5× 2^{2}= 20

m_{3} = ar^{3}= 5× 2^{3}= 40

last term =ar^{4}= 5× 2^{4}= 80

∴ number of means = 3

and last mean = 40 _{Ans}

The arithmetic and geometric mean are 25 and 20 respectively. What are the two numbers?

Let: the required number be a and b.

A.M. = \(\frac {a + b}{2}\) = 25

∴ a + b = 50 ..................(1)

G.M. = \(\sqrt {ab}\) = 20

Squaring on bot sides of the above

ab = 400 ....................(2)

We know,

(a - b)^{2} = (a + b)^{2} - 4ab

(a - b)^{2} = (50)^{2} - 4× 400

(a - b)^{2} = 2500 - 1600

(a - b)^{2} = 900

a - b = 30 ..................(3)

Adding equation (1) and (3)

a | + | b | = | 50 |

a | - | b | = | 30 |

2a | = | 80 |

or, a = \(\frac {80}{2}\)

∴ a = 40

Putting the value of a in equation (1)

a + b = 50

or, 40 + b = 50

or, b = 50 - 40

∴ b = 10

Hence, the required numbers are 40 and 10. _{Ans}

Find the sum of the first 6 terms of a geometric sequence in which 3^{rd} term is \(\frac 13\) and 6^{th} term is \(\frac {1}{81}\).

Here,

third term (t_{3}) = \(\frac 13\)

number of terms (n) = 3

t_{n}= ar^{n-1}

t_{3}= ar^{3}^{-1}

\(\frac 13\) = ar^{2}...........(1)

sixth term (t6) = \(\frac {1}{81}\)

no. of term (n) = 6

t_{n}= ar^{n-1}

t_{6}= ar^{6-1}

\(\frac {1}{81}\) = ar^{5} ............(2)

Dividing equation (2) by (1):

\(\frac {ar^5}{ar^2}\) = \(\cfrac{\frac{1}{81}}{\cfrac{1}{3}}\)

or, r^{3}= \(\frac {1}{81}\) × \(\frac 31\)

or, r^{3}= \(\frac {1} {27}\)

or, (r)^{3} = (\(\frac 13\))^{3}

∴ r = \(\frac 13\)

Putting the value of r in equation (1)

ar^{2} = \(\frac 13\)

or, a× (\(\frac 13\))^{2} = \(\frac 13\)

or, a× \(\frac 19\) = \(\frac 13\)

∴ a = \(\frac 93\) = 3

Again,

a = 3

r = \(\frac 13\)

n = 6

S_{6} = ?

We know,

S_{n} = \(\frac{a(r^n - 1)}{r - 1}\)

=\(\frac{3((\frac 13)^6 - 1)}{\frac 13 - 1}\)

=\(\frac{3(\frac {1}{729} - 1)}{\frac {1-3}{3}}\)

=\(\frac{3(\frac {1 - 729}{729})}{-\frac 23}\)

= 3× \(\frac {-728}{729}\)× \(\frac {-3}{2}\)

= \(\frac {6552}{1458}\)

= 4\(\frac {40}{81}\) _{Ans}

If the second and fifth terms of a geometric sequence are 4 and 32 respectively. Find the eight term.

Here,

second term (t_{2}) = 4

number of term (n) = 2

t_{n}= ar^{n-1}

t_{2}= ar^{2-1}

4 = ar ..................(1)

fifth term (t_{5}) = 32

no. of term (n) = 5

t_{n} = ar^{n-1}

t_{5} = ar^{5-1}

32 = ar^{4}....................(2)

Eq^{n} (1) is divided by eq^{n} (2)

\(\frac {ar^4}{ar}\) = \(\frac {32}{4}\)

r^{3} = 8

r = 2

Putting the value of r in eq^{n} (1)

ar = 4

or, a× 2 = 4

or, a = \(\frac 42\)

∴ a = 2

Again,

eight term (t_{8}) = ?

no. of term (n) = 8

first term (a) = 2

common ratio (r) = 2

t_{n}= ar^{n-1}

t_{8}= 2 × 2^{8-1} = 256

∴ The eight term = 256 _{Ans}

The sum of three consecutive terms in G.P. is 62 and their product is 1000, find the terms.

Let: three terms of G.P. are \(\frac ar\), a and ar.

From first condition:

\(\frac ar\) + a + ar = 62 .................... (1)

From second condition:

\(\frac ar\)× a× ar = 1000

or, a^{3} = 10^{3}

i.e. a = 10

Putting the value of a in eq^{n} (1),

\(\frac ar\) + a + ar = 62

or, \(\frac {10}{r}\) + 10 + 10r = 62

or, \(\frac {10 + 10r}{r}\) = 62 - 10

or, 10 + 10r^{2} = 52r

or, 10r^{2} - 52r + 10 = 0

or, 2(5r^{2} -26r + 5) = 0

or, 5r^{2} - 26r +5 = 0

or, 5r(r - 5) -1(r - 5) = 0

or, (r - 5) (5r - 1) = 0

Either: r - 5 = 0 ∴r = 5

Or, 5r - 1 = 0 ∴r = \(\frac 15\)

If r = 5, a = 10 then

\(\frac ar\) = \(\frac {10}{5}\) = 2

ar = 10× 5 = 50

If r = \(\frac 15\) and a = 10 then

\(\frac ar\) =\(\cfrac{10}{\cfrac{1}{5}}\) = 50

ar = 10× \(\frac 15\) = 2

∴ Required numbers are 2, 10 and 50 or 50, 10 and 2. _{Ans}

Three numbers are in A.P. and their sum is 15. If 1 , 3, 9 be added to them respectively they form a G.P. Find the number.

Let: the three number in A.P. be a - d, a, a + d

From first condition:

a - d + a + a + d = 15

or, 3a = 15

or, a = \(\frac {15}{3}\)

∴ a = 5

From second condition:

If 1, 3, 9 be added then a - d + 1, a + 3, a + d + 9 are in G.P.

5 - d + 1, 5 + 3, 5 + d + 9 are in G.P.

6 - d, 8, 14 + d are in G.P.

or, \(\frac {8}{6 - d}\) = \(\frac {14 + d}{8}\)

or, 64 = (6 - d) (14 + d)

or, 84 + 6d - 14d - d^{2} = 64

or, -d^{2} - 8d + 84 - 64 = 0

or, -(d^{2} + 8d - 20) = 0

or, d^{2} - 2d + 10d - 20 = 0

or, d(d - 2) + 10(d - 2) = 0

or, (d - 2) (d + 10) = 0

Either: d - 2 = 0∴ d = 2

Or: d + 10 = 0 ∴d = -10

When, a = 5 and d = 2 then

a - d, a, a + d

5 - 2, 5, 5 + 2

3, 5, 7

When, a = 5 and d = - 10

a - d, a, a + d

5 - (-10), 5, 5 + (-10)

5 + 10, 5, 5 - 10

15, 5, -5

∴ The three numbers are 3, 5, 7 or 15, 5, -5 _{Ans}

Find the ratio of two number when the ratio of their arithmetic mean is to their geometric mean is 5 : 3.

Let: the two number are a and b.

From question:

\(\frac {A.M}{G.M}\) = \(\frac 53\)

We know that,

A.M. = \(\frac {a + b}{2}\)

G.M. = \(\sqrt {ab}\)

\(\cfrac{\frac {a + b}{2}}{{\sqrt {ab}}}\) = \(\frac 53\)

or, \(\frac {3(a + b)}{2}\) = 5\(\sqrt {ab}\)

or, 3(a + b) = 10\(\sqrt {ab}\)

Squaring on both sides,

[3(a + b)]^{2} = (10\(\sqrt {ab}\))^{2}

or, 9(a^{2} + 2ab + b^{2}) = 100ab

or, 9a^{2} + 18ab + 9b^{2}- 100ab = 0

or, 9a^{2}- 82ab + 9b^{2} = 0

or, 9a^{2} - 81ab - ab + 9b^{2} = 0

or, 9a(a - 9b) -b(a - 9b) = 0

or, (a - 9b) (9a - b) = 0

Either: a - 9b = 0

a = 9b

\(\frac ab\) = \(\frac 91\)

Or: 9a - b = 0

9a = b

\(\frac ab\) = \(\frac 19\) (impossible)

∴ The required ratio is 9 : 1. _{Ans}

The sum of three consecutive terms in G.P. is 28 and their product is 512 find the terms.

Let: the three terms of the G.P. be \(\frac ar\), a and ar.

From first condition:

\(\frac ar\) + a + ar = 28 ..........................(1)

From second condition:

\(\frac ar\) × a × ar = 512

or, a^{3} = 8^{3}

∴ a = 8

Putting the value of a in eq^{n} (1)

\(\frac ar\) + a + ar = 28

or, \(\frac 8r\) + 8 + 8r = 28

or, \(\frac {8 + 8r^2}{r}\) = 28 - 8

or, 8 + 8r^{2} = 20r

or, 8r^{2} - 20r + 8 = 0

or, 4(2r^{2} - 5r + 2) = 0

or, 2r^{2} - 5r + 2 = 0

or, 2r^{2} - 4r - r + 2 = 0

or, 2r(r - 2) - 1(r - 2) = 0

or, (r - 2)(2r - 1) = 0

Either: r - 2 = 0 ∴r = 2

Or: 2r - 1 = 0 ∴r = \(\frac 12\)

When, r = \(\frac 12\) and a = 8 then:

\(\frac ar\), a, ar

=\(\cfrac{8}{\cfrac{1}{2}}\), 8, 8× \(\frac 12\)

= 16, 8, 4

When, r = 2 and a = 8 then:

\(\frac ar\), a, ar

= \(\frac 82\), 8, 8× 2

= 4, 8, 16

∴ The no. are 4, 8, 16 or 16, 8, 4. _{Ans}

The sum of the four consecutive terms of a G.P. is 30 and A.M. of the first and last number is 9. Find the common ratio.

Let: the four consecutive terms are a, ar, ar^{2} and ar^{3}.

From first condition:

a + ar + ar^{2} + ar^{3} = 30

or, a(1 + r) + ar^{2}(1 + r) = 30

or, (1 + r) (a + ar^{2}) = 30

or, a(1 + r) (1 + r^{2}) = 30 .........................(1)

From second condition:

\(\frac {a + ar^3}{2}\) = 9

or, a(1 + r^{3}) = 18

or, a(1 + r) (1 - r + r^{2}) = 18

or, a(1 + r) = \(\frac {18}{1 - r + r^2}\) ..........................(2)

Putting the value of a(1 + r) in eq^{n} (1)

\(\frac {18}{1 - r + r^2}\) (1 + r^{2}) = 30

or, 18 + 18r^{2} = 30 - 30r + 30r^{2}

or, 30r^{2} - 30r + 30 - 18 - 18r^{2} = 0

or, 12r^{2} - 30r + 12 = 0

or, 6(2r^{2} - 5r + 2) = 0

or, 2r^{2} - 5r + 2 = 0

or, 2r(r- 2) -1(r - 2) = 0

or, (r - 2)(2r - 1) = 0

Either: r - 2 = 0 ∴r = 2

Or: 2r - 1 = 0 ∴r = \(\frac 12\)

∴r = 2 or \(\frac 12\) _{Ans}

3 + 6 + 12 + ............ + 768 is the given series. (a) Find the number of terms of this series. (b) Find the sum of the series.

Here,

Given series is: 3 + 6 + 12 + .......... + 768

common ratio (r) = \(\frac 63\) = \(\frac {12}{6}\) = 2. So it is a geometric series.

First term (a) = 3

common ratio (r) = 2

last term (l) = 768

(a) no. of terms of the sequence (n) = ?

We know that,

l = ar^{n-1}

or, 768 = 3× 2^{n-1}

or, 2^{n-1} = \(\frac {768}{3}\)

or, 2^{n-1} = 256

or, 2^{n-1} = 2^{8}

or, n - 1 = 8

∴ n = 8 + 1 = 9

no. of terms (n) = 9

(b) The sum of the series (S_{9}) = ?

We know,

S_{n} = \(\frac {lr - a}{r - 1}\)

S_{9} = \(\frac {768 × 2 - 3}{2 - 1}\) = \(\frac {1536 - 3}{1}\) = 1533

∴ The sum of the series = 1533 _{Ans}

The 4^{th} term of G.S. is \(\frac 12\) and the 9^{th} term is \(\frac {16}{243}\). Find the fifth term of the series.

Here,

4^{th} term (t_{4}) = \(\frac 12\)

number of terms (n) = 4

We know that,

t_{n} = ar^{n-1}

t_{4} = ar^{4-1}

\(\frac 12\) = ar^{3}.............(1)

9^{th} term (t_{9}) = \(\frac {16}{243}\)

number of terms (n) = 5

We know,

t_{n} = ar^{n-1}

t_{9} = ar^{9-1}

\(\frac {16}{243}\) = ar^{8}..................(2)

Dividing equation (2) by (1): we get,

\(\cfrac{\frac{16}{243}}{\cfrac{1}{2}}\) = \(\frac {ar^8}{ar^3}\)

or, \(\frac {16× 2}{243}\) = r^{4}

or, r^{5} = \(\frac {2^5}{3^5}\)

or, (r)^{5} = (\(\frac 23\))^{5}

∴ r = \(\frac 23\)

Substituting the value of r in equation (1)

ar^{3} = \(\frac 12\)

or, a(\(\frac 23\))^{3} = \(\frac 12\)

or, a× \(\frac {8}{27}\) = \(\frac 12\)

∴ a = \(\frac {27}{16}\)

Again,

fifth term (t_{5}) = ar^{5-1}

= \(\frac {27}{16}\)× (\(\frac 23\))^{4}

= \(\frac {27}{16}\)× \(\frac {16}{27}\)× \(\frac 13\)

= \(\frac 13\) _{Ans}

Find the sum of 8^{th} term of a G.S. whose 3^{rd} and 4^{th} terms are \(\frac 14\) and \(\frac 18\) respectively.

Here,

third term (t_{3}) = \(\frac 14\)

number of terms (n) = 3

We know that,

t_{n} = ar^{n-1}

t_{3} = ar^{3-1}

\(\frac 13\) = ar^{2} ......................(1)

fourth term (t_{4}) = \(\frac 18\)

number of terms (n) = 4

t_{n} = ar^{n-1}

t_{4} = ar^{4-1}

\(\frac 18\) = ar^{3} ............................(2)

Dividing (2) by (1),

\(\frac {ar^3}{ar^2}\) =\(\cfrac{\frac{1}{8}}{\cfrac{1}{4}}\)

or, r = \(\frac 18\) ×\(\frac 41\) = \(\frac 12\)

Substituting the value of r in equation (1)

ar^{2} = \(\frac 14\)

a(\(\frac 12\))^{2} = \(\frac 14\)

a = \(\frac 14\) ×\(\frac 41\)

a = 1

Again,

S_{n} = \(\frac {a( r^n - 1)}{ r - 1}\)

S_{8} = \(\frac {1[(\frac {1}{2})^8 - 1)]}{\frac 12 - 1}\)

S_{8} =\(\cfrac{\frac{1}{256} - 1}{\cfrac{1}{2} - 1}\)

S_{8}=\(\cfrac{\frac{1 - 256}{256}}{\cfrac{1 - 2}{2}}\)

S_{8} = \(\frac {-255}{256}\)× \(\frac {2}{-1}\)

S_{8} = \(\frac {255}{128}\)

∴ The sum of eight terms = 1\(\frac {127}{128}\) _{Ans}

In a G.P. the sum of three terms is 21 and their product is 64, find the numbers.

Here,

Let: the three terms be \(\frac ar\), a and ar.

From first condition,

\(\frac ar\) + a + ar = 21 .................(1)

From second condition,

\(\frac ar\) ×a ×ar = 64 ............................(2)

or, a^{3} = 4^{3}

∴ a = 4

Substituting the value of a in equation (1)

\(\frac ar\) + a + ar = 21

or, \(\frac 4r\) + 4 + 4r = 21

or, \(\frac 4r\) + 4r = 21 - 4

or, \(\frac {4 + 4r^2}{r}\) = 17

or, 4r^{2} + 4 = 17r

or, 4r^{2} - 17r + 4 = 0

or, 4r^{2} - 16r - r + 4 = 0

or, 4r(r - 4) - 1(r - 4) = 0

or, (r - 4) (4r - 1) = 0

Either: r - 4 = 0 ∴r = 4

Or: 4r - 1 = 0 ∴ r = \(\frac 14\)

If a = 4 and r = 4

first term = \(\frac ar\) = \(\frac 44\) = 1

second term = a = 4

third term = ar = 4× 4 = 16

If a = 4 and r = \(\frac 14\)

first term =\(\cfrac{4}{\cfrac{1}{4}}\) = 16

second term = a = 4

third term = 4 ×\(\frac 14\) = 1

Hence, the number are 1, 4 and 16 or 16, 4 and 1. _{Ans}

Find the 4 geometric means between \(\frac {1}{16}\) and 64.

Here,

first term (a) = \(\frac {1}{16}\)

last term (b) = 64

number of means (n) = 4

We know that,

r = (\(\frac ba\))^{\(\frac{1}{n+1}\)}

r = (\(\cfrac{64}{\cfrac{1}{16}}\))^{\(\frac{1}{n+1}\)}= (64 × 16)^{\(\frac 15\)} = (2^{6}× 2^{4})^{\(\frac 15\)}=(2)^{10×\(\frac 15\)} = 2^{2}= 4

If G_{1}, G_{2}, G_{3} and G_{4}_{}are the 4 geometric means.

G_{1} = ar = \(\frac {1}{16}\)× 4 = \(\frac 14\)

G_{2} = ar^{2} = \(\frac {1}{16}\)× 4× 4 = 1

G_{3} = ar^{3} = \(\frac {1}{16}\)× 4× 4 × 4 =4

G_{4} = ar^{4} = \(\frac {1}{16}\)× 4× 4 × 4 × 4= 16

∴ The required means are \(\frac14\), 1, 4 and 16. _{Ans}

_{}

If a, b, c be in A.P. and x, y, z in G.P. prove that:

x^{b-c }× y^{c-a} × z^{a-b} = 1

Let: a, b and c are in A.P.

b - a = c - b

or, b + b = c + a

or, 2b = c + a

∴ b = \(\frac {c + a}{2}\) .......................(1)

Again,

x, y and z are in G.P.

\(\frac yx\) = \(\frac zy\)

or, y^{2} = xz ...........................(2)

L.H.S.

=x^{b-c}× y^{c-a}× z^{a-b}

= x^{a-b}× y^{c-a}× z^{a-b}[\(\because\) a - b = b - c]

= (xz)^{a-b}× y^{c-a}

= (y^{2})^{a-b}× y^{c-a}

= y^{2a-2b+c-a}

= y^{a+c-2b}

= y^{2b-2b}

= y^{0}

= 1

= R.H.S _{proved}

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