Sequence and Series
Sequence
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.
Series
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.
Progression
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.
Types of Progression
Progression is divided into following two types.
(i) Arithmetic progression
(ii) Geometric progression
Arithmetic progression or Sequence
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 = 1215 = 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_{n1}
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.
General term or n^{th} term of an A.P.
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 + (11)d
t_{2 }= a + d = a + (21)d
t_{3 }= a + 2d = a + (31)d
t_{4 }= a + 3d = a + (41)d
In general, t_{n }= a + (n1)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), ..........
Arithmetic Mean
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
ba = cb
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+21)d, where d is common difference
or, b =a + (n+1)d
or, (n+1)d = ba
∴ d= \(\frac{ba}{n+1}\)
Now, m_{1} = a+d = a + \(\frac{ba}{n+1}\)
m_{2} = a + 2d = a + \(\frac{2(ba)}{n+1}\)
m_{3} = a + \(\frac{3(ba)}{n+1}\)
.............................................
m_{n} = a + \(\frac{n(ba)}{n+1}\)
Sum of n terms of series in A.P.
Let us consider an arithmetic series
a + (a+d) + (a+2d) + (a+3d) + ...... + (l2d) +(ld) + 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 = ld
if the sum of n terms is denoted by S_{n}, then
S_{n} = a + (a+d) + (a+2d) + (a+3d) + ...... + (l2d) +(ld) + l .... (i)
Writing term in the reverse order,
S_{n} = l + (ld) + (l2d) + ...... + (a+3d) + (a+2d) + (a+d) + a .... (ii)
Adding the corresponding terms of (i) and (ii)
\(\frac{S_n \;= \;a \;+\; (a+d)\; +\; (a+2d)\; +\; (a+3d)\; + \;......\; + \;(l2d) \;+\; (ld)\; + \;l\\S_n\;=\;l\; +\;(ld)\;+\;(l2d)\;+\;......\;+\;(a+3d)\;+\;(a+2d)\;+\;(a+d)\;+a\:}{2S_n\;= \;(a+l) \;+ \;(a+l)\; + \;(a+l)\; +\; ............ \;+\; (a+l)\; + \;(a+l)\; +\; (a+l)}\)
= n times (a+l)
= n (a+l)
= \(\frac{n}{2}\)(a+l)
But, the last term l = a + (n1)d
So, S_{n} = \(\frac{n}{2}\)(a+l) = \(\frac{n}{2}\)[a+a+(n1)d] = \(\frac{n}{2}\)[2a+(n1)d]
∴ S_{n} = \(\frac{n}{2}\)[2a+(n1)d]
Thus, if d is unknown, S_{n} = \(\frac{n}{2}\)(a+l)
And, if l is unknown, S_{n} = \(\frac{n}{2}\)[2a+(n1)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) = 21 = 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+(n1)d] = \(\frac{n}{2}\)[2.1+(n1).1] = \(\frac{n}{2}\)[2+n1] = \(\frac{n}{2}\)(n+1)
2. Sum of first n odd numbers
1, 3, 5, 7, ......., (2n1) are the first n odd numbers.
Here, first term (a) = 1
Common difference (d) = 31 = 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+(n1)d] = \(\frac{n}{2}\)[2.1+(n1).2] = \(\frac{n}{2}\)(2+2n2) = \(\frac{n}{2}\) × 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) = 42 = 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+(n1)d] = \(\frac{n}{2}\)[2.2+(n1).2] = \(\frac{n}{2}\)(4+2n2) = \(\frac{n}{2}\)(2n2) = n(n+1)
Geometric Progression or Sequence
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, nonzero 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^{n1} gives us the n^{th} term or the last term of the geometric progression whose first term, common ratio and a number of terms are a, r and n respectively.
∴t_{n} = ar^{n1}
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\(^3\) + ........
Geometric Mean
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 nonzero 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 \(\begin{pmatrix}b\\a\\ \end{pmatrix}\)^{\(\frac{n}{n + 1}\)}
Relation between arithmetic mean and geometric mean
"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^{n1}
Here, first = a common ratio = r number of terms = n last term (l) = ar^{n1}
∴ S_{n} = \(\frac{lr  a}{r  1}\)
If the number of terms is odd, we take the middle term as aand 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$$ t_1, t2, t3, ........ tn$$ 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.,
\(\sum_{n=2}^5 (n^2 3)\)
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_{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}
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 + ......................
Here,
\(\sum_{n=4}^7 (3n 2)\)
where: n = 4, 5, 6, 7
\(\sum_{n=4}^7 (3n 2)\)
= (3×42) + (3×52) +(3×62) +(3×72)
= (122) + (152) + (182) + (212)
= 10 + 13 + 16 + 19
= 58 _{Ans}
\(\sum_{k=3}^7 (k^2 + 1)\)
= (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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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^{n1}
or, 128 = \(\frac 14\) (2)n1
or, 2^{7}× 2^{2} = 2^{n1}
or, 2^{n1} = 2^{9}
or, n 1 = 9
or, n = 9 + 1
∴ n = 10
∴ The number of term = 10 _{Ans}
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}
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^{n1}
t_{8} = (\(\frac 13\)) (3)^{81} = \(\frac 13\)× 3× 3^{6} = 729 _{Ans}
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^{n1}
or, \(\frac {1}{81}\) = 3 (\(\frac 13 \))^{n1}
or,  \(\frac {1}{3^4× 3}\) = (\(\frac 13\))^{n1}
or, (\(\frac 13\))^{5} =(\(\frac 13\))^{n1}
or, n  1 = 5
or, n = 5 + 1
∴ n = 6
∴ The number of term = 6 _{Ans}
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^{n1}
or, t_{5} = \(\frac 23\)(\(\frac 14\))^{51}
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}
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}
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}
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}
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}
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}
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^{n1}
or, 9 = \(\frac 19\)r^{51}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
Let:
first term = a  d
second term = a
third term = a + d
common difference = d
Let: the threeterms of A.S. are ad, 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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
Here,
third term (t_{3}) = \(\frac 13\)
number of terms (n) = 3
t_{n}= ar^{n1}
t_{3}= ar^{3}^{1}
\(\frac 13\) = ar^{2}...........(1)
sixth term (t6) = \(\frac {1}{81}\)
no. of term (n) = 6
t_{n}= ar^{n1}
t_{6}= ar^{61}
\(\frac {1}{81}\) = 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} = \(\frac{a(r^n  1)}{r  1}\)
=\(\frac{3((\frac 13)^6  1)}{\frac 13  1}\)
=\(\frac{3(\frac {1}{729}  1)}{\frac {13}{3}}\)
=\(\frac{3(\frac {1  729}{729})}{\frac 23}\)
= 3× \(\frac {728}{729}\)× \(\frac {3}{2}\)
= \(\frac {6552}{1458}\)
= 4\(\frac {40}{81}\) _{Ans}
Here,
second term (t_{2}) = 4
number of term (n) = 2
t_{n}= ar^{n1}
t_{2}= ar^{21}
4 = ar ..................(1)
fifth term (t_{5}) = 32
no. of term (n) = 5
t_{n} = ar^{n1}
t_{5} = ar^{51}
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^{n1}
t_{8}= 2 × 2^{81} = 256
∴ The eight term = 256 _{Ans}
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}
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}
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}
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}
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}
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^{n1}
or, 768 = 3× 2^{n1}
or, 2^{n1} = \(\frac {768}{3}\)
or, 2^{n1} = 256
or, 2^{n1} = 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}
Here,
4^{th} term (t_{4}) = \(\frac 12\)
number of terms (n) = 4
We know that,
t_{n} = ar^{n1}
t_{4} = ar^{41}
\(\frac 12\) = ar^{3}.............(1)
9^{th} term (t_{9}) = \(\frac {16}{243}\)
number of terms (n) = 5
We know,
t_{n} = ar^{n1}
t_{9} = ar^{91}
\(\frac {16}{243}\) = 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^{51}
= \(\frac {27}{16}\)× (\(\frac 23\))^{4}
= \(\frac {27}{16}\)× \(\frac {16}{27}\)× \(\frac 13\)
= \(\frac 13\) _{Ans}
Here,
third term (t_{3}) = \(\frac 14\)
number of terms (n) = 3
We know that,
t_{n} = ar^{n1}
t_{3} = ar^{31}
\(\frac 13\) = ar^{2} ......................(1)
fourth term (t_{4}) = \(\frac 18\)
number of terms (n) = 4
t_{n} = ar^{n1}
t_{4} = ar^{41}
\(\frac 18\) = 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}
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}
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}
_{}
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^{bc}× y^{ca}× z^{ab}
= x^{ab}× y^{ca}× z^{ab}[\(\because\) a  b = b  c]
= (xz)^{ab}× y^{ca}
= (y^{2})^{ab}× y^{ca}
= y^{2a2b+ca}
= y^{a+c2b}
= y^{2b2b}
= y^{0}
= 1
= R.H.S _{proved}

If f = {(1,2) , (2,3) , (4,5)} , find f (^1) and the domain and range of f (^1) .
f (^1) = {( 1,1) , (3,3) , (5,4)}
Domain of f (^1) = {(2 ,3, 3)}
Range of f (^1) = {1,2,3}
f (^1) = {( 2,3) , (1,2) , (5,4)}
Domain of f (^1) = {5,3,2)}
Range of f (^1) = {4,2,1}
f (^1) = {( 2,1) , (3,2) , (5,4)}
Domain of f (^1) = {(2 ,3, 5)}
Range of f (^1) = {1,2,4}
f (^1) = {( 2,1) , (1,2) , (3,4)}
Domain of f (^1) = {(5 ,3, 5)}
Range of f (^1) = {1,3,4}

If f = {(1,4) , (2,4) , (3,5)} and g = {(4,2) , (5,2) }, then show the function gof in arrow diagram and find it ordered pair form.
{(3,2) , (2,2) , (0,1)}
{(1,2) , (2,2) , (3,2)}
{(2,2) , (3,2) , (4,2)}
{(3,2) , (2,2) , (0,2)}

If f = {(b,q) , (a,p) , (c,r)} and gof = {(a,1) , (c,3) , (b,2)} , find the function g in terms of ordered pairs .
{(p,1) , (q,2) , (r,3)}
{(p,q) , (1,2) , (r,3)}
{(1,1) , (q,2) , (q,3)}
{(p,1) , (q,q) , (r,r)}

If f is an identity function and g= {(1,3) , (2,4) , (3,5)} then write the composite function gof in ordered pair form .
{(1,2) , (2,4) ,(4,5)}
{(1) , (2) ,(3,5)}
{(1,2) , (2,4) ,(4,5)}
{(1,3) , (2,4) ,(3,5)}

If gof = {(2,9) , (3,13) , (4,17)} and f (x) = 2x+1 , find the function 'g' in order pair form.
g = {(5,9) , (7,13) , (9,17)}
g = {(5,7) , (6,13) , (16,17)}
g = {(5,5) , (7,3) , (2,17)}
g = {(5,8) , (7,14) , (0,17)}

If f = {(1,5) , (2,4) , (3,6)} and fog = {(5,6) , (4,5) , (6,4)} then find g (^1) .
g (^1) = {(2,5) ,(4,4) (4,6)}
g (^1) = {(3,5) ,(1,4) (2,6)}
g (^1) = {(5,3) ,(4,4) (,6,2)}
g (^1) = {(1,2) ,(1,4) (2,6)}

If f = {(2,3) (4,5) , (5,6)} and g = {(4,3) , (2,5) , (5,8)} then find gof (^1) in ordered pair form .
{(5,5) , (3,3) , (8,8)}
{(3,5) , (5,3) , (6,8)}
{(3,5) , (5,3) , (5,3)}
{(1,5) , (2,5) , (,3,8)}

If f (x) = 2x 1 and g (x) = x+1 , find gof (x) and gof (2) .
x2 , 2
3x, 3
4x, 2
2x , 4

If f (x) = 4x , g(x) = x+1 and fog(x) = 20 , find the value of x.
80
2
4
8

If a function f (x) = 5 then find f (4) and ff (5) .
5,5
4,5
5
4,5

If f (x) = 3x1 and fg(x) = 9x7 , then find g(x) .
23  x
3x  2
x2
2x  3

If f (x) = 5  (frac{6}{x}) then what is the value of x at f (x) = f (^1) (1) ?
1
(frac{1}{5})
5
1

If f (x) = (frac{6}{x2}) , g(x) = k x (^2) 1 and gof (5) = 7 , what is the value of k ?
30
12
35
2

If f (x) = 4x+5 and fog (x) = 8x + 13 , find the value of x such that gof (x) = 28 .
1
16
2
2

If f (x) = 2x + 3 and fog (x) = 2x + 5 , find (gof) (^1) .
(frac{x4}{4})
(frac{x2}{2})
(frac{4x}{2})
(frac{x4}{2})

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AsmiFind the value of k if (k 2), (4k6), (3k6) are first term of an AP. 
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the sum of 3 terms of arithmetic series is 15 .the geometric mean of first and third terms is 4. find the numbers 
Feb 07, 2017 
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