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^{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^{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 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$$ t_1, t-2, t-3, ........ tn$$ 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.,
.
\(\sum_{n=2}^5 (n^2- 3)\)
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
\(\sum_{n=4}^7 (3n- 2)\)
= (3×4-2) + (3×5-2) +(3×6-2) +(3×7-2)
= (12-2) + (15-2) + (18-2) + (21-2)
= 10 + 13 + 16 + 19
= 58 _{Ans}
\(\sum_{k=3}^7 (k^2 + 1)\)
= (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^{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}
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^{n-1}
t_{8} = (\(\frac 13\)) (3)^{8-1} = \(\frac 13\)× 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^{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}
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}
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^{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}
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}
If f = {(1,2) , (2,3) , (4,5)} , find f (^-1) and the domain and range of f (^-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}
f (^-1) = {( 2,3) , (1,2) , (5,4)}
Domain of f (^-1) = {5,3,2)}
Range of f (^-1) = {4,2,1}
f (^-1) = {( 1,1) , (3,3) , (5,4)}
Domain of f (^-1) = {(2 ,3, 3)}
Range of f (^-1) = {1,2,3}
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) , (0,2)}
{(1,2) , (2,2) , (3,2)}
{(3,2) , (2,2) , (0,1)}
{(2,2) , (3,2) , (4,2)}
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,q) , (1,2) , (r,3)}
{(1,1) , (q,2) , (q,3)}
{(p,1) , (q,q) , (r,r)}
{(p,1) , (q,2) , (r,3)}
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) ,(3,5)}
{(1,2) , (2,4) ,(4,5)}
{(1,3) , (2,4) ,(3,5)}
{(1,2) , (2,4) ,(4,5)}
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) = {(1,2) ,(1,4) (2,6)}
g (^-1) = {(5,3) ,(4,4) (,6,2)}
If f = {(2,3) (4,5) , (5,6)} and g = {(4,3) , (2,5) , (5,8)} then find gof (^-1) in ordered pair form .
{(3,5) , (5,3) , (6,8)}
{(3,5) , (5,3) , (5,3)}
{(5,5) , (3,3) , (8,8)}
{(1,5) , (2,5) , (,3,8)}
If f (x) = 2x -1 and g (x) = x+1 , find gof (x) and gof (-2) .
3x, -3
2x , -4
4x, -2
x2 , -2
If f (x) = 4x , g(x) = x+1 and fog(x) = 20 , find the value of x.
8
4
2
80
If a function f (x) = 5 then find f (4) and ff (-5) .
4,5
5,5
-5
-4,-5
If f (x) = 3x-1 and fg(x) = 9x-7 , then find g(x) .
3x - 2
x-2
2x - 3
23 - x
If f (x) = 5 - (frac{6}{x}) then what is the value of x at f (x) = -f (^-1) (-1) ?
-5
-1
1
(frac{-1}{-5})
If f (x) = (frac{6}{x-2}) , g(x) = k x (^2) -1 and gof (5) = 7 , what is the value of k ?
30
35
12
2
If f (x) = 4x+5 and fog (x) = 8x + 13 , find the value of x such that gof (x) = 28 .
-2
1
16
2
If f (x) = 2x + 3 and fog (x) = 2x + 5 , find (gof) (^-1) .
(frac{x-4}{4})
(frac{x-2}{2})
(frac{x-4}{2})
(frac{4-x}{2})
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