CBSE Class 11 Mathematics Sequences And Series Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Sequences and Series (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( f \) is a function satisfying \( f(x + y) = f(x) \cdot f(y) \) such that \( f(1) = 3 \) and \( \sum_{x=1}^{n} f(x) = 120 \). Find the value of \( n \).
Answer:
We have, \( f(x + y) = f(x) \cdot f(y) \)
\( f(1) = 3 \)
\( \sum_{x=1}^{n} f(x) = 120 \)

Now, \( \sum_{x=1}^{n} f(x) = f(1) + f(2) + f(3) + \dots + f(n) = 120 \)
Also:
\( f(2) = f(1 + 1) = f(1) \cdot f(1) = (3)(3) = 9 \)
\( f(3) = f(1 + 2) = f(1) \cdot f(2) = (3)(9) = 27 \)
\( f(4) = f(1 + 3) = f(1) \cdot f(3) = (3)(27) = 81 \)

The series becomes:
\( 3 + 9 + 27 + 81 + \dots + f(n) = 120 \)

Clearly, this is a G.P. with \( a = 3 \) and \( r = 3 \).
\( 3 \left[ \frac{3^n - 1}{3 - 1} \right] = 120 \)
\( \Rightarrow 3 \frac{(3^n - 1)}{2} = 120 \)
\( \Rightarrow 3^n - 1 = \frac{2 \times 120}{3} \)
\( \Rightarrow 3^n - 1 = 80 \)
\( \Rightarrow 3^n = 81 \)
\( \Rightarrow 3^n = 3^4 \)
\( \Rightarrow n = 4 \) ans.

Question. If \( a, b, c \) are in G.P. and \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \), prove that \( x, y, z \) are in A.P.
Answer:
Given: \( a, b, c \) are in G.P.
\( \Rightarrow b^2 = ac \)

Given: \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} \)
Let \( a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k \)
\( \Rightarrow a^{\frac{1}{x}} = k \Rightarrow a = k^x \)
\( \Rightarrow b^{\frac{1}{y}} = k \Rightarrow b = k^y \)
\( \Rightarrow c^{\frac{1}{z}} = k \Rightarrow c = k^z \)

Substituting these into \( b^2 = ac \):
\( \Rightarrow (k^y)^2 = (k^x) \cdot (k^z) \)
\( \Rightarrow k^{2y} = k^{x+z} \)
\( \Rightarrow 2y = x + z \)
\( \therefore x, y, z \) are in A.P. ans.

Question. Find the value of \( n \) so that \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) is the G.M. between \( a \) and \( b \).
Answer:
We have, \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \text{G.M.} = \sqrt{ab} \)
\( \Rightarrow \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = a^{\frac{1}{2}} \cdot b^{\frac{1}{2}} \)
\( \Rightarrow a^{n+1} + b^{n+1} = a^{\frac{1}{2}} \cdot b^{\frac{1}{2}} (a^n + b^n) \)
\( \Rightarrow a^{n+1} + b^{n+1} = a^{n+\frac{1}{2}} \cdot b^{\frac{1}{2}} + b^{n+\frac{1}{2}} \cdot a^{\frac{1}{2}} \)
\( \Rightarrow a^{n+1} - a^{n+\frac{1}{2}} \cdot b^{\frac{1}{2}} = b^{n+\frac{1}{2}} \cdot a^{\frac{1}{2}} - b^{n+1} \)
\( \Rightarrow a^{n+\frac{1}{2}} \left( a^{\frac{1}{2}} - b^{\frac{1}{2}} \right) = b^{n+\frac{1}{2}} \left( a^{\frac{1}{2}} - b^{\frac{1}{2}} \right) \)
\( \Rightarrow a^{n+\frac{1}{2}} = b^{n+\frac{1}{2}} \)
\( \Rightarrow \left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1 \)
\( \Rightarrow \left(\frac{a}{b}\right)^{n+\frac{1}{2}} = \left(\frac{a}{b}\right)^0 \)
\( \Rightarrow n + \frac{1}{2} = 0 \)
\( \Rightarrow n = -\frac{1}{2} \) ans.

Question. Insert three numbers between 3 and 243 so that the resulting sequence is a G.P.
Answer:
Here, \( a = 3, b = 243 \) and \( n = 3 \).
Let the G.M.'s be \( G_1, G_2, G_3 \).
\( \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} = \left(\frac{243}{3}\right)^{\frac{1}{3+1}} = (81)^{\frac{1}{4}} = 3 \)
\( \therefore r = 3 \)

Now:
\( G_1 = ar^1 = (3)(3) = 9 \)
\( G_2 = ar^2 = (3)(9) = 27 \)
\( G_3 = ar^3 = (3)(27) = 81 \)
\( \therefore \) the required numbers are 9, 27, 81 ans.

Question. If the first and the \( n^{\text{th}} \) term of a G.P. are \( a \) and \( b \) respectively and if \( P \) is the product of \( n \) terms, prove that \( P^2 = (ab)^n \).
Answer:
Given, \( a_1 = a \)
\( \Rightarrow a_n = b \)
\( \Rightarrow ar^{n-1} = b \)
\( \Rightarrow r^{n-1} = \frac{b}{a} \)
\( \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n-1}} \)

Now, let \( P \) be the product of \( n \) terms:
\( \Rightarrow P = a \cdot ar \cdot ar^2 \cdot ar^3 \dots ar^{n-1} \)
\( \Rightarrow P = a^n \cdot r^{1+2+3+\dots+(n-1)} \)
\( \Rightarrow P = a^n \cdot r^{\frac{n(n-1)}{2}} \)

Substituting the value of \( r \):
\( \Rightarrow P = a^n \left[ \left(\frac{b}{a}\right)^{\frac{1}{n-1}} \right]^{\frac{n(n-1)}{2}} \)
\( \Rightarrow P = a^n \left(\frac{b}{a}\right)^{\frac{n}{2}} \)
\( \Rightarrow P = a^n \cdot \frac{b^{\frac{n}{2}}}{a^{\frac{n}{2}}} \)
\( \Rightarrow P = a^{n - \frac{n}{2}} \cdot b^{\frac{n}{2}} \)
\( \Rightarrow P = a^{\frac{n}{2}} \cdot b^{\frac{n}{2}} \)
\( \Rightarrow P = (ab)^{\frac{n}{2}} \)

Squaring both sides:
\( P^2 = (ab)^n \) (proved)

Question. If the \( p^{\text{th}} \), \( q^{\text{th}} \) and \( r^{\text{th}} \) terms of a G.P. are \( a, b \) and \( c \) respectively, prove that \( a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = 1 \).
Answer:
Let the first term of the G.P. be \( A \) and the common ratio be \( R \).
Given:
\( a_p = a = AR^{p-1} \)
\( a_q = b = AR^{q-1} \)
\( a_r = c = AR^{r-1} \)

Taking LHS: \( a^{q-r} \cdot b^{r-p} \cdot c^{p-q} \)
Substituting the values of \( a, b, c \):
\( = [AR^{p-1}]^{q-r} \cdot [AR^{q-1}]^{r-p} \cdot [AR^{r-1}]^{p-q} \)
\( = A^{q-r} \cdot R^{(p-1)(q-r)} \cdot A^{r-p} \cdot R^{(q-1)(r-p)} \cdot A^{p-q} \cdot R^{(r-1)(p-q)} \)
\( = A^{(q-r) + (r-p) + (p-q)} \cdot R^{(pq-pr-q+r) + (qr-qp-r+p) + (rp-rq-p+q)} \)
\( = A^0 \cdot R^0 \)
\( = (1)(1) = 1 = \text{R.H.S. (proved)} \)

Question. If \( A \) and \( G \) be A.M. and G.M. respectively between two positive numbers, prove that the numbers are \( A \pm \sqrt{(A + G)(A - G)} \).
Answer:
Let the two positive numbers be \( a \) and \( b \).
Then \( A = \frac{a+b}{2} \) and \( G = \sqrt{ab} \).

Consider, \( A + \sqrt{(A + G)(A - G)} \):
\( = A + \sqrt{A^2 - G^2} \)

Substituting the values of \( A \) and \( G \):
\( = \frac{a+b}{2} + \sqrt{\left(\frac{a+b}{2}\right)^2 - (\sqrt{ab})^2} \)
\( = \frac{a+b}{2} + \sqrt{\frac{a^2 + b^2 + 2ab}{4} - ab} \)
\( = \frac{a+b}{2} + \sqrt{\frac{a^2 + b^2 - 2ab}{4}} \)
\( = \frac{a+b}{2} + \sqrt{\left(\frac{a-b}{2}\right)^2} \)
\( = \frac{a+b}{2} + \frac{a-b}{2} \)
\( = a \)

Therefore, \( A + \sqrt{(A+G)(A-G)} = a \).
Similarly, \( A - \sqrt{(A+G)(A-G)} = b \).

Thus, the numbers are indeed \( A \pm \sqrt{(A + G)(A - G)} \) ans.

Question. Let \( S \) be the sum, \( P \) be the product and \( R \) be the sum of reciprocals of \( n \) terms in G.P. Prove that \( P^2 R^n = S^n \).
Answer:
Let the terms in G.P. be \( a, ar, ar^2, \dots, ar^{n-1} \).
Sum, \( S = a \left(\frac{r^n-1}{r-1}\right) \qquad (\text{assuming } r > 1) \)

Product, \( P = a \cdot ar \cdot ar^2 \dots ar^{n-1} \)
\( \Rightarrow P = a^n \cdot r^{1+2+3+\dots+(n-1)} \)
\( \Rightarrow P = a^n \cdot r^{\frac{n(n-1)}{2}} \)

Sum of reciprocals, \( R = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \dots + \frac{1}{ar^{n-1}} \)
This is also a G.P. with first term \( = \frac{1}{a} \) and common ratio \( = \frac{1}{r} \) (since \( r > 1 \), \( \frac{1}{r} < 1 \)):
\( \therefore R = \frac{1}{a} \left[ \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} \right] \)
\( R = \frac{1}{a} \left[ \frac{r^n - 1}{r - 1} \right] \cdot \frac{r}{r^n} \)

Taking LHS: \( P^2 \cdot R^n \)
Substituting \( P \) and \( R \):
\( \text{LHS} = \left[ a^n \cdot r^{\frac{n(n-1)}{2}} \right]^2 \left[ \frac{1}{a} \left( \frac{r^n-1}{r-1} \right) \cdot \frac{r}{r^n} \right]^n \)
\( = a^{2n} \cdot r^{n(n-1)} \cdot \frac{1}{a^n} \left( \frac{r^n-1}{r-1} \right)^n \cdot \frac{r^n}{r^{n^2}} \)
\( = a^{2n-n} \cdot r^{n^2 - n + n - n^2} \cdot \left( \frac{r^n-1}{r-1} \right)^n \)
\( = a^n \cdot r^0 \cdot \left( \frac{r^n-1}{r-1} \right)^n \)
\( = a^n \cdot \left( \frac{r^n-1}{r-1} \right)^n \)
\( = \left[ a \cdot \left( \frac{r^n-1}{r-1} \right) \right]^n \)
\( = S^n = \text{R.H.S.} \) (proved)

Question. Show that the ratio of the sum of the first \( n \) terms of a G.P. to the sum of terms from \( (n + 1)^{\text{th}} \) to \( (2n)^{\text{th}} \) term is \( \frac{1}{r^n} \).
Answer:
Let the G.P. consist of \( 2n \) terms:
\( a_1, a_2, \dots, a_n, a_{n+1}, \dots, a_{2n} \)

For the first \( n \) terms:
First term \( = a_1 = a \), common ratio \( = r \), number of terms \( = n \)
Sum, \( S_n = a \left( \frac{r^n - 1}{r - 1} \right) \)

For the terms from \( (n + 1)^{\text{th}} \) to \( (2n)^{\text{th}} \):
First term \( = a_{n+1} = ar^n \), common ratio \( = r \), number of terms \( = n \)
Sum, \( S'_n = ar^n \left( \frac{r^n - 1}{r - 1} \right) \)

Now, taking the ratio of the sums:
\( \frac{S_n}{S'_n} = \frac{a \left( \frac{r^n - 1}{r - 1} \right)}{ar^n \left( \frac{r^n - 1}{r - 1} \right)} = \frac{1}{r^n} \) (proved)

Question. If \( a, b, c, d \) and \( p \) are real numbers such that \( (a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0 \), then show that \( a, b, c \text{ & } d \) are in G.P.
Answer:
To show that \( a, b, c \text{ & } d \) are in G.P., we must prove that:
\( \frac{b}{a} = \frac{c}{b} = \frac{d}{c} \)

Given:
\( (a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0 \)
\( \Rightarrow a^2p^2 + b^2p^2 + c^2p^2 - 2abp - 2bcp - 2cdp + b^2 + c^2 + d^2 \le 0 \)
Rearranging into quadratic perfect square groups:
\( \Rightarrow (a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0 \)
\( \Rightarrow (ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0 \)

Since the sum of squares of real numbers cannot be less than 0:
\( \therefore (ap - b)^2 + (bp - c)^2 + (cp - d)^2 = 0 \)

This is possible only when each term is individually 0:
\( ap - b = 0 \Rightarrow ap = b \Rightarrow \frac{b}{a} = p \)
\( bp - c = 0 \Rightarrow bp = c \Rightarrow \frac{c}{b} = p \)
\( cp - d = 0 \Rightarrow cp = d \Rightarrow \frac{d}{c} = p \)

\( \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} \)
\( \Rightarrow a, b, c, d \) are in G.P. ans.

Question. If \( p, q, r \) are in G.P. and the equations \( px^2 + 2qx + r = 0 \) and \( dx^2 + 2ex + f = 0 \) have a common root, then show that \( \frac{d}{p}, \frac{e}{q}, \frac{f}{r} \) are in A.P.
Answer:
Given: \( p, q, r \) are in G.P.
\( \therefore q^2 = pr \)

To prove that \( \frac{d}{p}, \frac{e}{q}, \frac{f}{r} \) are in A.P., we need to show:
\( \frac{2e}{q} = \frac{d}{p} + \frac{f}{r} \)

Consider the equation: \( px^2 + 2qx + r = 0 \).
Using the quadratic formula:
\( x = \frac{-2q \pm \sqrt{4q^2 - 4pr}}{2p} \)
Substituting \( q^2 = pr \):
\( x = \frac{-2q \pm \sqrt{4pr - 4pr}}{2p} \)
\( x = \frac{-2q}{2p} = \frac{-q}{p} \)

Since \( x = -\frac{q}{p} \) is the common root, it must satisfy \( dx^2 + 2ex + f = 0 \):
\( \therefore d \left(-\frac{q}{p}\right)^2 + 2e \left(-\frac{q}{p}\right) + f = 0 \)
\( \Rightarrow \frac{dq^2}{p^2} - \frac{2eq}{p} + f = 0 \)
Substituting \( q^2 = pr \):
\( \Rightarrow \frac{dpr}{p^2} - \frac{2eq}{p} + f = 0 \)
\( \Rightarrow \frac{dr}{p} - \frac{2eq}{p} + f = 0 \)
\( \Rightarrow dr - 2eq + fp = 0 \)
\( \Rightarrow 2eq = dr + fp \)

Dividing both sides by \( q^2 \) (and replacing with \( pr \) on the RHS):
\( \Rightarrow \frac{2eq}{q^2} = \frac{dr}{pr} + \frac{fp}{pr} \)
\( \Rightarrow \frac{2e}{q} = \frac{d}{p} + \frac{f}{r} \)
\( \therefore \frac{d}{p}, \frac{e}{q}, \frac{f}{r} \) are in A.P. (proved)

Question. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, \( \frac{1}{2} \).
Answer:
\( 1^{\text{st}} \) sequence: \( 2, 4, 8, 16, 32 \)
\( 2^{\text{nd}} \) sequence: \( 128, 32, 8, 2, \frac{1}{2} \)

New sequence of products: \( 256, 128, 64, 32, 16 \).
This is a G.P. where:
\( a_1 = 256 \)
\( r = \frac{128}{256} = \frac{1}{2} \)
Number of terms, \( n = 5 \).

Sum, \( S_5 = a \left( \frac{1 - r^5}{1 - r} \right) \)
\( = 256 \left( \frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}} \right) \)
\( = 256 \left( \frac{1 - \frac{1}{32}}{\frac{1}{2}} \right) \)
\( = 2 \times 256 \left( \frac{31}{32} \right) \)
\( = 16 \times 31 \)
\( = 496 \) ans.

Question. Find four numbers forming a G.P. in which the third term is greater than the first term by 9 & the second term is greater than fourth term by 18.
Answer:
Let the four numbers be \( a, ar, ar^2, ar^3 \).

We are given:
\( a_3 = a_1 + 9 \)
\( \Rightarrow ar^2 = a + 9 \)
\( \Rightarrow ar^2 - a = 9 \)
\( \Rightarrow a(r^2 - 1) = 9 \) ................ (i)

And:
\( a_2 = a_4 + 18 \)
\( \Rightarrow ar = ar^3 + 18 \)
\( \Rightarrow ar - ar^3 = 18 \)
\( \Rightarrow -ar(r^2 - 1) = 18 \) ................ (ii)

Dividing equation (ii) by equation (i):
\( \frac{-ar(r^2 - 1)}{a(r^2 - 1)} = \frac{18}{9} \)
\( \Rightarrow -r = 2 \Rightarrow r = -2 \)

Substituting \( r = -2 \) in equation (i):
\( a((-2)^2 - 1) = 9 \)
\( \Rightarrow a(4 - 1) = 9 \)
\( \Rightarrow 3a = 9 \Rightarrow a = 3 \)

Thus, the numbers are:
\( a = 3 \)
\( ar = 3(-2) = -6 \)
\( ar^2 = 3(4) = 12 \)
\( ar^3 = 3(-8) = -24 \)
\( \therefore \) the numbers are \( 3, -6, 12, -24 \) ans.

Question. Evaluate \( \sum_{k=1}^{11} (2 + 3^k) \).
Answer:
\( \sum_{k=1}^{11} (2 + 3^k) = (2 + 3^1) + (2 + 3^2) + (2 + 3^3) + \dots + (2 + 3^{11}) \)
\( = (2 + 2 + 2 + \dots 11 \text{ terms}) + (3^1 + 3^2 + 3^3 + \dots + 3^{11}) \)
The second bracket is a G.P. with \( a = 3, r = 3, n = 11 \):
\( = (2 \times 11) + 3 \left( \frac{3^{11} - 1}{3 - 1} \right) \)
\( = 22 + 3 \left( \frac{3^{11} - 1}{2} \right) \)
\( = 22 + \frac{3^{12} - 3}{2} \)
\( = \frac{44 + 3^{12} - 3}{2} \)
\( = \frac{41 + 3^{12}}{2} \) ans.

Question. If \( p^{\text{th}} \), \( q^{\text{th}} \), \( r^{\text{th}} \) and \( s^{\text{th}} \) terms of an A.P. are in G.P., then show that \( (p - q), (q - r), (r - s) \) are also in G.P.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the A.P.
\( a_p = a + (p - 1)d \)
\( a_q = a + (q - 1)d \)
\( a_r = a + (r - 1)d \)
\( a_s = a + (s - 1)d \)

Since \( a_p, a_q, a_r, a_s \) are in G.P.:
\( \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_s}{a_r} \) ................ (i)

Consider, \( \frac{a_q}{a_p} = \frac{a_r}{a_q} \):
Using ratio properties \( \left( \text{if } \frac{x}{y} = \frac{w}{z} = \frac{x - w}{y - z} \right) \):
\( \Rightarrow \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_q - a_r}{a_p - a_q} \)
\( = \frac{[a+(q-1)d] - [a+(r-1)d]}{[a+(p-1)d] - [a+(q-1)d]} \)
\( = \frac{(q - r)d}{(p - q)d} = \frac{q - r}{p - q} \) ................ (ii)

Now, consider \( \frac{a_r}{a_q} = \frac{a_s}{a_r} \):
\( \Rightarrow \frac{a_r}{a_q} = \frac{a_s}{a_r} = \frac{a_r - a_s}{a_q - a_r} \)
\( = \frac{[a+(r-1)d] - [a+(s-1)d]}{[a+(q-1)d] - [a+(r-1)d]} \)
\( = \frac{(r - s)d}{(q - r)d} = \frac{r - s}{q - r} \) ................ (iii)

From equations (i), (ii) and (iii):
\( \Rightarrow \frac{q-r}{p-q} = \frac{r-s}{q-r} \)
\( \Rightarrow (q - r)^2 = (p - q)(r - s) \)
\( \Rightarrow (p - q), (q - r), (r - s) \) are in G.P.

Question. If the \( 4^{\text{th}} \), \( 10^{\text{th}} \) and \( 16^{\text{th}} \) term of a G.P. are \( x, y, z \) respectively. Prove that \( x, y, z \) are in G.P.
Answer:
Let \( a \) be the first term and \( r \) be the common ratio of the G.P.
\( a_4 = x \Rightarrow ar^3 = x \)
\( a_{10} = y \Rightarrow ar^9 = y \)
\( a_{16} = z \Rightarrow ar^{15} = z \)

To prove \( x, y, z \) are in G.P., we must show \( y^2 = xz \):
LHS: \( y^2 = (ar^9)^2 = a^2 r^{18} \)
RHS: \( xz = (ar^3)(ar^{15}) = a^2 r^{18} \)

Clearly, LHS = RHS \( \Rightarrow y^2 = xz \).
Thus, \( x, y, z \) are in G.P.