CBSE Class 11 Mathematics Sequences And Series Worksheet Set 07

CBSE Class 11 Mathematics Worksheet - Sequences and Series (6). 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. The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, find the number of terms.
Answer: Given, \( a_1 + a_2 + a_3 + a_4 = 56 \) and \( a_1 = 11 \)
\( \Rightarrow a + (a + d) + (a + 2d) + (a + 3d) = 56 \)
\( \Rightarrow 4a + 6d = 56 \)
\( \Rightarrow 44 + 6d = 56 \)
\( \Rightarrow 6d = 12 \)
\( \Rightarrow d = 2 \)

Now, sum of the last four terms is 112:
\( \Rightarrow a_{n} + a_{n-1} + a_{n-2} + a_{n-3} = 112 \)
\( \Rightarrow [a + (n - 1)d] + [a + (n - 2)d] + [a + (n - 3)d] + [a + (n - 4)d] = 112 \)
\( \Rightarrow 4a + d(n - 1 + n - 2 + n - 3 + n - 4) = 112 \)
\( \Rightarrow 4(11) + 2(4n - 10) = 112 \)
\( \Rightarrow 44 + 8n - 20 = 112 \)
\( \Rightarrow 24 + 8n = 112 \)
\( \Rightarrow 8n = 88 \)
\( \Rightarrow n = 11 \) ans.

Question. Find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Answer: The numbers which are divisible by 2 or 5 from 1 to 100 are \( 2, 4, 5, 6, 8, 10, 12, \dots, 100 \).
There are two sequences in the above set of numbers:
\( 1^{\text{st}} \text{ sequence: } 2, 4, 6, 8, 10, 12, \dots, 100 \)
\( 2^{\text{nd}} \text{ sequence: } 5, 15, 25, \dots, 95 \)

The \( 1^{\text{st}} \) sequence is an A.P. with \( a = 2 \), \( d = 2 \), and \( n = 50 \):
\( \therefore \text{its sum} = \frac{50}{2}[2(2) + (50 - 1)2] \)
\( = 25(4 + 98) \)
\( = 25(102) \)
\( = 2550 \)

The \( 2^{\text{nd}} \) sequence is also an A.P. with \( a = 5 \), \( d = 10 \), and \( n = 10 \):
\( \therefore \text{its sum} = \frac{10}{2}[2(5) + (10 - 1)10] \)
\( = 5(10 + 90) \)
\( = 500 \)

\( \therefore \text{Total sum} = 2550 + 500 = 3050 \) ans.

G.P.

Question. The sum of the first three terms of a G.P. is \( \frac{13}{12} \) & their product is \( -1 \). Find the common ratio & their terms.
Answer: Let the terms be \( \frac{a}{r}, a, ar \).
Product \( = -1 \)
\( \Rightarrow \frac{a}{r} \cdot a \cdot ar = -1 \)
\( \Rightarrow a^3 = -1 \)
\( \Rightarrow a = -1 \)

Sum \( = \frac{13}{12} \)
\( \Rightarrow \frac{a}{r} + a + ar = \frac{13}{12} \)
\( \Rightarrow a \left(\frac{1}{r} + 1 + r\right) = \frac{13}{12} \)
\( \Rightarrow (-1) \left[\frac{1 + r + r^2}{r}\right] = \frac{13}{12} \)
\( \Rightarrow \frac{1 + r + r^2}{r} = -\frac{13}{12} \)
\( \Rightarrow 12 + 12r + 12r^2 = -13r \)
\( \Rightarrow 12r^2 + 25r + 12 = 0 \)
\( \Rightarrow 12r^2 + 16r + 9r + 12 = 0 \)
\( \Rightarrow 4r(3r + 4) + 3(3r + 4) = 0 \)
\( \Rightarrow (3r + 4)(4r + 3) = 0 \)
\( \Rightarrow r = \frac{-4}{3} \text{ and } r = \frac{-3}{4} \)

For \( a = -1 \) and \( r = \frac{-4}{3} \), the terms are \( \frac{3}{4}, -1, \frac{4}{3} \).
For \( a = -1 \) and \( r = \frac{-3}{4} \), the terms are \( \frac{4}{3}, -1, \frac{3}{4} \).
\( \therefore \) required terms are \( \frac{3}{4}, -1, \frac{4}{3} \) or \( \frac{4}{3}, -1, \frac{3}{4} \) ans.

Question. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers, we obtain an A.P. Find the numbers.
Answer: Let the numbers in G.P. be \( a, ar, ar^2 \).
Given, \( a + ar + ar^2 = 56 \)
\( \Rightarrow a(1 + r + r^2) = 56 \) ................ (i)

We are given that \( a - 1, ar - 7, ar^2 - 21 \) are in A.P.
\( \Rightarrow 2(ar - 7) = (a - 1) + (ar^2 - 21) \)
\( \Rightarrow 2ar - 14 = a + ar^2 - 22 \)
\( \Rightarrow ar^2 - 2ar + a = 8 \)
\( \Rightarrow a(r^2 - 2r + 1) = 8 \) ................ (ii)

Dividing (i) by (ii):
\( \Rightarrow \frac{a(1 + r + r^2)}{a(r^2 - 2r + 1)} = \frac{56}{8} = 7 \)
\( \Rightarrow 1 + r + r^2 = 7(r^2 - 2r + 1) \)
\( \Rightarrow 1 + r + r^2 = 7r^2 - 14r + 7 \)
\( \Rightarrow 6r^2 - 15r + 6 = 0 \)
\( \Rightarrow 2r^2 - 5r + 2 = 0 \)
\( \Rightarrow 2r^2 - 4r - r + 2 = 0 \)
\( \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \)
\( \Rightarrow (2r - 1)(r - 2) = 0 \)
\( \Rightarrow r = \frac{1}{2} \text{ & } r = 2 \)

Put \( r = \frac{1}{2} \) in equation (i):
\( \therefore a \left(1 + \frac{1}{2} + \frac{1}{4}\right) = 56 \)
\( \Rightarrow a\left(\frac{7}{4}\right) = 56 \)
\( \Rightarrow a = \frac{4 \times 56}{7} \)
\( \Rightarrow a = 32 \)

For \( r = 2 \) in equation (i):
\( \Rightarrow a(1 + 2 + 4) = 56 \)
\( \Rightarrow a(7) = 56 \)
\( \Rightarrow a = 8 \)

\( \therefore \) For \( a = 8 \text{ & } r = 2 \), the numbers are \( 8, 16, 32 \).
For \( a = 32 \text{ & } r = \frac{1}{2} \), the numbers are \( 32, 16, 8 \).
\( \therefore \) required numbers are \( 8, 16, 32 \) or \( 32, 16, 8 \) ans.

Question. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places. Find the common ratio.
Answer: Let the G.P. contain \( 2n \) number of terms.
We have \( a_1 + a_2 + a_3 + \dots + a_{2n} = 5(a_1 + a_3 + a_5 + \dots + a_{2n-1}) \)
\( \Rightarrow a + ar + ar^2 + \dots + ar^{2n-1} = 5(a + ar^2 + ar^4 + \dots + ar^{2n-2}) \)

LHS G.P.: first term \( = a \), common ratio \( = r \), number of terms \( = 2n \)
RHS G.P.: first term \( = a \), common ratio \( = r^2 \), number of terms \( = n \)
\( \Rightarrow a \left( \frac{r^{2n} - 1}{r - 1} \right) = 5a \left( \frac{(r^2)^n - 1}{r^2 - 1} \right) \)
\( \Rightarrow \frac{r^{2n} - 1}{r - 1} = 5 \left[ \frac{r^{2n} - 1}{(r + 1)(r - 1)} \right] \)
\( \Rightarrow 1 = \frac{5}{r + 1} \)
\( \Rightarrow r + 1 = 5 \)
\( \Rightarrow r = 4 \) ans.

Question. If \( \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} \), then show that \( a, b, c \text{ & } d \) are in G.P.
Answer: Consider,
\( \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} \)
\( \Rightarrow (a + bx)(b - cx) = (a - bx)(b + cx) \)
\( \Rightarrow ab - acx + b^2x - bcx^2 = ab + acx - b^2x - bcx^2 \)
\( \Rightarrow 2b^2x = 2acx \)
\( \Rightarrow b^2 = ac \)
\( \therefore a, b, c \) are in G.P. ................ (i)

Now consider,
\( \frac{b+cx}{b-cx} = \frac{c+dx}{c-dx} \)
\( \Rightarrow (b + cx)(c - dx) = (b - cx)(c + dx) \)
\( \Rightarrow bc - bdx + c^2x - cdx^2 = bc + bdx - c^2x - cdx^2 \)
\( \Rightarrow 2c^2x = 2bdx \)
\( \Rightarrow c^2 = bd \)
\( \dots b, c, d \) are in G.P. ................ (ii)

From (i) & (ii), \( a, b, c, d \) are in G.P.

Question. If \( a, b, c, d \) are in G.P. then show that \( (a^n + b^n), (b^n + c^n), (c^n + d^n) \) are in G.P.
Answer: Given, \( a, b, c, d \) are in G.P.
Let \( a = a, b = ar, c = ar^2, d = ar^3 \).
To prove \( (a^n + b^n), (b^n + c^n), (c^n + d^n) \) are in G.P.,
i.e., \( (b^n + c^n)^2 = (a^n + b^n)(c^n + d^n) \)

Taking LHS:
\( (b^n + c^n)^2 = [(ar)^n + (ar^2)^n]^2 \)
\( = [a^n r^n + a^n r^{2n}]^2 \)
\( = a^{2n} r^{2n} [1 + r^n]^2 \)

Taking RHS:
\( (a^n + b^n)(c^n + d^n) = (a^n + (ar)^n)((ar^2)^n + (ar^3)^n) \)
\( = (a^n + a^n r^n)(a^n r^{2n} + a^n r^{3n}) \)
\( = a^n(1 + r^n) \cdot a^n r^{2n}(1 + r^n) \)
\( = a^{2n} r^{2n}(1 + r^n)^2 \)

\( \therefore \text{LHS} = \text{RHS} \)
\( \therefore (a^n + b^n), (b^n + c^n), (c^n + d^n) \) are in G.P.

Question. If \( a \) and \( b \) are the roots of \( x^2 - 3x + p = 0 \) and \( c, d \) are the roots of \( x^2 - 12x + q = 0 \), where \( a, b, c, d \) form a G.P. show that \( (q + p) : (q - p) = 17 : 15 \).
Answer: Given, \( a \text{ & } b \) are roots of \( x^2 - 3x + p = 0 \)
\( \Rightarrow a + b = 3 \) and \( ab = p \) \( \left\{ \text{since } \alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} \right\} \)

Also, \( c \text{ & } d \) are the roots of \( x^2 - 12x + q = 0 \)
\( \Rightarrow c + d = 12 \) and \( cd = q \)

Since \( a, b, c, d \) are in G.P., let:
\( a = a, b = ar, c = ar^2, d = ar^3 \)

To prove: \( \frac{q + p}{q - p} = \frac{17}{15} \)
Taking LHS:
\( \frac{q + p}{q - p} = \frac{cd + ab}{cd - ab} \)
\( = \frac{(ar^2)(ar^3) + (a)(ar)}{(ar^2)(ar^3) - (a)(ar)} \)
\( = \frac{a^2 r^5 + a^2 r}{a^2 r^5 - a^2 r} \)
\( = \frac{a^2 r(r^4 + 1)}{a^2 r(r^4 - 1)} \)
\( = \frac{r^4 + 1}{r^4 - 1} \) ................ (i)

Now we have:
\( \Rightarrow a + b = 3 \Rightarrow a + ar = 3 \Rightarrow a(1 + r) = 3 \) ................ (iii)
\( \Rightarrow c + d = 12 \Rightarrow ar^2 + ar^3 = 12 \Rightarrow ar^2(1 + r) = 12 \) ................ (iv)

Dividing (iv) by (iii):
\( \frac{ar^2(1 + r)}{a(1 + r)} = \frac{12}{3} \)
\( \Rightarrow r^2 = 4 \)

Put \( r^2 = 4 \) in equation (i):
\( \therefore \frac{q + p}{q - p} = \frac{(r^2)^2 + 1}{(r^2)^2 - 1} = \frac{(4)^2 + 1}{(4)^2 - 1} = \frac{17}{15} \)
\( \therefore (q + p) : (q - p) = 17 : 15 \) ans.

Question. The ratio of the A.M. and G.M. of two possible numbers \( a \) and \( b \) is \( m : n \). Show that \( a : b = (m + \sqrt{m^2 - n^2}) : (m - \sqrt{m^2 - n^2}) \).
Answer: Given, \( \frac{\text{A.M.}}{\text{G.M.}} = \frac{m}{n} \)
\( \Rightarrow \frac{a+b}{2\sqrt{ab}} = \frac{m}{n} \)

Applying componendo and dividendo \( \left(\frac{N+D}{N-D}\right) \):
\( \Rightarrow \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{m+n}{m-n} \)
\( \Rightarrow \frac{(\sqrt{a})^2+(\sqrt{b})^2+2\sqrt{a}\sqrt{b}}{(\sqrt{a})^2+(\sqrt{b})^2-2\sqrt{a}\sqrt{b}} = \frac{m+n}{m-n} \)
\( \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = \frac{m+n}{m-n} \)
\( \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{m+n}}{\sqrt{m-n}} \)

Applying componendo and dividendo again:
\( \Rightarrow \frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})} = \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \)
\( \Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \)

Squaring both sides:
\( \Rightarrow \frac{a}{b} = \frac{(m+n)+(m-n)+2\sqrt{m+n}\sqrt{m-n}}{(m+n)+(m-n)-2\sqrt{m+n}\sqrt{m-n}} \)
\( \Rightarrow \frac{a}{b} = \frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}} \)
\( \Rightarrow \frac{a}{b} = \frac{2(m+\sqrt{m^2-n^2})}{2(m-\sqrt{m^2-n^2})} \)
\( \therefore a : b = (m + \sqrt{m^2 - n^2}) : (m - \sqrt{m^2 - n^2}) \) ans.

Question. If \( a, b, c \) are in A.P., \( b, c, d \) are in G.P. and \( \frac{1}{c}, \frac{1}{d}, \frac{1}{e} \) are in A.P. prove that \( a, c, e \) are in G.P.
Answer: Given, \( a, b, c \) are in A.P.
\( \Rightarrow 2b = a + c \) ................ (i)

Given, \( b, c, d \) are in G.P.
\( \Rightarrow c^2 = bd \) ................ (ii)

Given, \( \frac{1}{c}, \frac{1}{d}, \frac{1}{e} \) are in A.P.
\( \Rightarrow \frac{2}{d} = \frac{1}{c} + \frac{1}{e} \)
\( \Rightarrow \frac{2}{d} = \frac{e+c}{ce} \)
\( \Rightarrow \frac{d}{2} = \frac{ec}{e+c} \)
\( \Rightarrow d = \frac{2ec}{e+c} \) ................ (iii)

To prove \( a, c, e \) are in G.P., i.e., \( c^2 = ae \).
From (ii), we have:
\( c^2 = bd \)

Substituting the values of \( b \) and \( d \) from equation (i) and (iii):
\( \Rightarrow c^2 = \left(\frac{a+c}{2}\right) \left(\frac{2ec}{e+c}\right) \)
\( \Rightarrow c^2(e + c) = (a + c)(ec) \)
\( \Rightarrow c^2e + c^3 = aec + ec^2 \)
\( \Rightarrow c^3 = aec \)
\( \Rightarrow c^2 = ae \)
\( \therefore a, c, e \) are in G.P. (proved)

Question. Find the sum to \( n \) terms of given series \( 5 + 55 + 555 + \dots......... \)
Answer: Let \( S_n = 5 + 55 + 555 + \dots n \text{ terms} \)
\( S_n = 5[1 + 11 + 111 + \dots n \text{ terms}] \)

Multiplying and dividing by 9:
\( = \frac{5}{9} [9 + 99 + 999 + \dots n \text{ terms}] \)
\( = \frac{5}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots n \text{ terms}] \)
\( = \frac{5}{9} [(10 + 10^2 + 10^3 + \dots n \text{ terms}) - (1 + 1 + 1 + \dots n \text{ terms})] \)
\( \left\{ \text{using G.P. sum with } a = 10, r = 10 \right\} \)
\( = \frac{5}{9} \left[ 10 \left( \frac{10^n - 1}{10 - 1} \right) - n \right] \)
\( = \frac{5}{9} \left[ \frac{10^{n+1} - 10}{9} - n \right] \)
\( \therefore S_n = \frac{5}{81} [10^{n+1} - 10 - 9n] \) ans.

Question. Find the sum of the series to \( n \) terms \( 0.6 + 0.66 + 0.666 + \dots.........n \text{ terms.} \)
Answer: Let \( S_n = 0.6 + 0.66 + 0.666 + \dots n \text{ terms} \)
\( S_n = 6[0.1 + 0.11 + 0.111 + \dots n \text{ terms}] \)

Multiplying and dividing by 9:
\( = \frac{6}{9} [0.9 + 0.99 + 0.999 + \dots n \text{ terms}] \)
\( = \frac{2}{3} [(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots n \text{ terms}] \)
\( = \frac{2}{3} [(1 + 1 + 1 + \dots n \text{ terms}) - (0.1 + 0.01 + 0.001 + \dots n \text{ terms})] \br /> \( = \frac{2}{3} \left[ n - \left( \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \dots n \text{ terms} \right) \right] \)
\( \left\{ \text{using G.P. sum with } a = \frac{1}{10}, r = \frac{1}{10} \right\} \)
\( = \frac{2}{3} \left[ n - \frac{1}{10} \left( \frac{1 - \frac{1}{10^n}}{1 - \frac{1}{10}} \right) \right] \)
\( = \frac{2}{3} \left[ n - \frac{1}{10} \frac{\left( 1 - \frac{1}{10^n} \right)}{\frac{9}{10}} \right] \)
\( = \frac{2}{3} \left[ \frac{9n - 1 + \frac{1}{10^n}}{9} \right] \)
\( \therefore S_n = \frac{2}{27} \left[ 9n - 1 + \frac{1}{10^n} \right] \) ans.