Maharashtra Board Class 11 Maths Part 1 Chapter 4 Sequences and Series 4.2 Solutions

Question 1. For the following G.P.'s, find Sn.
(i) 3, 6, 12, 24, ....
(ii) p, q, \( \frac{q^2}{p} \), \( \frac{q^3}{p^2} \), ....
Answer: Solution:
i. 3, 6, 12, 24, ...
Here, a = 3, r = \( \frac{6}{3} \) = 2 > 1
Sn = \( \frac{a(r^n-1)}{r-1} \), for r > 1
\( \therefore \) Sn = \( \frac{3(2^n-1)}{2-1} \)
\( \therefore \) Sn = \( 3(2^n-1) \)
ii. p, q, \( \frac{q^2}{p} \), \( \frac{q^3}{p^2} \), ...
Here, a = p, r = \( \frac{q}{p} \)
Let \( \frac{q}{p} \) < 1
Sn = \( \frac{a(1-r^n)}{1-r} \), for r < 1
\( \therefore \) Sn = \( p \left[ \frac{1-\left(\frac{q}{p}\right)^n}{1-\frac{q}{p}} \right] \)
\( \therefore \) Sn = \( p \left[ \frac{1-\left(\frac{q}{p}\right)^n}{\frac{p-q}{p}} \right] \)
\( \therefore \) Sn = \( \frac{p^2}{p-q} \left[ 1-\left(\frac{q}{p}\right)^n \right] \)
Let \( \frac{q}{p} \) > 1
Sn = \( \frac{a(r^n-1)}{r-1} \), for r > 1
\( \therefore \) Sn = \( p \left[ \frac{\left(\frac{q}{p}\right)^n-1}{\frac{q}{p}-1} \right] \)
= \( p \left[ \frac{\left(\frac{q}{p}\right)^n-1}{\frac{q-p}{p}} \right] \)
= \( \frac{p^2}{q-p} \left[ \left(\frac{q}{p}\right)^n-1 \right] \)
In simple words: The sum of n terms for a G.P. is found using different formulas depending on whether the common ratio 'r' is greater than or less than 1. For r > 1, the formula is \( \frac{a(r^n-1)}{r-1} \), and for r < 1, it's \( \frac{a(1-r^n)}{1-r} \).

🎯 Exam Tip: Remember to choose the correct sum formula based on the value of the common ratio (r) to avoid negative terms in the numerator or denominator, which can complicate calculations but not change the final result.

Question 2. For a G.P., if
(i) a = 2, r = \( -\frac{2}{3} \), find S6.
(ii) S5 = 1023, r = 4, find a.
Answer: Solution:
i. a = 2, r = \( -\frac{2}{3} \)
Sn = \( \frac{a(1-r^n)}{1-r} \), for r < 1
\( \therefore \) S6 = \( 2 \left[ \frac{1-\left(-\frac{2}{3}\right)^6}{1-\left(-\frac{2}{3}\right)} \right] \)
= \( 2 \left[ \frac{1-\frac{64}{729}}{1+\frac{2}{3}} \right] \)
= \( 2 \left[ \frac{\frac{729-64}{729}}{\frac{3+2}{3}} \right] \)
= \( 2 \left[ \frac{\frac{665}{729}}{\frac{5}{3}} \right] \)
= \( 2 \times \frac{665}{729} \times \frac{3}{5} \)
= \( 2 \times \frac{133}{243} \)
\( \therefore \) S6 = \( \frac{266}{243} \)
ii. r = 4, S5 = 1023
Sn = \( a \left( \frac{r^n-1}{r-1} \right) \), for r > 1
\( \therefore \) S5 = \( a \left( \frac{4^5-1}{4-1} \right) \)
1023 = \( a \left( \frac{1024-1}{3} \right) \)
1023 = \( a \left( \frac{1023}{3} \right) \)
\( \therefore \) a = 3
In simple words: This question demonstrates how to find the sum of a G.P. given the first term 'a' and common ratio 'r', or how to find 'a' when the sum of 'n' terms and 'r' are known, by applying the appropriate G.P. sum formula.

🎯 Exam Tip: Pay close attention to the sign of the common ratio 'r' and whether it is greater than or less than 1, as this determines which form of the sum formula to use. Accurate calculation of powers is crucial for correct results.

Question 3. For a G. P., if
(i) a = 2, r = 3, Sn = 242, find n.
(ii) sum of the first 3 terms is 125 and the sum of the next 3 terms is 27, find the value of r.
Answer: Solution:
(i) a = 2, r = 3, Sn = 242
Sn = \( a \left( \frac{r^n-1}{r-1} \right) \), for r > 1
\( \therefore \) 242 = \( 2 \left( \frac{3^n-1}{3-1} \right) \)
\( \therefore \) 242 = \( 2 \left( \frac{3^n-1}{2} \right) \)
\( \therefore \) 242 = \( 3^n-1 \)
\( \therefore \) \( 3^n = 243 \)
\( \therefore \) \( 3^n = 3^5 \)
\( \therefore \) n = 5
Thus, our answer is correct.
(ii) S3 = 125, S6 = 125 + 27 = 152
Sn = \( a \left( \frac{1-r^n}{1-r} \right) \)
\( \therefore \) S3 = \( a \left( \frac{1-r^3}{1-r} \right) \)
125 = \( a \left( \frac{1-r^3}{1-r} \right) \) ...(i)
Also, S6 = \( a \left( \frac{1-r^6}{1-r} \right) \)
152 = \( a \left( \frac{1-r^6}{1-r} \right) \) ...(ii)
Dividing (ii) by (i), we get
\( \frac{152}{125} = \frac{a \left( \frac{1-r^6}{1-r} \right)}{a \left( \frac{1-r^3}{1-r} \right)} \)
\( \therefore \frac{152}{125} = \frac{1-r^6}{1-r^3} \)
\( \therefore \frac{152}{125} = \frac{(1+r^3)(1-r^3)}{1-r^3} \)
\( \therefore 1+r^3 = \frac{152}{125} \)
\( \therefore r^3 = \frac{152}{125} - 1 \)
\( \therefore r^3 = \frac{152-125}{125} \)
\( \therefore r^3 = \frac{27}{125} \)
\( \therefore r^3 = \left( \frac{3}{5} \right)^3 \)
\( \therefore r = \frac{3}{5} \)
In simple words: Part (i) shows how to find the number of terms 'n' when the first term, common ratio, and sum are given. Part (ii) demonstrates finding the common ratio 'r' by setting up a ratio of sums of different numbers of terms, using the formula for sum of G.P.

🎯 Exam Tip: When dealing with sums of 'n' terms and '2n' terms, remember the identity \( \frac{1-r^{2n}}{1-r^n} = 1+r^n \). This simplification is key for solving problems like part (ii) efficiently.

Question 4. For a G. P.,
(i) if t3 = 20, t6 = 160, find S7.
(ii) if t4 = 16, t9 = 512, find S10.
Answer: Solution:
(i) t3 = 20, t6 = 160
tn = \( ar^{n-1} \)
\( \therefore \) t3 = \( ar^{3-1} = ar^2 \)
\( \therefore ar^2 = 20 \)
\( \therefore a = \frac{20}{r^2} \) ......(i)
Also, t6 = \( ar^5 \)
\( ar^5 = 160 \)
\( \left( \frac{20}{r^2} \right) r^5 = 160 \) ...[From (i)]
\( 20r^3 = 160 \)
\( r^3 = \frac{160}{20} = 8 \)
r = 2
Substituting the value of r in (i), we get
\( a = \frac{20}{2^2} = \frac{20}{4} = 5 \)
Now, Sn = \( \frac{a(r^n-1)}{r-1} \), for r > 1
S7 = \( \frac{5(2^7-1)}{2-1} \)
= \( 5(128-1) = 635 \)
(ii) t4 = 16, t9 = 512
tn = \( ar^{n-1} \)
\( \therefore \) t4 = \( ar^{4-1} = ar^3 \)
\( ar^3 = 16 \)
\( \therefore a = \frac{16}{r^3} \) ...(i)
Also, t9 = \( ar^8 \)
\( ar^8 = 512 \)
\( \left( \frac{16}{r^3} \right) r^8 = 512 \)
\( 16r^5 = 512 \)
\( r^5 = \frac{512}{16} = 32 \)
\( r = 2 \)
Substituting r = 2 in (i), we get
\( a = \frac{16}{2^3} = \frac{16}{8} = 2 \)
Now, Sn = \( \frac{a(r^n-1)}{r-1} \) for r > 1
S10 = \( \frac{2(2^{10}-1)}{2-1} \)
= \( 2(1024-1) \)
= \( 2046 \)
In simple words: This problem shows how to find the first term 'a' and common ratio 'r' of a G.P. given two non-consecutive terms, and then use these values to calculate the sum of a specified number of terms.

🎯 Exam Tip: When given two terms of a G.P. (e.g., t_x and t_y), remember that \( \frac{t_y}{t_x} = r^{y-x} \). This shortcut can quickly help find the common ratio 'r' before finding 'a' and then the sum.

Question 5. Find the sum to n terms:
(i) 3 + 33 + 333 + 3333 + ....
(ii) 8 + 88 + 888 + 8888 + ....
Answer: Solution:
(i) Sn = 3 + 33 + 333 +..... upto n terms
= 3(1 + 11 + 111 +..... upto n terms)
= \( \frac{3}{9} \)(9 + 99 + 999 + upto n terms)
= \( \frac{3}{9} \)[(10 - 1) + (100 - 1) + (1000 - 1) +... upto n terms]
= \( \frac{3}{9} \)[(10 + 100 + 1000 + upto n terms) - (1 + 1 + 1 + ... n times)]
But 10, 100, 1000, n terms are in G.P. with a = 10, r = \( \frac{100}{10} \) = 10
\( \therefore \) Sn = \( \frac{3}{9} \left[ 10 \left( \frac{10^n-1}{10-1} \right) - n \right] \)
= \( \frac{3}{9} \left[ \frac{10}{9} (10^n-1) - n \right] \)
\( \therefore \) Sn = \( \frac{1}{3} \left[ \frac{10}{9} (10^n-1) - n \right] \)
= \( \frac{1}{27} [10(10^n-1)-9n] \)
(ii) Sn = 8 + 88 + 888 + upto n terms
= 8(1 + 11 + 111 + upto n terms)
= \( \frac{8}{9} \)(9 + 99 + 999 + upto n terms)
= \( \frac{8}{9} \)[(10 - 1) + (100 - 1) + (1000 - 1) +... upto n terms]
= \( \frac{8}{9} \)[(10 + 100 + 1000 + upto n terms) - (1 + 1 + 1 + ... n times)]
But 10, 100, 1000, n terms are in G.P. with a = 10, r = \( \frac{100}{10} \) = 10
\( \therefore \) Sn = \( \frac{8}{9} \left[ 10 \left( \frac{10^n-1}{10-1} \right) - n \right] \)
= \( \frac{8}{9} \left[ \frac{10}{9} (10^n-1) - n \right] \)
\( \therefore \) Sn = \( \frac{8}{81} [10(10^n-1)-9n] \)
In simple words: To find the sum of 'n' terms for sequences like 3, 33, 333..., or 8, 88, 888..., convert them into a form involving powers of 10 minus 1, then apply the G.P. sum formula for powers of 10.

🎯 Exam Tip: For problems involving sums of repeating digits (e.g., 3, 33, 333), the trick is to factor out the digit, multiply by 9/9, and express terms as \( (10^k-1) \). This transforms the series into a G.P. and a simple arithmetic series.

Question 6. Find the sum to n terms:
(i) 0.4 + 0.44 + 0.444 + ....
(ii) 0.7 + 0.77 + 0.777 + ....
Answer: Solution:
(i) Sn = 0.4 + 0.44 + 0.444 + upto n terms
= 4(0.1 + 0.11 + 0.111 + upto n terms)
= \( \frac{4}{9} \)(0.9 + 0.99 + 0.999 + upto n terms)
= \( \frac{4}{9} \)[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) ... upto n terms]
= \( \frac{4}{9} \)[(1 + 1 + 1 + ...n times) - (0.1 + 0.01 + 0.001 +... upto n terms)]
But 0.1, 0.01, 0.001, n terms are in G.P.
with a = 0.1, r = \( \frac{0.01}{0.1} \) = 0.1
\( \therefore \) Sn = \( \frac{4}{9} \left[ n - 0.1 \left( \frac{1-(0.1)^n}{1-0.1} \right) \right] \)
\( \therefore \) Sn = \( \frac{4}{9} \left[ n - 0.1 \left( \frac{1-(0.1)^n}{0.9} \right) \right] \)
\( \therefore \) Sn = \( \frac{4}{9} \left[ n - \frac{1}{9} (1-(0.1)^n) \right] \)
\( \therefore \) Sn = \( \frac{4}{81} \left[ 9n - (1-\frac{1}{10^n}) \right] \)
(ii) Sn = 0.7 + 0.77 + 0.777 + upto n terms
= 7(0.1 + 0.11 + 0.111 + upto n terms)
= \( \frac{7}{9} \)(0.9 + 0.99 + 0.999 + upto n terms)
= \( \frac{7}{9} \)[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) +... upto n terms]
= \( \frac{7}{9} \)[(1 + 1 + 1 +... n times) - (0.1 + 0.01 + 0.001 +... upto n terms)]
But 0.1, 0.01, 0.001, n terms are in G.P.
with a = 0.1, r = \( \frac{0.01}{0.1} \) = 0.1
\( \therefore \) Sn = \( \frac{7}{9} \left[ n - 0.1 \left( \frac{1-(0.1)^n}{1-0.1} \right) \right] \)
\( \therefore \) Sn = \( \frac{7}{9} \left[ n - \frac{0.1}{0.9} (1-(0.1)^n) \right] \)
\( \therefore \) Sn = \( \frac{7}{9} \left[ n - \frac{1}{9} (1-(0.1)^n) \right] \)
\( \therefore \) Sn = \( \frac{7}{81} \left[ 9n - (1-\frac{1}{10^n}) \right] \)
In simple words: To sum decimal sequences with repeating digits, factor out the repeating digit, convert the terms to (1 - powers of 0.1), and then apply the G.P. sum formula for the decimal part.

🎯 Exam Tip: Similar to integer repeating digit sums, for decimal repeating digit sums (e.g., 0.4, 0.44), convert terms to \( \frac{\text{digit}}{9} \times (1 - 10^{-k}) \). This again breaks the series into an arithmetic progression and a geometric progression, making summation straightforward.

Question 7. Find the nth terms of the sequences:
(i) 0.5, 0.55, 0.555,.....
(ii) 0.2, 0.22, 0.222,.....
Answer: Solution:
(i) Let t1 = 0.5, t2 = 0.55, t3 = 0.555 and so on.
t1 = 0.5
t2 = 0.55 = 0.5 + 0.05
t3 = 0.555 = 0.5 + 0.05 + 0.005
\( \therefore \) tn = 0.5 + 0.05 + 0.005 + upto n terms
But 0.5, 0.05, 0.005, ... upto n terms are in G.P. with a = 0.5 and r = 0.1
\( \therefore \) tn = the sum of first n terms of the G.P.
\( \therefore \) tn = \( 0.5 \left( \frac{1-(0.1)^n}{1-0.1} \right) \)
\( \therefore \) tn = \( \frac{0.5}{0.9} \{1 - (0.1)^n\} \)
\( \therefore \) tn = \( \frac{5}{9} \{1 - (0.1)^n\} \)
(ii) Let t1 = 0.2, t2 = 0.22, t3 = 0.222 and so on
t1 = 0.2
t2 = 0.22 = 0.2 + 0.02
t3 = 0.222 = 0.2 + 0.02 + 0.002
\( \therefore \) tn = 0.2 + 0.02 + 0.002 + upto n terms
But 0.2, 0.02, 0.002, ... upto n terms are in G.P. with a = 0.2 and r = 0.1
\( \therefore \) tn = the sum of first n terms of the G.P.
\( \therefore \) tn = \( 0.2 \left( \frac{1-(0.1)^n}{1-0.1} \right) \)
\( \therefore \) tn = \( \frac{0.2}{0.9} \{1-(0.1)^n\} \)
\( \therefore \) tn = \( \frac{2}{9} \{1-(0.1)^n\} \)
In simple words: To find the nth term of these sequences, recognize that each term is a sum of a geometric progression of decimals, where the first term is the initial digit/10 and the common ratio is 0.1.

🎯 Exam Tip: For sequences like 0.5, 0.55, 0.555..., note that the nth term, \( t_n \), is actually the sum of a G.P. up to 'n' terms (e.g., \( t_n = 0.5 + 0.05 + ... + 0.5 \times (0.1)^{n-1} \)). Apply the G.P. sum formula directly to find \( t_n \).

Question 8. For a sequence, if Sn = 2(3n-1), find the nth term, hence showing that the sequence is a G.P.
Answer: Solution:
Sn = \( 2(3^n-1) \)
Sn-1 = \( 2(3^{n-1}-1) \)
But tn = Sn - Sn-1
= \( 2(3^n-1) - 2(3^{n-1}-1) \)
= \( 2(3^n-1 - 3^{n-1} + 1) \)
= \( 2(3^n - 3^{n-1}) = 2(3 \times 3^{n-1} - 3^{n-1}) \)
\( \therefore \) tn = \( 2 \times 3^{n-1}(3-1) = 4 \times 3^{n-1} \)
tn-1 = \( 4 \times 3^{(n-1)-1} = 4 \times 3^{n-2} \)
The sequence (tn) is a G. P., if \( \frac{t_n}{t_{n-1}} \) = constant for all n \( \in \) N
\( \frac{t_n}{t_{n-1}} = \frac{4 \times 3^{n-1}}{4 \times 3^{n-2}} \)
= \( \frac{3^{n-1}}{3^{n-2}} = 3^{(n-1)-(n-2)} \)
= \( 3^1 = 3 \) = constant for all n \( \in \) N
\( \therefore \) r = 3
\( \therefore \) the sequence is a G.P. with tn = \( 4 \times 3^{n-1} \).
In simple words: The nth term of a sequence can be found by subtracting the sum of (n-1) terms from the sum of 'n' terms. If the ratio of consecutive terms ( \( \frac{t_n}{t_{n-1}} \) ) is a constant, the sequence is a Geometric Progression.

🎯 Exam Tip: To prove a sequence is a G.P. given \( S_n \), always calculate \( t_n = S_n - S_{n-1} \) and then find the ratio \( \frac{t_n}{t_{n-1}} \). If this ratio is a constant, it confirms a G.P., and that constant is the common ratio.

Question 9. If S, P, R are the sum, product and sum of the reciprocals of n terms of a G.P. respectively, then verify that \( \left(\frac{S}{R}\right)^n = P^2 \).
Answer: Solution:
Let a be the 1st term and r be the common ratio of the G.P.
\( \therefore \) the G.P. is a, ar, \( ar^2 \), \( ar^3 \), ..., \( ar^{n-1} \)
S = \( a + ar + ar^2 + ... + ar^{n-1} = a \left( \frac{r^n-1}{r-1} \right) \)
P = a(ar) (\( ar^2 \)) ... (\( ar^{n-1} \))
= \( a^n r^{1+2+3+...+(n-1)} \)
= \( a^n r^{\frac{n(n-1)}{2}} \)
\( \therefore P^2 = a^{2n} r^{n(n-1)} \) ...(i)
R = \( \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + ... + \frac{1}{ar^{n-1}} \)
= \( \frac{r^{n-1}+r^{n-2}+r^{n-3} + ... + r^2+r+1}{ar^{n-1}} \)
= \( \frac{1+r+r^2 +...+r^{n-2}+r^{n-1}}{ar^{n-1}} \)
1, r, \( r^2 \), ..., \( r^{n-1} \) are in G.P., with a = 1, r = r
\( \therefore 1+r+r^2+...+ r^{n-1} = 1 \cdot \left( \frac{r^n-1}{r-1} \right) \)
\( \therefore R = \frac{\frac{r^n-1}{r-1}}{ar^{n-1}} = \frac{1}{a r^{n-1}} \times \frac{r^n-1}{r-1} \)
\( \therefore R = \frac{1}{ar^{n-1}} \cdot \frac{S}{a} \)
\( \therefore a^2 r^{n-1} R = S \)
\( \therefore a^2 r^{n-1} = \frac{S}{R} \)
\( \therefore (a^2 r^{n-1})^n = \left( \frac{S}{R} \right)^n \)
\( \therefore a^{2n} r^{n(n-1)} = \left( \frac{S}{R} \right)^n \)
\( \therefore P^2 = \left( \frac{S}{R} \right)^n \) ...[From (i)]
In simple words: This problem verifies a relationship between the sum, product, and sum of reciprocals of a G.P.'s terms. By expressing S, P, and R in terms of the first term 'a' and common ratio 'r', we can algebraically show that \( \left(\frac{S}{R}\right)^n = P^2 \).

🎯 Exam Tip: This is a classic G.P. identity problem. Remember the formula for the sum of an arithmetic progression (for the powers in P) and the sum of a geometric progression (for S and R). Careful algebraic manipulation is key to reaching the desired proof.

Question 10. If Sn, S2n, S3n are the sum of n, 2n, 3n terms of a G.P. respectively, then verify that Sn (S3n – S2n) = (S2n – Sn)².
Answer: Solution:
Let a and r be the 1st term and common ratio of the G.P. respectively.
\( \therefore S_n = a \left( \frac{r^n-1}{r-1} \right) \), \( S_{2n} = a \left( \frac{r^{2n}-1}{r-1} \right) \), \( S_{3n} = a \left( \frac{r^{3n}-1}{r-1} \right) \)
\( \therefore S_{2n}-S_n = a \left( \frac{r^{2n}-1}{r-1} \right) - a \left( \frac{r^n-1}{r-1} \right) \)
= \( \frac{a}{r-1} (r^{2n}-1 - r^n+1) \)
= \( \frac{a}{r-1} (r^{2n}-r^n) \)
= \( \frac{ar^n}{r-1} (r^n-1) \)
\( \therefore S_{2n}-S_n = r^n \cdot a \left( \frac{r^n-1}{r-1} \right) \) ...(1)
\( \therefore S_{3n}-S_{2n} = a \left( \frac{r^{3n}-1}{r-1} \right) - a \left( \frac{r^{2n}-1}{r-1} \right) \)
= \( \frac{a}{r-1} (r^{3n}-1 - r^{2n}+1) \)
= \( \frac{a}{r-1} (r^{3n}-r^{2n}) \)
= \( \frac{ar^{2n}}{r-1} (r^n-1) \)
\( \therefore S_{3n}-S_{2n} = r^{2n} \cdot a \left( \frac{r^n-1}{r-1} \right) \)
Now, consider LHS:
\( S_n(S_{3n}-S_{2n}) = \left[ a \left( \frac{r^n-1}{r-1} \right) \right] \left[ r^{2n} \cdot a \left( \frac{r^n-1}{r-1} \right) \right] \)
= \( r^{2n} \cdot \left[ a \left( \frac{r^n-1}{r-1} \right) \right]^2 \)
Consider RHS:
\( (S_{2n}-S_n)^2 = \left[ r^n \cdot a \left( \frac{r^n-1}{r-1} \right) \right]^2 \) ...[From (1)]
= \( (r^n)^2 \cdot \left[ a \left( \frac{r^n-1}{r-1} \right) \right]^2 \)
= \( r^{2n} \cdot \left[ a \left( \frac{r^n-1}{r-1} \right) \right]^2 \)
Since LHS = RHS, the identity is verified.
In simple words: This problem asks to verify an identity relating sums of 'n', '2n', and '3n' terms of a G.P. By writing out the expressions for \( S_n \), \( S_{2n} \), and \( S_{3n} \) and performing algebraic manipulations, the given equation can be proven.

🎯 Exam Tip: This type of problem often involves factoring expressions like \( r^{2n}-r^n = r^n(r^n-1) \) and \( r^{3n}-r^{2n} = r^{2n}(r^n-1) \). Recognizing these common factors and carefully substituting the formula for \( S_n \) will lead to the solution.