Maharashtra Board Class 11 Maths Part 1 Chapter 4 Sequences and Series 4.3 Solutions

Question 1. Determine whether the sum to infinity of the following G.P's exist. If exists, find it.
(i) \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots \)
(ii) \( 2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots \)
(iii) \( -3, 1, -\frac{1}{3}, \frac{1}{9}, \ldots \)
(iv) \( \frac{1}{5}, -\frac{2}{5}, \frac{4}{5}, -\frac{8}{5}, \frac{16}{5}, \ldots \)
Answer:
(i) The given G.P. is \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots \)
Here, \( a = \frac{1}{2}, r = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \)
Since, \( |r| = |\frac{1}{2}| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \)
(ii) The given G.P. is \( 2, \frac{4}{3}, \frac{8}{9}, \frac{16}{27}, \ldots \)
Here, \( a = 2, r = \frac{\frac{4}{3}}{2} = \frac{2}{3} \)
Since, \( |r| = |\frac{2}{3}| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{2}{1-\frac{2}{3}} = \frac{2}{\frac{1}{3}} = 6 \)
(iii) The given G.P. is \( -3, 1, -\frac{1}{3}, \frac{1}{9}, \ldots \)
Here, \( a = -3, r = \frac{1}{-3} = -\frac{1}{3} \)
Since, \( |r| = |-\frac{1}{3}| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{-3}{1-(-\frac{1}{3})} = \frac{-3}{1+\frac{1}{3}} = \frac{-3}{\frac{4}{3}} = -\frac{9}{4} \)
(iv) The given G.P. is \( \frac{1}{5}, -\frac{2}{5}, \frac{4}{5}, -\frac{8}{5}, \frac{16}{5}, \ldots \)
Here, \( a = \frac{1}{5}, r = \frac{-\frac{2}{5}}{\frac{1}{5}} = -2 \)
Since, \( |r| = |-2| = 2 > 1 \)
Sum to infinity does not exist.
In simple words: The sum to infinity of a Geometric Progression (G.P.) exists only if the absolute value of its common ratio (r) is less than 1. If it exists, the sum is calculated using the formula \( S_\infty = \frac{a}{1-r} \), where 'a' is the first term.

🎯 Exam Tip: Always check the condition \( |r| < 1 \) before attempting to calculate the sum to infinity of a G.P. Showing this condition is crucial for full marks.

Question 2. Express the following recurring decimals as a rational number.
(i) \( 0.\overline{32} \)
(ii) \( 3.\overline{5} \)
(iii) \( 4.\overline{18} \)
(iv) \( 0.3\overline{45} \)
(v) \( 3.4\overline{56} \)
Answer:
(i) \( 0.\overline{32} = 0.323232 \ldots \)
\( = 0.32 + 0.0032 + 0.000032 + \ldots \)
Here, \( 0.32, 0.0032, 0.000032, \ldots \) are in G.P. with \( a = 0.32 \) and \( r = 0.01 \)
Since, \( |r| = |0.01| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{0.32}{1-(0.01)} = \frac{0.32}{0.99} \)
\( \implies 0.\overline{32} = \frac{32}{99} \)
(ii) \( 3.\overline{5} = 3.555 \ldots = 3 + 0.5 + 0.05 + 0.005 + \ldots \)
Here, \( 0.5, 0.05, 0.005, \ldots \) are in G.P. with \( a = 0.5 \) and \( r = 0.1 \)
Since, \( |r| = |0.1| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{0.5}{1-(0.1)} = \frac{0.5}{0.9} = \frac{5}{9} \)
\( \implies 3.\overline{5} = 3 + \frac{5}{9} = \frac{27+5}{9} = \frac{32}{9} \)
(iii) \( 4.\overline{18} = 4.181818 \ldots \)
\( = 4 + 0.18 + 0.0018 + 0.000018 + \ldots \)
Here, \( 0.18, 0.0018, 0.000018, \ldots \) are in G.P. with \( a = 0.18 \) and \( r = 0.01 \)
Since, \( |r| = |0.01| < 1 \)
Sum to infinity exists.
Sum of infinity \( = \frac{a}{1-r} = \frac{0.18}{1-(0.01)} = \frac{0.18}{0.99} = \frac{18}{99} = \frac{2}{11} \)
\( \implies 4.\overline{18} = 4 + \frac{2}{11} = \frac{44+2}{11} = \frac{46}{11} \)
(iv) \( 0.3\overline{45} = 0.3454545 \ldots \)
\( = 0.3 + 0.045 + 0.00045 + 0.0000045 + \ldots \)
Here, \( 0.045, 0.00045, 0.0000045, \ldots \) are in G.P. with \( a = 0.045 \) and \( r = 0.01 \)
Since, \( |r| = |0.01| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{0.045}{1-0.01} = \frac{0.045}{0.99} = \frac{45}{990} \)
\( \implies 0.3\overline{45} = 0.3 + \frac{45}{990} \)
\( = \frac{3}{10} + \frac{1}{22} \)
\( = \frac{33+5}{110} = \frac{38}{110} = \frac{19}{55} \)
Alternate method:
Let \( x = 0.3\overline{45} \)
\( 10x = 3.\overline{45} = 3 + 0.\overline{45} \)
Now, \( 0.\overline{45} = 0.454545 \ldots = 0.45 + 0.0045 + 0.000045 + \ldots \)
Here, \( a = 0.45, r = 0.01 \)
Sum to infinity \( = \frac{0.45}{1-0.01} = \frac{0.45}{0.99} = \frac{45}{99} = \frac{5}{11} \)
So, \( 10x = 3 + \frac{5}{11} = \frac{33+5}{11} = \frac{38}{11} \)
\( \implies x = \frac{38}{110} = \frac{19}{55} \)
(v) \( 3.4\overline{56} = 3.4565656 \ldots \)
\( = 3.4 + 0.056 + 0.00056 + 0.0000056 + \ldots \)
Here, \( 0.056, 0.00056, 0.0000056, \ldots \) are in G.P. with \( a = 0.056 \) and \( r = 0.01 \)
Since, \( |r| = |0.01| < 1 \)
Sum to infinity exists.
Sum to infinity \( = \frac{a}{1-r} = \frac{0.056}{1-0.01} = \frac{0.056}{0.99} = \frac{56}{990} \)
\( \implies 3.4\overline{56} = 3.4 + \frac{56}{990} \)
\( = \frac{34}{10} + \frac{56}{990} \)
\( = \frac{3366+56}{990} = \frac{3422}{990} = \frac{1711}{495} \)
In simple words: To convert a recurring decimal into a rational number, express the repeating part as the sum to infinity of a Geometric Progression (G.P.) and then combine it with the non-repeating part.

🎯 Exam Tip: When converting recurring decimals, carefully identify the first term 'a' and the common ratio 'r' of the G.P. formed by the repeating digits. Remember that \( r \) will often be \( 0.1, 0.01, 0.001 \), etc., depending on the number of repeating digits.

Question 3. If the common ratio of a G.P. is \( \frac{2}{3} \) and sum of its terms to infinity is 12. Find the first term.
Answer:
Given: \( r = \frac{2}{3} \), sum to infinity \( = 12 \)
Sum to infinity \( = \frac{a}{1-r} \)
\( \implies 12 = \frac{a}{1-\frac{2}{3}} \)
\( \implies 12 = \frac{a}{\frac{1}{3}} \)
\( \implies a = 12 \times \frac{1}{3} \)
\( \implies a = 4 \)
In simple words: Using the formula for the sum to infinity of a G.P., \( S_\infty = \frac{a}{1-r} \), we can find the first term 'a' when the sum to infinity and common ratio 'r' are known.

🎯 Exam Tip: Ensure you correctly substitute the given values into the sum to infinity formula and perform algebraic operations carefully to solve for the unknown variable.

Question 4. If the first term of a G.P. is 16 and sum of its terms to infinity is \( \frac{176}{5} \), find the common ratio.
Answer:
Given: \( a = 16 \), sum to infinity \( = \frac{176}{5} \)
Sum to infinity \( = \frac{a}{1-r} \)
\( \implies \frac{176}{5} = \frac{16}{1-r} \)
\( \implies 176(1-r) = 16 \times 5 \)
\( \implies 176 - 176r = 80 \)
\( \implies 176r = 176 - 80 \)
\( \implies 176r = 96 \)
\( \implies r = \frac{96}{176} = \frac{12 \times 8}{22 \times 8} = \frac{12}{22} = \frac{6}{11} \)
In simple words: With the first term and sum to infinity given, the common ratio 'r' of a G.P. can be found by rearranging the sum to infinity formula \( S_\infty = \frac{a}{1-r} \) and solving for 'r'.

🎯 Exam Tip: Double-check your algebraic steps, especially when isolating 'r', to avoid calculation errors. Always verify that the calculated 'r' satisfies \( |r| < 1 \).

Question 5. The sum of the terms of an infinite G.P. is 5 and the sum of the squares of those terms is 15. Find the G.P.
Answer:
Let the required G.P. be \( a, ar, ar^2, ar^3, \ldots \)
Sum to infinity of this G.P. \( = 5 \)
\( \implies 5 = \frac{a}{1-r} \)
\( \implies a = 5(1-r) \ldots(i) \)
Also, the sum of the squares of the terms is 15.
The G.P. of squares is \( a^2, a^2r^2, a^2r^4, \ldots \)
\( \implies (a^2 + a^2r^2 + a^2r^4 + \ldots) = 15 \)
This is a G.P. with first term \( a^2 \) and common ratio \( r^2 \). Since \( |r|<1 \), \( |r^2|<1 \).
So, \( 15 = \frac{a^2}{1-r^2} \)
\( \implies 15(1-r^2) = a^2 \)
Substitute \( a = 5(1-r) \) from (i):
\( \implies 15(1-r)(1+r) = [5(1-r)]^2 \)
\( \implies 15(1-r)(1+r) = 25(1-r)^2 \)
Since \( 1-r \neq 0 \) (as sum to infinity exists), we can divide both sides by \( (1-r) \):
\( \implies 15(1+r) = 25(1-r) \)
Divide by 5:
\( \implies 3(1+r) = 5(1-r) \)
\( \implies 3+3r = 5-5r \)
\( \implies 3r+5r = 5-3 \)
\( \implies 8r = 2 \)
\( \implies r = \frac{2}{8} = \frac{1}{4} \)
Substitute \( r = \frac{1}{4} \) into (i):
\( a = 5(1-\frac{1}{4}) \)
\( \implies a = 5(\frac{3}{4}) \)
\( \implies a = \frac{15}{4} \)
Required G.P. is \( a, ar, ar^2, ar^3, \ldots \)
i.e., \( \frac{15}{4}, \frac{15}{4} \times \frac{1}{4}, \frac{15}{4} \times (\frac{1}{4})^2, \ldots \)
i.e., \( \frac{15}{4}, \frac{15}{16}, \frac{15}{64}, \ldots \)
In simple words: This problem involves two infinite G.P. sums: one for the terms themselves and another for their squares. By setting up equations based on the sum to infinity formula for both G.P.s, we can solve them simultaneously to find the first term 'a' and common ratio 'r' of the original G.P.

🎯 Exam Tip: When dealing with sums of squares of G.P. terms, remember that the new G.P. has \( a' = a^2 \) and \( r' = r^2 \). Careful algebraic manipulation and substitution are key to solving such problems accurately.