Q102 / 180 · KCET 2014 · Mathematics
Consider an infinite geometric series with first term ' \( a \) ' and common ratio ' \( r \) '. If the sum is \( 4 \) and
the second term is \( \frac{3}{4} \), then
the second term is \( \frac{3}{4} \), then
- A\( a=\frac{4}{7}, r=\frac{3}{7} \)
- Correct answerB\( a=3, r=\frac{1}{4} \)
- C\( a=2, r=\frac{3}{8} \)
- D\( a=\frac{3}{2}, r=\frac{1}{2} \)
Solution
Given that, first term is a and common ratio is \( r \).
and \( t_{2}=\frac{3}{4} \) \( \Rightarrow a r=\frac{3}{4} \rightarrow(1) \)
Now, \( 4=\frac{a}{1-r} \Rightarrow a=4-4 r \)
\( \Rightarrow 4 r=4-a \rightarrow(2) \)
From Eqs. (1) and (2), we get
\( a=3, r=\frac{1}{4} \)
and \( t_{2}=\frac{3}{4} \) \( \Rightarrow a r=\frac{3}{4} \rightarrow(1) \)
Now, \( 4=\frac{a}{1-r} \Rightarrow a=4-4 r \)
\( \Rightarrow 4 r=4-a \rightarrow(2) \)
From Eqs. (1) and (2), we get
\( a=3, r=\frac{1}{4} \)
Hence, the correct answer is option: 'B'.