KCET · Maths · Straight Lines
If the \( 2^{\text {nd }} \) and \( 5^{\text {th }} \) terms of G.P. are \( 24 \) and \( 3 \) respectively, then the sum of \( 1^{\text {st }} \) six terms
is
- A \( \frac{189}{2} \)
- B \( \frac{189}{5} \)
- C \( \frac{179}{2} \)
- D \( \frac{2}{189} \)
Answer & Solution
Correct Answer
(A) \( \frac{189}{2} \)
Step-by-step Solution
Detailed explanation
Given that, \(T_{2}=24\) and \(T_{5}=3\)
Now, \(a r=24 \rightarrow(1)\)
and \(a r^{4}=3 \rightarrow(2)\)
Dividing Eq. (2) by Eq. (1), we have
\(\frac{a r^{4}}{a r}=\frac{3}{24} \Rightarrow r^{3}=\frac{1}{8} \Rightarrow r=\frac{1}{2}\)
So, from Eq. (1), we have
\(a=24 \times 2=48\)
Therefore,
\(S_{6}=\frac{a\left(1-r^{5}\right)}{1-r}=\frac{48\left(1-(1 / 2)^{6}\right)}{1-1 / 2}=\frac{189}{2}\)
Now, \(a r=24 \rightarrow(1)\)
and \(a r^{4}=3 \rightarrow(2)\)
Dividing Eq. (2) by Eq. (1), we have
\(\frac{a r^{4}}{a r}=\frac{3}{24} \Rightarrow r^{3}=\frac{1}{8} \Rightarrow r=\frac{1}{2}\)
So, from Eq. (1), we have
\(a=24 \times 2=48\)
Therefore,
\(S_{6}=\frac{a\left(1-r^{5}\right)}{1-r}=\frac{48\left(1-(1 / 2)^{6}\right)}{1-1 / 2}=\frac{189}{2}\)
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