Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
is
Solution:
1656 Upvotes
Verified Answer
The correct answer is:
\( \frac{189}{2} \)
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,
$S_{6}=\frac{a\left(1-r^{5}\right)}{1-r}=\frac{48\left(1-(1 / 2)^{6}\right)}{1-1 / 2}=\frac{189}{2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.