The third term of a geometric progression is square of its first term and the fifth term of it is 64. Find the sum of the first six terms of the Geometric Progression.

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Solution

$a_{n} = a r^{(n - 1)}$ , where $r$ is the common ratio of the GP.
According to question, we have
$a_{3} = (a)^{2}$
$\Rightarrow a r^{2} = a^{2}$
$\Rightarrow r^{2} = a$
Also given, $a_{5} = 64$
$\Rightarrow a r^{4} = 64$
$\Rightarrow r^{6} = 64$
Thus, $r = 2$ and $a = 4$
$S_{n} = \frac{{a (r}^{n} - 1)}{r - 1}$
So, $S_{6} = \frac{4 \times (64 - 1)}{2 - 1}$
$\Rightarrow S_{6} = 252$

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The third term of a geometric progression is square of its first term and the fifth term of it is 64. Find the sum of the first six terms of the Geometric Progression.

an=ar(n−1), where r is the common ratio of the GP.According to question, we havea3=(a)2⇒ar2=a2⇒r2=aAlso given, a5=64⇒ar4=64⇒r6=64Thus, r=2 and a=4Sn=a(rn−1)r−1So, S6=4×(64−1)2−1⇒S6=252