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14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

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Let GP be $a,ar,ar^2,ar^3,ar^4,ar^5,ar^6................................$

Given : The sum of first three terms of a G.P. is 16

$a+ar+ar^2=16$

$\Rightarrow a(1+r+r^2)=16...............................(1)$

Given : the sum of the next three terms is128.

$ar^3+ar^4+ar^5=128$

$\Rightarrow ar^3(1+r+r^2)=128...............................(2)$

Dividing equation (2) by (1), we have

$\Rightarrow \frac{ar^3(1+r+r^2)}{a(1+r+r^2)}=\frac{128}{16}$

$\Rightarrow r^3=8$

$\Rightarrow r^3=2^3$

$\Rightarrow r=2$

Putting value of r =2 in equation 1,we have

$\Rightarrow a(1+2+2^2)=16$

$\Rightarrow a(7)=16$

$\Rightarrow a=\frac{16}{7}$

$S_n=\frac{a(1-r^n)}{1-r}$

$S_n=\frac{\frac{16}{7}(1-2^n)}{1-2}$

$S_n=\frac{16}{7}(2^n-1)$

Posted by

seema garhwal

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