Question

In a geometric progression with common ratio 'q', the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. If the sum of the first nine terms of the progression is $\frac{\lambda }{q^{100}}$ then the value of $\lambda$ equals to

• A. $10$
• B. $14$
• C. $12$
• D. $22$

Answer 1

The correct option is C $12$

$r=q$ (common ratio)

$S_n=\frac{a(r^n-1)}{(r-1)}{S}_{109}=S_{100}+12\frac{a(q^{109}-1)}{(q-1)}=\frac{a(q^{100}-1)}{(q-1)}+12\left(1\right)\frac{a(q^9-1)}{(q-1)}=\frac{\lambda }{q^{100}}\lambda =\frac{a(q^{109}-q^{100})}{(q-1)}\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\lambda =12$