Question

In a geometric progression with common ration $q$ 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. $22$
• E. $13$

Answer 1

The correct option is C $12$

Let a be the first term, then
$a\left(\frac{q^{109}-1}{q-1}\right)-a\frac{q^{100}-1}{q-1}=12$
$\Rightarrow \frac{a}{q-1}(q^{109}-q^{100})=12$
$\Rightarrow a.q^{100}\frac{(q^9-1)}{q-1}=12$ ........(i)
$\Rightarrow q^{100}\frac{\lambda }{q^{100}}=12$ $\therefore \lambda =12$