Question

In an increasing geometric progression, the sum of the first and the last term is $99,$ the product of the second and the last but one term is $288$ and the sum of all the terms is $189.$ Then, the number of terms in the progression is equal to

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (B)

G.P. is increasing, i.e. $r>1.$
Given, $a+ar^{n - 1}=99,ar\cdot ar^{n - 2}=288$ and $\frac{a \left(1 - r^{n}\right)}{1 - r}=189$ .
$a\left(1 + r^{n - 1}\right)=99$ and $a^{2}r^{n - 1}=288$
$\Rightarrow a\left(1 + \frac{288}{a^{2}}\right)=99$
$\Rightarrow a^{2}+288=99a\Rightarrow a^{2}-99a+288=0$
$\Rightarrow a=3,96\Rightarrow r^{n - 1}=\frac{288}{a^{2}}=32,\frac{1}{32}$
As $r>1\Rightarrow r^{n - 1}=32fora=3$
Now, $\frac{a \left(1 - r^{n}\right)}{1 - r}=189\Rightarrow \frac{3 \left(1 - r \cdot r^{n - 1}\right)}{1 - r}=189$
$\Rightarrow \left(\frac{1 - r \cdot 32}{1 - r}\right)=63\Rightarrow 1-32r=63-63r$
$\Rightarrow 31r=62\Rightarrow r=2.$
Now, $2^{n - 1}=32\Rightarrow n=6.$