Question

The sum of first three terms of a GP is $\frac{61}{5}$ and their product is 64. The fourth term of the GP can be

• A. $\frac{8}{5}$
• B. $\frac{64}{5}$
• C. $\frac{64}{25}$
• D. 1

Answer 1

The correct option is C $\frac{64}{25}$

Let the first three terms of be GP be $\frac{a}{r},a,ar$.
Given $\frac{a}{r}+a+ar=\frac{61}{5}$........(i)
and $\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=64$
$\Rightarrow a^3=64$
$\Rightarrow a=4$
Substituting a = 4 in (i), we get
$\frac{4}{r}+4+4r=\frac{61}{5}$
$\Rightarrow 5(4+4r+4r^2)=61r$
$\Rightarrow 20+20r+20r^2=61r$
$\Rightarrow 20r^2-41r+20=0$
$\Rightarrow 20r^2-25r-16r+20=0$
$\Rightarrow 5r(4r-5)-4(4r-5)=0$
$\Rightarrow (4r-5)(5r-4)=0$
$\Rightarrow r=\frac{5}{4}$ or $r=\frac{4}{5}$
Now, the fourth term of the GP will be $a{r}^2$
Thus, if $r=\frac{5}{4},a_4=4(\frac{5}{4}{)}^2=\frac{25}{4}$
or, if $r=\frac{4}{5},a_4=4(\frac{4}{5}{)}^2=\frac{64}{5}$

Hence, the correct answer is option (3).