Question

First, second and seventh terms of an A.P. (all the terms are distinct), whose sum is $93$, are in G.P. Fourth term of this G.P. is

• A. $21$
• B. $31$
• C. $75$
• D. $375$

Answer 1

The correct option is D $375$

Since,

$\begin{array}{cc}& \\ a,a+d,a+6d\to areinA.P.\to \left(i\right)& \\ a+a+d+a+6d=93& \left[Given\right]\\ 3a+7d=93& \\ a,a+d,a+6d\to InG.P,& \\ So,& \\ {\left(a+d\right)}^2=a\left(a+6d\right)& \\ a^2+d^2+2ad=a^2+6ad& \\ d^2=4a& \\ d=4a& \end{array}$

Put the value of $d$ in equation (i)

$a,5a,25a\to G.P$

Then, from above

$\begin{array}{c}3a+7\left(4a\right)=93\\ 31a=93\\ a=3\\ Soterms\\ 3,15,75,375\\ using\\ a_4=a{r}^{n-1}\\ a_4=3{\left(5\right)}^{4-1}\\ a_4=3{\left(5\right)}^3\\ a_4=375\end{array}$

Hence, this is the answer.