Question

If there are $3A.M^{\prime }s$ between $4$ and $84$, find their sum.

• A. $88$
• B. $132$
• C. $166$
• D. $336$

Answer 1

Answer: (B)

$132$

Let the 3 AMs be $A_1,A_2,A_3$

then $4,A_1,A_2,A_3,84$ forms an arithmetic progression

$a_n=a_1+(n-1)d{S}_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}(a+a_n)$

here, $n=5,a=4,a_n=84S_n=\frac{5}{2}(4+84)=220$

$220=4+A_1+A_2+A_3+84A_1+A_2+A_3=220-88A_1+A_2+A_3=132$