Class 11

Math

Algebra

Sequences and Series

If there are $3A.M_{′}s$ between $4$ and $84$, find their sum.

- $88$
- $132$
- $166$
- $336$

Let the 3 AMs be $A_{1},A_{2},A_{3}$

then $4,A_{1},A_{2},A_{3},84$ forms an arithmetic progression

$a_{n}=a_{1}+(n−1)dS_{n}=2n [2a+(n−1)d]=2n (a+a_{n})$

here, $n=5,a=4,a_{n}=84S_{n}=25 (4+84)=220$

$220=4+A_{1}+A_{2}+A_{3}+84A_{1}+A_{2}+A_{3}=220−88A_{1}+A_{2}+A_{3}=132$