Question

An A.P. consist of $37$ term. the sum of the three middle most terms is $225$ then find the A.P.

Answer 1

An AP consists of $37$ terms. The sum of three middle most of terms is $225$ and sum of the last three is $429$. Find the A.P.

Sol:- Let the first term be 'a' and common difference be 'd' respectively.

Since the A.P. contains $37$ terms. So, the middle most term is $(37+1)/2$th term $=19th$ term.

Thus, three middle most terms of A.P are $18th$, $19th$ and $20th$ terms.

Given $a+(18-1)d+a(19-1)d+a(20-1)d=225$

$\Rightarrow a+17d+a+18d+a+19d=225$

$\Rightarrow 3a+54d=225$

$\Rightarrow 3(a+18d)=225$

$\Rightarrow a+18d=75$

$\Rightarrow a=75-18d$ ……….$\left(1\right)$

According to given information

$a+(35-1)d+a+(36-1)d+a+(37-1)d=429$

$\Rightarrow a+34d+a+35d+a+36d=429$

$\Rightarrow 3a+105d=429$

$\Rightarrow 3(a+35d)=429$

$\Rightarrow (75-18d+35d)=143$

$\Rightarrow 17d=143-75=68$

$\Rightarrow 17d=68$

$\Rightarrow d=4$ $\therefore$ Thus, the A.P. $3,7,11,15,....$

Put $d=4$ in $\left(1\right)$

$a=75-\left(8\right(4\left)\right)$

$a=75-72$

$a=3$.