Question

There are $37$ terms in an A.P. The sum of the middle three term is $225$. And the sum of the last three term is $429$. Find the A.P.

Answer 1

No. of terms of an $A.P.=37$

Let first term be 'a' and common difference 'd'

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

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

There middle terms are $a+17d,a+18d,a+19d$

Then $a+17d+a+18d+a+19d=225$

$3a+54d=225$ ...(1)

Last three terms are $a+34d,a+35d,,a+36d$

Then $a+34d+a+35d+a+36d=429$

$3a+105d=429$ ...(2)

from (1) and (2) we get

$51d=204\Rightarrow d=4$

$3a+420=429$

$a=3$

AP is $3,7,11,15,....147$.