Question

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

Answer 1

$n=37$

Middle term$=\frac{37+1}{2}=19$

$3$ Middle term $={18}^{th},{19}^{th},{20}^{th}$

$T_{18}+T_{19}+T_{20}=225$

Last $3$ terms $={35}^{th},{36}^{th},{37}^{th}$

$T_{35}+T_{36}+T_{37}=429$

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

$\Rightarrow 3a+54d=225$

$\Rightarrow a+8d=75$

$\Rightarrow a=75-18d$

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

$\Rightarrow 3a+105d=429$

$\Rightarrow a+35d=143$

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

$\Rightarrow 17d=68$

$d=4$

$a=75-18\left(4\right)$

$=3$

$AP=3,7,11,15,19,....$.