Question

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

Answer 1

Let a,d and n be the first term, common difference and the no. of terms of given AP.

Here $n=37$

so middle most term is $\frac{n+1}{2}$ $\Rightarrow \frac{37+1}{2}\Rightarrow {18}^{th}$

$\therefore$ three middle most terms are ${18}^{th},{19}^{th},{20}^{th}$

$a_{18}+a_{19}+a_{20}=225\Rightarrow a+17d+a+18d+a+19d=225$

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

last three terms are ${35}^{th},{36}^{th},{37}^{th}$

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

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

subtracting (1) from (2), we get

$51d=204$

$\Rightarrow d=4$

putting $d$ in (1), we get

$3a+54\times 4=225$

$3a=225-216\Rightarrow a=3$

Therefore the AP is $a,a+d,a+2d,a+3d,....$

i.e $3,7,11,\mathrm{15......}$