Question

There are $37$ terms in an $A.P.,$ the sum of three terms placed exactly at the middle is $225$ and the sum of last three terms is $429$. Write the $A.P.$

Answer 1

Given that: Sum of three middle terms $=225$

Let the three middle most terms of the AP be: $a-d,a,a+d$

$\Rightarrow (a-d)+a+(a+d)=225$

$\Rightarrow 3a=225$

$\Rightarrow a=75$

Now, the A.P. is $a-18d,\dots ,a–2d,a-d,a,a+d,a+2d,\dots ,a+18d$

Also given that the sum of last three terms $=429$

$\Rightarrow (a+18d)+(a+17d)+(a+16d)=429$

$\Rightarrow 3a+51d=429$

$\Rightarrow a+17d=143$

$\Rightarrow 75+17d=143$

$\Rightarrow d=4$

Now, first term $=a-18d=75-18\left(4\right)=3$

$\therefore$ The A.P. is $3,7,11,\dots ,147$