Question

If the 8th term of an A.P, is 37 and the 15th term is 15 more than the 12th term, find the A.P.
Also, find the sum of first 20 terms of this A.P.

Answer 1

Given that ${15}^{th}$ term is $15$ more than the ${17}^{th}$ term

$a+14d=15+a+11d$

$14d-11d=15$

$3d=15$

$d=5$

Substitute the value of d in $t_8$ term we get

$a+7\left(5\right)=37$

$a+35=37$

$a=2$

$a=2$, $d=5$

Now, AP terms will be

First term $a=2$

$2^{nd}$ term $a+d=7$

$3^{rd}$ term $-a+2d=12$

$4^{th}$ term $=a+3d=17$ and so on

AP $=2,7,12,17$….

Sum of $1^{st}$ $20$ term in AP$=s_n=n/2(2a+(n-1\left)d\right)$

$=20/2[2\times 2+(20-1)\times 5]$

$=10[4+19\times 5]=10[4+95]=10\times 99=990$

Sum of the $1^{st}$ $20$ terms in A.P is $990$.