Question

In an AP if $a_3$ = 8 and the ninth term exceeds 3 times third term by 2 then find the sum to 19 terms.

Answer 1

Let 'a' be the first term and 'd' be the common difference of the given A.P.
Then the nth term of the AP is
$a_n=a+(n-1)d$
Now, it is given that the third term is 8, then
$a_3=a+(3-1)d\Rightarrow a+2d=8....\left(1\right)$
Also, the ninth term exceeds 3 times third term by 2, then
$a_9=3a_3+2$
$\Rightarrow a+(9-1)d=3\left(8\right)+2$ $(As,a_3=8)$
$\Rightarrow a+8d=26.....\left(2\right)$
Subtracting equation (1) from equation (2), we get
$6d=18\Rightarrow d=3$
Put d = 3 in equation (1), we have
$a+2\left(3\right)=8\Rightarrow a=2$
Now, the sum of first n terms of the AP is given by
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Therefore, the sum of first 19 terms of the given AP is
$S_{19}=\frac{19}{2}\left[2\right(2)+(19-1\left)\right(3\left)\right]=\frac{19}{2}\times 58=551.$