Question

The third term of an A.P. is 8 and the ninth term of the A.P. exceeds three times the third term by 2. Find the sum of its first 19 terms.

Answer 1

We know that the $n$th term of an A.P with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$.

Here, it is given that the third term of an A.P is $8$, therefore,

$\Rightarrow$$T_3=a+(3-1)d$
$\Rightarrow 8=a+2d$
$\Rightarrow a+2d=8......\left(1\right)$

It is also given that the ninth term of an A.P exceeds three times the third term by $2$, therefore,

$\Rightarrow$$T_9=3T_3+2=(3\times 8)+2=24+2=26$

But

$\Rightarrow$$T_9=a+(9-1)d=a+8d$, thus,

$\Rightarrow$$a+8d=\mathrm{26.........}\left(2\right)$

Now, subtract equation 1 from equation 2 as follows:

$\Rightarrow$$(a-a)+(8d-2d)=26-8$

$\Rightarrow 6d=18$
$\Rightarrow d=\frac{18}{6}=3$

Substitute $d=3$ in equation 1:

$a+(2\times 3)=8\Rightarrow a+6=8\Rightarrow a=8-6=2$

We also know that the sum of $n$ terms of an A.P with first term $a$ and common difference $d$ is:

$\Rightarrow$$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow$Substitute $n=19$, $a=2$ and $d=3$ in $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ as follows:

$\Rightarrow$$S_{19}=\frac{19}{2}\left[\right(2\times 2)+(19-1\left)3\right]=\frac{19}{2}[4+(18\times 3\left)\right]=\frac{19}{2}(4+54)=\frac{19}{2}\times 58=19\times 29=551$

Hence, the sum of the first $19$ terms of an A.P is $S_{19}=551$.