Question

Find the sum of the first $19$ terms of A.P, whose nth terms $t_n=7-4n$.

Answer 1

Since, $t_n=a+(n-1)d$

$\therefore a+(n-1)d=7-4d=7-4-4d+4=(7-4)+(n-1)\times 4=3+(n-1)\times 4$

So, $a=3$ and $d=4$

Thus, the sum of first $19$ terms $=S_{19}=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{19}{2}[6+18\times 4]=741$