Question

Find the number of terms of the AP $18,15\frac{1}{2},13,....,-49\frac{1}{2}$
and find the sum of all its terms.

Answer 1

The Ap is given as $18,15\frac{1}{2},13,....,-49\frac{1}{2}.$
First term a = 18, common difference $d=15\frac{1}{2}-18=-2\frac{1}{2}$ and the last term of the $AP=-49\frac{1}{2}.$
Let the AP has n terms
$a_n=a+(n-1)d$
$-\left(\frac{99}{2}\right)=18-\left(\frac{5}{2}\right)(n-1)$
$5(n-1)=135$
$n=28$
$\therefore n=27+1=28$
Thus, the given AP has 28 terms.
Now, the sum of all the terms $\left(S_n\right)$ is given by,
$S_n=\frac{n}{2}[2a+(n-1)d|=\frac{28}{2}[2\times 18+(28-1)\times (-\frac{5}{2})]=14[36-27\times \frac{5}{2}]=-441$
Thus, the sum of all the terms of the AP is -441