**Question:**

The first term of an AP is – 5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

**Solution:**

Let the first term, common difference and the number of terms of an AP are a, d and n respectively.

Given that, first term (a) = – 5 and last term (l) = 45

Sum of the terms of the AP = 120 ⇒ Sn = 120

We know that, if last term of an AP is known, then sum of n terms of an AP is,

$S_{n}=\frac{n}{2}(a+l)$

$\Rightarrow$ $120=\frac{n}{2}(-5+45) \Rightarrow 120 \times 2=40 \times n$

$\Rightarrow$ $n=3 \times 2 \Rightarrow n=6$

$\therefore$ Number of terms of an AP is known, then the $n$th term of an AP is,

$l=a+(n-1) d \Rightarrow 45=-5+(6-1) d$

$\Rightarrow \quad 50=5 d \Rightarrow d=10$

So, the common difference is 10.

Hence, number of terms and the common difference of an AP are 6 and 10 respectively.