Question:
The $2^{\text {nd }}, 31^{\text {st }}$ and the last terms of an AP are $7 \frac{3}{4}, \frac{1}{2}$ and $-6 \frac{1}{2}$ respectively.
Find the first term and the number of terms.
Solution:
To Find: First term and number of terms.
Given: $a_{2}=\frac{31}{4}, a_{31}=\frac{1}{2}$, and an $=\frac{-13}{2}$
Formula Used: $a_{n}=a+(n-1) d$
(Where $a=a_{1}$ is first term, $a_{2}$ is second term, $a_{n}$ is nth term and $d$ is common difference of given $\mathrm{AP}$ )
By using above formula, we have
$a_{2}=\frac{31}{4}=a+d$ and $a_{31}=\frac{1}{2}=a+(31-1) d$
On solving both equation, we get
$a=8$ and $d=-0.25$
Now an $=\frac{-13}{2}=8+(n-1)(-0.25)$
On solving the above equation, we get
N = 59
So the First term is equal to 8 and the number of terms is equal to 59.