Find the number of terms in each of the following APs:$18, 15\frac{1}{2}, 13, …, -47$
Given:
Given A.P. is $18, 15\frac{1}{2}, 13, …, -47$.
To do:
We have to find the number of terms in the given A.P.
Solution:
Here,
$a_1=18, a_2=15\frac{1}{2}, a_3=13$
Common difference $d=a_2-a_1=15\frac{1}{2}-18=\frac{15\times2+1}{2}-18=\frac{31-2\times18}{2}=\frac{31-36}{2}=\frac{-5}{2}$
Let $-47$ be the nth term.
We know that,
nth term $a_n=a+(n-1)d$
Therefore,
$a_{n}=18+(n-1)(\frac{-5}{2})$
$-47=18+\frac{-5(n-1)}{2}$
$-47-18=\frac{-5n+5}{2}$
$-65=\frac{-5n+5}{2}$
$2(-65)=-5n+5$ (On cross multiplication)
$5n=130+5$
$5n=135$
$n=\frac{135}{5}$
$n=27$
Hence, there are 27 terms in the given A.P.       
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