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Find the number of terms in each of the following APs:
(i) $7, 13, 19, …, 205$
(ii) $18, 15\frac{1}{2}, 13, …, -47$
To do:
We have to find the number of terms in each of the given APs.
Solution:
(i) Given A.P. is $7, 13, 19, …, 205$.
Here,
$a_1=7, a_2=13, a_3=19$
Common difference $d=a_2-a_1=13-7=6$
Let $205$ be the nth term.
We know that,
nth term $a_n=a+(n-1)d$
Therefore,
$a_{n}=7+(n-1)(6)$
$205=7+n(6)-1(6)$
$205-7=6n-6$
$198+6=6n$
$6n=204$
$n=\frac{204}{6}$
$n=34$
Hence, there are 34 terms in the given A.P.     
(ii) Given A.P. is $18, 15\frac{1}{2}, 13, …, -47$.
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.