Find the number of terms in each of the following APs:$18, 15\frac{1}{2}, 13, …, -47$

AcademicMathematicsNCERTClass 10

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.       

raja
Updated on 10-Oct-2022 13:20:19

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