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If the sum of the first $n$ terms of an AP is $\frac{1}{2}(3n^2+7n)$ then find its $n^{th}$ term. Hence, find its $20^{th}$ term.
Given: the sum of the first $n$ terms of an AP is $\frac{1}{2}(3n^2+7n)$
To do: To find its $n^{th}$ term and its $20^{th}$ term.
Solution:
$S_n=\frac{1}{2}( 3n^2+7n)$
$S_1=\frac{1}{2}(3+7)=5$
$S_2=\frac{1}{2}(3\times2^2+7\times2)=\frac{26}{2}=13$
We know
$S_1=a_1=5$
$S_2=a_1+a_2=13$
$\Rightarrow S_2-S_1=a_1+a_2-a_1$
$\Rightarrow 13-5=a_2$
$\Rightarrow a_2=8$
And we also know that common difference $d=a_2-a_1$
$\Rightarrow d=8-5=3$
$n^{th}$ term of AP $=a_n=5+( n-1)3$
$a_n= 2+3n$
Therefore $20^{th}$ term $=a_{20}=2+3(20)=62$
Hence $20^{th}$ term of AP is $62$.
Updated on: 10-Oct-2022
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