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If the sum of $n$ terms of an A.P. is $2 n^2+5 n$, then the $n^{\text {th }}$ term will be
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Verified Answer
The correct answer is:
$4 n+3$
Given that $S_n=2 n^2+5 n$
Putting $n=1,2,3, \ldots \ldots \ldots . ., S_1=2 \times 1+5 \times 1=7$
So, $T_1=S_1=a=7, T_2=S_2-S_1=18-7=11$
$T_3=S_3-S_2=33-18=15$
Therefore series is $7,11,15, \ldots . . . .$.
Now, $n^{\text {th }}$ term $=a+(n-1) d=7+(n-1) 4=4 n+3$
Aliter : As we know $T_n=S_n-S_{n-1}$
$=\left(2 n^2+5 n\right)-\left\{2(n-1)^2+5(n-1)\right\}$
$=2 n^2+5 n-2 n^2+4 n-2-5 n+5=4 n+3$.
Putting $n=1,2,3, \ldots \ldots \ldots . ., S_1=2 \times 1+5 \times 1=7$
So, $T_1=S_1=a=7, T_2=S_2-S_1=18-7=11$
$T_3=S_3-S_2=33-18=15$
Therefore series is $7,11,15, \ldots . . . .$.
Now, $n^{\text {th }}$ term $=a+(n-1) d=7+(n-1) 4=4 n+3$
Aliter : As we know $T_n=S_n-S_{n-1}$
$=\left(2 n^2+5 n\right)-\left\{2(n-1)^2+5(n-1)\right\}$
$=2 n^2+5 n-2 n^2+4 n-2-5 n+5=4 n+3$.
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