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Sum of n Terms of an AP - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Sum of n terms of an AP is considered one the most difficult concept.
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60 Questions around this concept.
Solve by difficulty
Let be terms of an A.P. If
then
equals:
Find the sum of the first 15 terms of series $1+3+7+13+\ldots$
Let $S_n$ denote the sum of the first n terms of an A.P. If then $S_{3 \mathrm{n}}: S_{\mathrm{n}}$ is equal to
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The sum of n terms of the series
$
\left(\frac{1}{2}\right)+\left(\frac{3}{4}\right)+\left(\frac{7}{8}\right)+\left(\frac{15}{16}\right)+\left(\frac{31}{32}\right)+\cdots{\text {will be }}
$
Concepts Covered - 1
Sum of n terms of an AP
The sum, $\mathrm{S}_{\mathrm{n}}$ of n terms of an AP with the first term ' a ' and common difference ' d ' is given by
$
S_n=\frac{n}{2}[2 a+(n-1) d]
$
OR
$
S_n=\frac{n}{2}[a+l]
$
$a \rightarrow$ first term
$d \rightarrow$ common difference
$n \rightarrow$ number of terms
$
S_n=a+(a+d)+(a+2 d)+\ldots . .+(a+(n-2) d)+(a+(n-1) d) \ldots \ldots(i)
$
Rewrite
$
\begin{aligned}
& S_n=(a+(n-1) d)+(a+(n-2) d)+\ldots \ldots \ldots+(a-2 d)+(a-d)+a \ldots \ldots \\
& (i)+(i i) \\
& 2 S_n=(2 a+(n-1) d)+(2 a+(n-1) d+\ldots \ldots \ldots \ldots \ldots \text { upto } n \text { term } \\
& S_n=\frac{n}{2}[(2 a+(n-1) d)]
\end{aligned}
$
Also, $a+(n-1) d=l=$ last term
we can also write
$
S_n=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}[a+l]
$
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