Name: Sequence and series
Uploaded: 2019-07-19
Duration: 23 min 56 s
Description: In this video, the basic concepts of Sequence and series and sum of n terms of series are discussed. After watching this video, you will be able to solve most of the questions related to this topic.

Question

Is the $n^{th}$ term of a progression is equal to the difference between sum of the first $n$ terms $(S_{n})$ and sum of the first $(n - 1)$ terms, $(S_{n - 1})$ ?

Accepted Answer

We know that, $n^{th}$ term of an AP is $a_{n} = a + (n - 1) d . . . . (i)$

Sum of first $n$ terms of an AP is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Therefore, sum of first $(n - 1)$ terms of an AP is $S_{n - 1} = \frac{(n - 1)}{2} [2 a + (n - 1 - 1) d]$

$\Rightarrow S_{n - 1} = \frac{(n - 1)}{2} [2 a + (n - 2) d]$

Now, difference between sum of first $(n - 1)$ terms and sum of first $n$ terms

$\Rightarrow S_{n - 1} - S_{n}$

$= \frac{n}{2} [2 a + (n - 1) d] - \frac{(n - 1)}{2} [2 a + (n - 2) d]$

$= \frac{2 a n + n^{2} d - n d}{2} - [\frac{2 a n + n^{2} d - 2 n d - 2 a - n d + 2 d}{2}]$

$= \frac{2 a n + n^{2} d - n d - 2 a n - n^{2} d + 2 n d + 2 a + n d - 2 d}{2}$

$= \frac{2 n d - 2 d + 2 a}{2}$

$= \frac{2 [a + d (n - 1)]}{2}$

$= a + (n - 1) d . . . . (ii)$

From $(i)$ and $(ii)$ we can say that $S_{n - 1} - S_{n} = a_{n}$

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