If $S_{n}$ denotes the sum of $n$ terms of an AP whose common differences is $d$ , show that $d = S_{n} - 2 S_{n - 1} + S_{n - 2}$

Open in App

Solution

Let $a$ be the first term of the given $A . P .$
Then $S_{n} = \frac{n}{2} {2 a + (n - 1) d}$ ......(1).
Now,
$S_{n} + S_{n - 2} - 2 S_{n - 1}$
$= \frac{n}{2} {2 a + (n - 1) d}$ $+ \frac{n - 2}{2} {2 a + (n - 3) d}$ $- 2 \times \frac{n - 1}{2} {2 a + (n - 2) d}$
$= \frac{1}{2} [n {2 a + (n - 1) d} + (n - 2) {2 a + (n - 3) d} - 2 (n - 1) {2 a + (n - 2) d}]$
$= \frac{1}{2} [n {2 a + (n - 1) d} + (n) {2 a + (n - 3) d} - 2 {2 a + (n - 3) d} - 2 n {2 a + (n - 2) d} + 2 {2 a + (n - 2) d}]$
$= \frac{1}{2} [2 a n + n^{2} d - n d + 2 a n + n^{2} d - 3 n d - 4 a - 2 n d + 6 d - 4 n a - 2 n^{2} d + 4 n d + 4 a + 2 n d - 4 d]$
$= \frac{1}{2} [2 d]$
$= d$ .

Suggest Corrections

Similar questions

Q. If

S_{n}

denotes the sum of

n

terms of an

A P

whose common difference is

d

and first term is

a

.
Find

S_{n} - 2 S_{n} + S_{n + 2}

Q. The common difference of an A.P., the sum of whose n terms is S_n, is

(a) S_n − 2S_n−1+ S_n−2

(b) S_n − 2S_n−1− S_n−2

(c) S_n − S_n−2

(d) S_n − S_n−1

Q. Let,

S_{n}

denote the sum of first

n

terms of an arithmetic progression whose first term is

- 4

and common difference is

1

. If

V_{n} = 2 S_{n + 2} - 2 S_{n + 1} + S_{n} (n \in N)

, then the minimum value of

V_{n}

Let $S_{n}$ denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by $d = S_{n} - k S_{n - 1} + S_{n - 2}$ , then k =

Q. If

S_{n} = n P + \frac{n (n - 1) Q}{2}

, where

S n

denotes the sum of the first

n

terms of an

A P

, then the common difference is

Join BYJU'S Learning Program

If Sn denotes the sum of n terms of an AP whose common differences is d, show that d=Sn−2Sn−1+Sn−2

If $S_{n}$ denotes the sum of $n$ terms of an AP whose common differences is $d$ , show that $d = S_{n} - 2 S_{n - 1} + S_{n - 2}$