Q 1= let SN DENOTE THE SUM OF N TERMS OF AN A P WHOSE FIRST TERM - Maths - Arithmetic Progressions

Question

Q 1= let SN DENOTE THE SUM OF N TERMS OF
AN A P WHOSE FIRST TERM IS A IF THE COMMON DIFFERENCE IS D THEN
PROVE THAT
D=SN -2SN-1
+SN-2

Aakash Sharma · Accepted Answer

Dear Student!

Here is the answer to your query.

Given : The first term and the common difference of an A.P. is A and D respectively.

Then Sum of N term (S_N) = $\frac{N}{2} [2 A + (N - 1) D]$

$= \frac{2 AN}{2} + \frac{(N - 1) ND}{2} = AN+ \frac{N^{2} D - ND}{2} \Rightarrow S_{N - 1} = \frac{N - 1}{2} [2 A + ((N - 1) - 1) D] = \frac{2 A (N - 1)}{2} + \frac{(N - 2) (N - 1) D}{2} = AN - A + \frac{N^{2} D - 3 ND + 2 D}{2}$

$and S_{N - 2} = \frac{N - 2}{2} [2 A + ((N - 2) - 1) D] = \frac{2 A (N - 2)}{2} + \frac{(N - 3) (N - 2) D}{2} = AN - 2 A + \frac{N^{2} D - 5 ND + 6 D}{2} ∴ S_{N} - 2 S_{N - 1} + S_{N - 2} = AN + \frac{N^{2} D - ND}{2} - 2 [AN - A + \frac{N^{2} D - 3 ND}{2}] + AN - 2 A + \frac{N^{2} D - SND + 6 D}{2} = AN - 2 AN + AN + \frac{N^{2} D}{2} - N^{2} D + \frac{N^{2} D}{2} - \frac{ND}{2} + 3 ND - \frac{5}{2} ND - 2 D + 3 D = 2 AN - 2 AN + N^{2} D - N^{2} D + \frac{5}{2} ND - \frac{5}{2} ND + D = D ∴ D = S_{N} - 2 S_{N - 1} + S_{N - 2}$

Cheers!

Q.1= let SN DENOTE THE SUM OF N TERMS OF AN A.P .WHOSE FIRST TERM IS A .IF THE COMMON DIFFERENCE IS D .THEN PROVE THAT D=SN -2SN-1 +SN-2

Q.1= let S_N DENOTE THE SUM OF N TERMS OF AN A.P .WHOSE FIRST TERM IS A .IF THE COMMON DIFFERENCE IS D .THEN PROVE THAT

D=S_N -2S_N-1 +S_N-2