Question

QUESTION.
The sum of first n terms of three arithmetic progression are S1, S2 and S3 respectively. The first term of each A.P is 1 and common difference are 1,2and 3 respectively. Prove that S1+S2=2S2​

Answer 1

$\huge{\underline{Explanation:-}}$

Sn = 2n

[2a+(n−1)d]

Using the above formula, we get:

S1 = 2n+n 2

Similarly,

S2 =n^2

and,

S3= 3n^2-n/2

S1+S3

=n+n^2/2 + 3n^2-n/2

=n+n^2+3n^2-n/2

=4n^2

=2n^2

=2S2

Therefore,

S1+S3 = S2

Hence, proved...

Answer 2

Answer:

Sn = 2n

[2a+(n−1)d]

Using the above formula, we get:

S1 = 2n+n 2

Similarly,

S2 =n^2

and,

S3= 3n^2-n/2

S1+S3

=n+n^2/2 + 3n^2-n/2

=n+n^2+3n^2-n/2

=4n^2

=2n^2

=2S2

Therefore,

S1+S3 = S2

Hence, proved...

Step-by-step explanation:

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