Question

If the sum of the first n, 2n and 3n terms of an AP be S1, S2and S3 respectively then prove that S3=3(S2-S1).​

Answer 1

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Answer:

Let a be the first term and d be the common difference of the given A.P. Then,

S

1

= Sum of n terms ⟹S

1

=

2

n

{2a+(n−1)d} ...(i)

S

2

= Sum of 2n terms ⟹S

2

=

2

2n

{2a+(2n−1)d} ...(ii)

and, S

3

= Sum of 3n terms ⟹S

3

=

2

3n

{2a+(3n−1)d} ...(iii)

Now,

S

2

−S

1

=

2

2n

{2a+(2n−1)d}−

2

n

{2a+(n−1)d}

⟹S

2

−S

1

=

2

n

[2{2a+(2n−1)d}−{2a+(n−1)d}]=

2

n

{2a+(3n−1)d}

∴3(S

2

−S

1

)=

2

3n

{2a+(3n−1)d}=S

3

[Using (iii)]

Hence, S

3

=3(S

2

−S

1

)