Question

Sum of first nterms of three Ap are S1 ,S2 ,S3.the frist term is 1and the common differences are 1,2,3 .prove that S1+S2=2S2

Answer 1

Sum of three A.P's = S1, S2, S3

given a for all the sum is 1

And d ie. common  difference of S1= 1

                                                       S2= 2

                                                       S3= 3

T.P : S1 + S3= 2S2

IE.

n/2 {2a+(n-1)d} + n/2 {2a+(n-1)d} = 2*n/2 {2a+(n-1)d}

=> n/2 {2*1 +(n-1) * 1} + n/2 {2 * 1 +(n-1) * 3] = 2* n/2{2 * 1 +(n-1) * 2}

LHS

n/2 { 2 + n - 1} + n/2 { 2+ 3n -3}

n/2 { 1+n}+ n/2 { 3n - 1}

n/2 {1+n+3n-1}                  [taking n/2 common]

n/2{4n}

n * 2n

=>2n^2

RHS

2 * n/2 (2 + 2n - 2)

n (2n)

=> 2n^2

Therefore LHS = RHS

ie. S1+S3 = 2S2

hope you understand

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Answer 2

Substituting in the formula,

S1=n/2(2a+(n-1)d)...........(i)

S2=2n/2(2a+(2n-1)d)................ (ii)

S3=3n/2(2a+(3n-1)d),................ (iii)

So,

S2-S1

=2n/2(2a+(2n-1)d)-[n/2(2a+(n-1)d]

_by proper subtraction we get

=n/2[2a + (3n-1)d]

So, 3(S1-S2)

=3n/2(2a + (3n-1)d)

=S3