Question

The sum of first n terms of three APs are S1 , S2 and S3 the first term of each Ap is unity and their common difference are 1,2 and 3 , respectively.

prove that S1 + S2 = 2S2.

Answer 1

Hey mate here is your answer☺️✌️

s1=> [2a +(n-1)d]n/2

=>[2a +n-1)n/2

s2=>[2a+(n-1)2]n/2

=>[2a+2n-2]n/2
=>(2an +2n2 -2n)/2

s3=>[2a+(n-1)3]n/2
=>(2a+3n-3)n/2

s3 +s1=>[2an+3n2 -3n +2an+n2-n]/2

=>(4an+4n2-4n)/2

=>2an+2n2 -2

=>2s2

Answer 2

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HERE IS YOUR ANSWER :-

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S1 = sum of first n terms of an Ap whose first term is 1 and common difference is 1.

S1 = n/2 [2 * 1 + (n - 1) * 1 ]

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

S2 = sum of first n terms of an Ap whose first term is 1 and. common difference is 2

so,

S2 = n/2 [ 2 * 1 + ( n – 1 ) * 2 ] = n^2

and S3 = sum of first n terms of an Ap whose first term is 1 and common difference is 3

S3 = n/2 [ 2 * 1 + ( n – 1) 3]

= n/2 ( 3n– 1)

Now,

S1 + S3 = n/2 ( n + 1 ) + n/2 (3n – )

= n/2 (4n) = 2 n^2


=> S1 + S2 = 2S2. [ S2 = n^2]

PROVED


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