The sums of first n terms of three A.P. are S₁,S₂and S₃. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S₁+ S₃= 2S₂

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] F o r S_{1}, w h e n a = 1 a n d d = 1 . S_{1} = \frac{n}{2} [2 (1) + (n - 1)] = \frac{n}{2} [2 + n - 1] = \frac{n}{2} [n + 1] F o r S_{2}, w h e n a = 1 a n d d = 2 . S_{2} = \frac{n}{2} [2 (1) + (n - 1) (2)] = \frac{n}{2} [2 + 2 n - 2] = \frac{n}{2} [2 n] = n^{2} F o r S_{3}, w h e n a = 1 a n d d = 3 . S_{3} = \frac{n}{2} [2 (1) + (n - 1) (3)] = \frac{n}{2} [2 + 3 n - 3] = \frac{n}{2} [3 n - 1] T h e r e f o r e, S_{1} + S_{3} = \frac{n}{2} [n + 1] + \frac{n}{2} [3 n - 1] = \frac{n}{2} [4 n] = 2 n^{2} S_{1} + S_{3} = 2 S_{2}$
Hence proved.

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The sums of first n terms of three A.P.s are $S_{1}, S_{2}$ and $S_{3}$ . The first term of each). their common differences are 2, 4 and 6 respectively. Prove that $S_{1} + S_{3} = 2 S_{2}$ .

Q. The sums of first n terms of three arithmetic progressions are

S_{1}, S_{2}

and

S_{3}

respectively. The first term of each A.P. is 1 and their common difference are 1, 2 and 3 respectively. Prove that

S_{1} + S_{3} = 2 S_{2}

Q. Sum of

n

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respectively. If first term of each progression is

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and common differences are

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respectively, then prove that:

S_{1} + S_{3} = 2 S_{2}

Q. The sums of n terms of three arithmetical progressions are

S_{1}, S_{2}

and

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. The first term of each is unity and the common differences are

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respectively. Prove that

S_{1} + S_{3} = 2 S_{2}

The sums of n terms of three A.P.'s whose first term is 1 and common differences are 1, 2, 3 are $s_{1}, s_{2}, s_{3}$ respectively. The true relation is

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The sums of first n terms of three A.P. are S1,S2 and S3. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S1 + S3 = 2S2

Sn=n22a+n-1dFor S1, when a=1 and d=1.S1=n221+n-1=n22+n-1=n2n+1For S2, when a=1 and d=2.S2=n221+n-12=n22+2n-2=n22n=n2For S3, when a=1 and d=3.S3=n221+n-13=n22+3n-3=n23n-1Therefore, S1+S3=n2n+1+n23n-1=n24n=2n2S1+S3=2S2 Hence proved.

The sums of first n terms of three A.P. are S₁,S₂and S₃. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S₁+ S₃= 2S₂