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The sums of n terms of three arithmetic...

The sums of `n` terms of three arithmetical progressions are `S_1, S_2`and `S_3dot` The first term of each unity and the common differences are `1,2` and `3` respectively. Prove that `S_1+S_3=2S_2dot`

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To prove that \( S_1 + S_3 = 2S_2 \) for the given arithmetic progressions, let's start by calculating the sums of the first \( n \) terms of each progression. ### Step 1: Define the first terms and common differences We have three arithmetic progressions (APs): - For \( S_1 \): First term \( A_1 = 1 \), common difference \( D_1 = 1 \) - For \( S_2 \): First term \( A_2 = 1 \), common difference \( D_2 = 2 \) - For \( S_3 \): First term \( A_3 = 1 \), common difference \( D_3 = 3 \) ...
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The sums of n terms of three arithmetical progressions are S_(1), S_(2) and S_(3) . The first term of each is unity and the common differences are 1, 2 and 3 respectively. Then S_(1) + S_(3) = 3S_(2)

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  • Let S_(n) denotes the sum of n terms of S30 an arithmetic progression, then (S _(30))/( S _(20) - S _(10)) is equal to

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