If in an AP, Sn(subscript)=(n^2)p and Sm(subscript)=(m^2)p, where Sr(subscript) denotes the sum of r terms of the AP, then Sp(subscript) is equal to what?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Since the sum of n terms of an arithmetic progression is given by the formula `S_n = ((a_1 + a_n)*n)/2`  yields:

`S_n = ((2a_1 + (n-1)*d)*n)/2 = n^2*p`

`((a_1 + a_n)*n) = 2n^2*p =>2a_1 + (n-1)*d = 2n*p `

`S_m =((2a_1 + (m-1)*d)*m)/2 = m^2*p =>2a_1 + (m-1)*d = 2m*p`

You need to solve the system of equations such that:

`{(2a_1 + (n-1)*d = 2n*p),(2a_1 + (m-1)*d = 2m*p):}``=> (n-1)*d - (m-1)*d = 2p(n - m)`

`d(n - 1 - m + 1) = 2p(n - m) => d*(n - m) = 2p(n - m)`

Reducing duplicate factors yields:

`d = 2p`

`2a_1 = 2np - 2p(n-1) => a_1 = np - np + p => a_1 = p`

You may evaluate the sum `S_p`  such that:

`S_p = (2a_1 + (p-1)*2p)/2 = p^2*p`

Substituting `p`  for `a_1`  and `2p`  for `d`  yields:

`S_p = (2p + (p-1)*2p)/2 `

Reducing by 2 yields:

`S_p = p +p(p - 1) => S_p = p(1 + p - 1) = p^2`

Hence, evaluating the sum `S_p` , under the given conditions, yields `S_p = p^2` .

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