If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

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In the given problem, we are given an A.P whose S_n = n^2 p and S_m = m^2 p We need to find SP Now, as we know, `S_n = n /2 [ 2a + ( n - 1 ) d]` Where, first term = a Common difference = d Number of terms = n So, S_n = n/2 [ 2a + ( n-1) d ] n^2 p = n/2 [ 2a + (n-1)d] p = 1/(2n) [2a + nd - d] .............(1) Similarly, S_n = m/2 [2a + (m-1)d] m^2 p = m/2 [2a + (m + 1)d] p = 1/(2m)[2a + md -d] ...............(2) Equating (1) and (2), we get, $\frac{1}{2}$n\left( 2a + nd - d \right) = $\frac{1}{2}$m\left( 2a + md - d \right) \Rightarrow m\left( 2a + nd - d \right) = n\left( 2a + md - d \right) \Rightarrow 2am + mnd - md = 2an + mnd - nd Solving further, we get, 2am - 2an = - nd + md 2a ( m - n) = d (m - n) 2a = d ..............(3) Further, substituting (3) in (1), we get, S_n = n/2 [d + ( n-1) d] n^2 p = n/2 [d + nd - d ] p = 1/(2n)[nd] p = d/2 ..............(4) Now, S_p = p/2 [2a + ( p - 1) d ] S_p = p/2 [ d +pd - d] ( Using 3) S_p = p/2 [ p(2 p)] ( Using 4 ) S_p = p^3 Thus, S_p = p^3

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