If Sn = n2p and Sm = m2p,m≠n, in an A.P., prove that Sp = p3

1.	If Sn = n2p and Sm = m2p,m≠n, in an A.P., prove that Sp = p3
Answer» Given an AP whose sum of n terms is n²p and same AP with m terms whose sum is m²p To prove : S_p = p³ We know that the sum of AP is given by the formula : s = \(\frac{n}{2}\)(2a + (n-1)d) Substituting the values in the above equation, we get \(\frac{n}{2}\)(2a + (n-1)d) = n²p …. (i) Similarly, For series with m terms \(\frac{m}{2}\)(2a + (m-1)d) = m²p ...(ii) Subtracting (ii) from (i) we get, d = 2p Substituting d in (i) we get, a = p Now, Using the sum formula for AP consisting of p terms we get S_p = \(\frac{p}{2}\)(2a + (n-1)d) Substituting the values in the above equation, S_p = \(\frac{p}{2}\)(2p + (p-1)2p) S_P = p³

Answer»

Given an AP whose sum of n terms is n²p and same AP with m terms whose sum is m²p

To prove : S_p = p³

We know that the sum of AP is given by the formula :

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above equation, we get

\(\frac{n}{2}\)(2a + (n-1)d) = n²p …. (i)

Similarly,

For series with m terms \(\frac{m}{2}\)(2a + (m-1)d) = m²p ...(ii)

Subtracting (ii) from (i) we get,

d = 2p

Substituting d in (i) we get,

a = p

Now,

Using the sum formula for AP consisting of p terms we get

S_p = \(\frac{p}{2}\)(2a + (n-1)d)

Substituting the values in the above equation,

S_p = \(\frac{p}{2}\)(2p + (p-1)2p)

S_P = p³

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