If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r te

1.	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 toA. \(\frac{1}{2} p^3\)B. mn p C. p 3 D. (m + n) p2
Answer» Correct answer is C. P³ Let first term = a and Common difference = d ∴ According to the question, S_n = n² p S_n = n/2 (2a + (n–1) d) = n²p 2a + (n–1) d = 2np……………….(1) And S_m = m²p S_m= m/2 (2a + (m–1) d) = m²p 2mp = (2a + (m–1) d)……………….(2) Subtracting 2 from 1 2a + (n–1) d – 2a – (m–1) d = 2np – 2mp d (n–1 –m + 1) = 2p (n– m) d = 2p putting value of d in (1) 2a + (n–1) 2p = 2np a + (n–1)p = np a = p now S_p = p/2 (2a + (p–1) d) putting value of a = p and d = 2p S_p = p/2 (2p + (p–1) 2p) S_p = p³

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 toA. \(\frac{1}{2} p^3\)B. mn p C. p 3 D. (m + n) p2

Answer»

Correct answer is C. P³

Let first term = a and Common difference = d

∴ According to the question, S_n = n² p

S_n = n/2 (2a + (n–1) d) = n²p

2a + (n–1) d = 2np……………….(1)

And S_m = m²p

S_m= m/2 (2a + (m–1) d) = m²p

2mp = (2a + (m–1) d)……………….(2)

Subtracting 2 from 1

2a + (n–1) d – 2a – (m–1) d = 2np – 2mp

d (n–1 –m + 1) = 2p (n– m) d = 2p

putting value of d in (1)

2a + (n–1) 2p = 2np

a + (n–1)p = np

a = p

now S_p = p/2 (2a + (p–1) d)

putting value of a = p and d = 2p

S_p = p/2 (2p + (p–1) 2p)

S_p = p³