Question

if in an arithmetic progression , Sn = n.n.p and Sm = m.m.p , where Sr denotes the sum of r terms of the A.P , then Sp = ?

Answer 1

The answer is given below :

Let us consider that the nth term of Sn is tn and mth term of Sm is tm.

Let, a be the first term of the AP and d be the common difference.

Given,

Sn = n²p

=> (n/2)(a + tn) = n²p

=> a + tn = 2np .....(i)

and

Sm = m²p

=> (m/2)(a + tm) = m²p

=> a + tm = 2mp .....(ii)

Subtracting (ii) from (i), we get

tn - tm = 2p(n - m) .....(iii)

From (i), we get

a + a + (n - 1)d = 2np

=> 2a + (n - 1)d = 2np .....(iv)

From (iii), we get

a + (n - 1)d - a - (m - 1)d = 2p(n - m)

=> (n - m)d = 2p(n - m)

=> d = 2p

From (iv), we get

2a + (n - 1)(2p) = 2np

=> 2a - 2p = 0

=> a = p

Therefore,

Sp = (p/2)[ a + (p - 1)d ]

= (p/2)[ 2p + (p - 1)(2p) ]

= (p/2)[2p + 2p² - 2p ]

= (p/2)(2p²)

= p³ [Answer]

Rule :

n-th term of the AP

= a + (n - 1)d,

where a is the first term of the AP and d is the common difference.

Thank you for your question.