Question

1. Let Sn denote the sum to n terms of an AP and let Sn = nép and
Sm = m2p. Then prove Sp = p3​

Answer 1

ANSWER
Given that

⇒S
n
​
=n
2
p

⇒S
n
​
=
2
n
​
(2a+(n−1)d)

⇒
2
n
​
(2a+(n−1)d)=n
2
p

⇒2a+(n−1)d=2np (1)


Also given that

⇒S
m
​
=m
2
p

⇒S
m
​
=
2
m
​
(2a+(m−1)d)

⇒
2
m
​
(2a+(m−1)d)=m
2
p

⇒2a+(m−1)d=2mp (2)

Now, (1)−(2)

⇒2a+(n−1)d−2a−(m−1)d=2p(n−m)

⇒nd−d−md+d=2pn−2pm

⇒d(n−m)=2p(n−m)

⇒d=2p

Substitute this in (1), we get

⇒2a+(n−1)2p=2np

⇒2a+2np−2p=2np

⇒2a=2p

⇒a=p

Now, consider S
p
​


⇒S
p
​
=
2
p
​
(2a+(p−1)d)

Substitute the values of a,d in the above equation, we get

⇒S
p
​
=
2
p
​
(2p+(p−1)2p)

⇒S
p
​
=p(p+(p−1)p)

⇒S
p
​
=p(p+p
2
−p)

⇒S
p
​
=p(p
2
)

⇒S
p
​
=p
3



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Answer 2

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