If in an A.P., Sn= n 2 p and Sm=m 2 p, then Sp=?
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In an A.P.,
We have S n = n 2 p
⇒ 2a + ( n – 1 )d = 2np ... (1)
Also, S m = m 2 p
⇒ 2a + ( m – 1) d = 2mp ... (2)
Subtracting, (1) – (2)
2a + ( n – 1 )d – (2a + ( m – 1) d) = 2np – 2mp
⇒ nd – d – md + d = 2p ( n – m)
⇒ (n – m) d = 2p ( n – m)
⇒ d = 2p
Put value of d in equation (1):
2a + ( n – 1 ) 2p = 2np
⇒ a + ( n –1 )p = 2np
⇒ a = np – ( n – 1 ) p
= np – np + p
⇒ a = p
Now, s p = p 2( p) = p 3
Thus , S p = p 3.
We have S n = n 2 p
⇒ 2a + ( n – 1 )d = 2np ... (1)
Also, S m = m 2 p
⇒ 2a + ( m – 1) d = 2mp ... (2)
Subtracting, (1) – (2)
2a + ( n – 1 )d – (2a + ( m – 1) d) = 2np – 2mp
⇒ nd – d – md + d = 2p ( n – m)
⇒ (n – m) d = 2p ( n – m)
⇒ d = 2p
Put value of d in equation (1):
2a + ( n – 1 ) 2p = 2np
⇒ a + ( n –1 )p = 2np
⇒ a = np – ( n – 1 ) p
= np – np + p
⇒ a = p
Now, s p = p 2( p) = p 3
Thus , S p = p 3.
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