If in an A P , Sn= n2p and Sm=m2p, then Sp= - Maths - Sequences and Series

Question

If in an A P , Sn= n2p and
Sm=m2p, then Sp=?

Ishan Goyal · Accepted Answer

Here is the answer to your question In an A P , We have S n = n 2 p $rn rn \Rightarrow rn rn \frac{rn}{rn n rn} rn rn [rn 2 rn a rn â¢ rn + rn rn (rn n rn â¢ rn - rn 1 rn) rn rn d rn] rn = rn rn {rn}_{rn {rn}^{rn n rn rn rn 2 rn rn} rn p rn rn rn rn rn} rn rn rn$ ⇒ 2a + ( n – 1 )d = 2np (1) Also, S m = m 2 p $rn rn \Rightarrow rn rn \frac{rn}{rn m rn} rn rn [rn 2 rn a rn + rn â¢ rn (rn m rn â¢ rn - rn 1 rn â¢ rn) rn rn d rn] rn rn = rn rn {rn}^{rn m rn rn rn 2 rn rn} rn p rn rn rn$ ⇒ 2a + ( m – 1) d = 2mp (2) (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 $rn rn = rn rn \frac{rn}{rn p rn} rn rn [rn 2 rn a rn â¢ rn + rn rn (rn rn p rn â¢ rn rn - rn 1 rn) rn rn d rn] rn rn rn = rn rn \frac{rn}{rn p rn} rn rn [rn 2 rn p rn â¢ rn + rn rn (rn p rn â¢ rn - rn 1 rn) rn rn 2 rn p rn] rn rn â¢ rn ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½(Putting values of rn rn rn rn rn rn rn rn a rn rn rn and ï¿½ rn rn rn d rn rn rn) rn rn = rn rn \frac{rn}{rn p rn} rn â¢ rn (rn 2 rn p rn) rn rn [rn 1 rn â¢ rn + rn rn (rn rn p rn â¢ rn - rn 1 rn) rn] rn rn$ = p 2( p) = p 3 Thus , S p = p 3

If in an A.P., Sn= n2p and Sm=m2p, then Sp=?

If in an A.P., S_n= n²p and S_m=m²p, then S_p=?