If in an A.P., S_{n}= n^{2}p and S_{m}=m^{2}p, then S_{p}=?

Here is the answer to your question.

In an A.P.,

We have S* _{n} *=

*n*

^{2}

*p*

⇒ 2*a* + ( *n* *– *1 )*d* = 2*np* ... (1)

Also, S* _{m} *=

*m*

^{2}

*p*

⇒ 2*a* + ( *m* *–* 1) *d* = 2*mp* ... (2)

(1) *–* (2)

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

⇒* nd – d – md + d* = 2*p* ( *n – m*)

*⇒ *(*n *–* m*) *d* = 2*p* ( *n *–* m*)

⇒ *d* = 2*p*

Put value of *d* in equation (1):

2*a* + ( *n* – 1 ) 2*p* = 2*np*

⇒ *a* + ( *n* –1 )p = 2*np*

⇒ *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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