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if the sum of n terms of an A.P. is (pn+qn2), where p&q are constants, find the common difference.

It is known that,

According to the given condition,

Comparing the coefficients of n2 on both sides, we obtain

=> d = 2 q

Thus, the common difference of the A.P. is 2q.

  • 28
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I can show the above in a more solid way...

we know,

an=Sn-Sn-1

an-1=Sn-1-S(n-1)-1

=Sn-1-Sn-2

We also can get,

d=an-an-1

=(Sn-Sn-1)-(Sn-1-Sn-2)

=[(pn + qn2) - {p(n-1) + q(n-1)2}] - [{p(n-1) + q(n-1)2} - {p(n-2) + q(n-2)2}]

=p{n - (n-1) - (n-1) + (n-2)} + q{n2 - (n-1)2 - (n-1)2 + (n-2)2}

=p(0) + q(2)

=2q


SEE... ITIS MORE SOLID THAN COMPARING...

and guess what...

in general, you can also say that,

d= Sn+Sn-2-2Sn-1

This is called the "PSK theorem", well... it is newly made... used a lot...

and... I made it... easier to prove now ha !



















  • 14

wow... I didn't know it was already made...)

  • 8
Why do we compare coefficient of n^2 on both sides
  • 2
Thanks for answer
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