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Question
If the sumk of n terms of an A.P. is (pn+qn2), Where pa and q are constants, find the common difference.
If the sumk of n terms of an A.P. is (pn+qn2), Where pa and q are constants, find the common difference.
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Solution
Here Sn=pn+qn2
Repalcing n by n-1 in Sn
∴Sn−1=p(n−1)+q(n−1)2
= pn−p+qn2−2pn+q
We know that an=Sn−Sn−1
= pn+qn2−(pn−p+qn2−2qn+q)
= pn+qn2−pn−p+qn2−2qn+q)
= p + 2qn -q
Replacing n by (n-1) in an
∴an−1=p+2q(n−1)−q
= p+2qn−2q−q=p+2qn−3q
We know that d = a_n - a_{n-1}
= p+2qn−q−(p+2qn−3q)
=p+2qn−q−p−2qn+3q=2q
Here Sn=pn+qn2
Repalcing n by n-1 in Sn
∴Sn−1=p(n−1)+q(n−1)2
= pn−p+qn2−2pn+q
We know that an=Sn−Sn−1
= pn+qn2−(pn−p+qn2−2qn+q)
= pn+qn2−pn−p+qn2−2qn+q)
= p + 2qn -q
Replacing n by (n-1) in an
∴an−1=p+2q(n−1)−q
= p+2qn−2q−q=p+2qn−3q
We know that d = a_n - a_{n-1}
= p+2qn−q−(p+2qn−3q)
=p+2qn−q−p−2qn+3q=2q
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