IIT JEE , AIEEE Forum - Mathematics , Algebra

prashant_prakhar

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

deedee

is d answer 2q.......

loyalgooner

Yup.....i am also getting common difference as 2q.

prashant_prakhar

how sdid u solve it............
can u plz tell me step wise..........

deedee

i didn't solved it dear ..

there 's a trick ..

if sn of an ap.............is =An^2 + Bn

then .....d common difference is.........

2 * coefficient of n^2

cheers ...........do rate ma reply.......

loyalgooner

see, Sum of n terms = pn + qn^2

Thus sum of n-1 terms will be p(n-1) + q(n-1)^2.

now nth term = sum of n terms - sum of n-1 terms

so u get tn = p - q^2 + 2nq

Now put n=1 and you get 1st term as p - q^2 + 2q and you put n=2 and you get 2nd term as p - q^2 + 4q

common difference = 2nd term - 1st term = 2q

Plz rate me for my efforts! Cheers!

spideyunlimited

pn + qn^2 = n/2 ( 2a + (n-1) d )
p + qn = ( 2a + (n-1) d ) / 2
p + qn = a + (n-1)d/2
p + qn = a - d/2 + nd/2

Comparing like terms:
p = a - (d/2)
and
qn = nd/2
q = d/2
d = 2q

Thus common difference d = 2q

and also p = a - (d/2) = a - q
so a = p + q

aditi_g

take n=1
sum of 1 term ie. the first term will be a1 = p + q
take n=2
a1+a2 = 2p + 4q
==> a2= p+ 3q
common diff = a2 - a1
=p + 3q - p - q=2q
= rate me if it was useful......

krishna.gopal

Prashant it is 2q only and all the explanations of loyal,, spiedy and aditi are all right.

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