The sum of n terms of an AP is 4pn(n+3).The sum of squares of these terms is: Options: A . &nbsp643p2(n3+6n2+12) B . &nbsp323p2(2n3+9n2+13n) C . &nbsp643p2(4n3+12n2+2n) D . &nbsp1283p2(n3+6n2+12.)

The sum of n terms of an AP is 4pn(n+3).The sum of squares of&

Question

The sum of n terms of an AP is 4pn(n+3).The sum of squares of these terms is:

Options:

A . 643p2(n3+6n2+12)

B . 323p2(2n3+9n2+13n)

C . 643p2(4n3+12n2+2n)

D . 1283p2(n3+6n2+12.)

Answer: Option B
:
B
Suppose the AP is 1,2,3
At n=1, sum =1
4(p)(1)(4)=1

\Rightarrow

p =

\frac{1}{16}

Sum of squares =

1^{2}

= 1 itself
Look for 1 in the answer options
Only at option b

\frac{32}{3} \times \frac{1}{256} \times (24) = 1

Answer is option (b)

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