Question

The sum of n terms of an A.P is \[an\left( n-1 \right)\]. The sum of squares of these terms is
A. \[{{a}^{2}}{{n}^{2}}\left( n-{{1}^{2}} \right)\]
B. \[\dfrac{{{a}^{2}}}{6}n\left( n-1 \right)\left( 2n-1 \right)\]
C. \[\dfrac{2{{a}^{2}}}{3}n\left( n-1 \right)\left( 2n-1 \right)\]
D. \[\dfrac{2{{a}^{2}}}{3}n\left( n+1 \right)\left( 2n+1 \right)\]

Answer 1

Hint: To solve this type of problem we have to know the formulas like sum of n terms in A.P. We should also know the sum of n natural numbers. By solving using these formulas we will arrive at the final solution. Sum of n natural numbers is \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].

Complete step-by-step answer:
Given the sum of n terms of an A.P is \[{{S}_{1}}\]=\[an\left( n-1 \right)\].
The sum of n-1 terms of A.P is \[{{S}_{2}}=a\left( n-1 \right)\left( n-1-1 \right)\]
\[{{S}_{2}}=a\left( n-1 \right)\left( n-2 \right)\]
We know that the nth term of an A.P is \[{{S}_{1}}-{{S}_{2}}\].
\[{{S}_{1}}-{{S}_{2}}=an\left( n-1 \right)-a\left( n-1 \right)\left( n-2 \right)\]
\[=a\left( n-1 \right)\left( n-n+2 \right)\]
\[=2a\left( n-1 \right)\].
Putting n=1,2,3,4,5,6, . . . . . . . . . . . . . . .
We get 0, 2a, 4a, 6a, 8a, . . . . . . . . .
The above sequence is in A.P,
Squaring the A.P terms we get,
\[0,4{{a}^{2}},16{{a}^{2}},36{{a}^{2}},64{{a}^{2}}\]. . . . . . . . . .
Now writing the sum for the above sequence we get,
\[0+4{{a}^{2}}+16{{a}^{2}}+36{{a}^{2}}+64{{a}^{2}}................\]
\[{{S}_{n}}=4{{a}^{2}}\left[ 0+1+4+9+16+25+........ \right]\]
Now writing in the form of squares we get,
\[{{S}_{n}}=4{{a}^{2}}\left[ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+........ \right]\]
\[{{S}_{n}}=4{{a}^{2}}\left[ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+........{{\left( n-1 \right)}^{2}} \right]\] . . . . . . . . . . . . . . . . (a)
We know that Sum of n natural numbers is \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].
Then sum of (n-1) natural numbers is \[\dfrac{n-1\left( n+1-1 \right)\left( 2n-2+1 \right)}{6}\] . . . . . . . . . (b)
= \[\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}\]
Now the sum of the sequence (a) is
\[{{S}_{n}}=\dfrac{4{{a}^{2}}n\left( n-1 \right)\left( 2n-1 \right)}{6}\]
\[{{S}_{n}}=\dfrac{2{{a}^{2}}n\left( n-1 \right)\left( 2n-1 \right)}{3}\]
Therefore the answer is option C.

Note: (b) is the value which we got by substituting the value (n-1) in the sum of n natural numbers to get the sum of (n-1) natural numbers. After finding the sum of (n-1) natural numbers we just wrote the expression for (3).