Let the sum of the first n terms of an A.P.

Question

Let the sum of the first $n$ terms of an A.P. be $3 n^2+5 n$. Then the sum of squares of the first 10 terms of the A.P. is:

Accepted Answer

Given,

${S}_{n}=3{n}^{2}+5n$

where ${S}_{n}$ is the sum of the first $n$ terms of the A.P.

To get the $n$th term, we use

${a}_{n}={S}_{n}-{S}_{n-1}$

So,

${S}_{n-1}=3(n-1{)}^{2}+5(n-1)$

$=3({n}^{2}-2n+1)+5n-5$

$=3{n}^{2}-6n+3+5n-5$

$=3{n}^{2}-n-2$

Now,

${a}_{n}=(3{n}^{2}+5n)-(3{n}^{2}-n-2)$

${a}_{n}=6n+2$

So the terms of the A.P. are:

$8,14,20,\ldots$

Now we need the sum of squares of the first $10$ terms:

$\sum _{10}^{n=1}(6n+2{)}^{2}$

Expand:

$(6n+2{)}^{2}=36{n}^{2}+24n+4$

So,

$\sum _{10}^{n=1}(6n+2{)}^{2}=36\sum _{10}^{n=1}{n}^{2}+24\sum _{10}^{n=1}n+4\sum _{10}^{n=1}1$

Now use the standard sums:

$\sum _{10}^{n=1}{n}^{2}=\frac{10\cdot 11\cdot 21}{6}=385$

$\sum _{10}^{n=1}n=\frac{10\cdot 11}{2}=55$

$\sum _{10}^{n=1}1=10$

Substitute:

$36(385)+24(55)+4(10)$

$=13860+1320+40$

$=15220$

So the correct answer is

$15220$

Option C

Solutions · 1