Question

If the sum of first n positive integers is $\frac{1}{5}$ times the sum of their squares, then n equals

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

7

Sum of first $n$ positive integers $=\frac{\left(n\right)(n+1)}{2}.$

Sum of the squares of first $n$ positive integers $=\frac{\left(n\right)(n+1)(2n+1)}{2}.$

Now,

$\frac{\left(n\right)(n+1)}{2}=\frac{1}{5}\frac{n(n+1)(2n+1)}{6}.$

$2n+1=15$

$n=7$