Question

If ${\sum }_{i=1}^{10}{x}_1=60$ and ${\sum }_{i=1}^{10}{x}_i^2=360$ then ${\sum }_{i=1}^{10}{x}_i^3$ is

• A. 2160
• B. 3250
• C. 3360
• D. None of these

Answer 1

Answer: (A)

2160

Given,

${\sum }_{i=1}^{10}{x}_i=60$

$⟹x_1+x_2+.....+x_{1}0=60$

$⟹\frac{x_1+x_2+....+x_{1}0}{10}=6-\left(1\right)$

Also ${\sum }_{i=1}^{10}{x}_i^2=360$

$⟹x_1^2+x_2^2+.....+x_1{0}^2=360$

$⟹\frac{x_1^2+x_2^2+....+x_1{0}^2}{10}=36=(6{)}^2-(2)$

then,

${\sum }_{i=1}^{10}{x}_i^3$

$=x_1^3+x_2^3+.....+x_1{0}^3$

$=\frac{x_1^3+x_2^3+....+x_1{0}^3}{10}=(6{)}^3=216=216\times 10=2160$