Question

If $x_1,x_2,......,x_n$ are n variables such that ${\sum }_{i=1}^n(2x_i+5)=1600$ and ${\sum }_{i=1}^n(x_i^2-4)=900.$ If ${\sum }_{i=1}^n(x_i+1{)}^2=250$ then the value of n is

• A. 15
• B. 20
• C. 10
• D. 5

Answer 1

The correct option is C 10

Given: ${\sum }_{i=1}^n(2x_i+5)=1600$, ${\sum }_{i=1}^n(x_i^2-4)=900$ and ${\sum }_{i=1}^n(x_i+1{)}^2=250$

Consider ${\sum }_{i=1}^n(2x_i+5)=1600$
$\Rightarrow 2{\sum }_{i=1}^n{x}_i+5n=1600$
( ${\sum }_{i=1}^{n}a=a_n$, where a is any constant)
$\Rightarrow 2{\sum }_{i=1}^n{x}_i=1600-5n$..........(i)

Also ${\sum }_{i=1}^n(x_i^2-4)=900$
$\Rightarrow {\sum }_{i=1}^n{x}_i^2-4n=900$
${\sum }_{i=1}^{n}a=a_n$, where a is any constant)
$\Rightarrow {\sum }_{i=1}^n{x}_i^2=900+4n$........(ii)

And, ${\sum }_{i=1}^n(x_i+1{)}^2=250$
$\Rightarrow {\sum }_{i=1}^n(x_i^2+2x_1+1)=250n$
$\Rightarrow {\sum }_{i=1}^n{x}_i^2+2{\sum }_{i=1}^n{x}_1+n=250n$
${\sum }_{i=1}^{n}a=a_n$, where a is any constant)
$\Rightarrow (900+4n)+(1600-5n)+n=250n$ [from (i) and (ii)]
$\Rightarrow 2500=250n$
$\Rightarrow n=10$

Hence, the correct answer is option c.