Question

Solve the equation: 1+3+5+7+.....X = 400

Answer 1

$\text{Clearly 1,3,5.......are in A.P }$

$\text{Here, a=1, d=2 and l=x}$

$\text{Then, the number of terms}$

$\bf\,n=\frac{l-a}{d}+1$

$n=\frac{x-1}{2}+1$

$\implies\bf\,n=\frac{x+1}{2}$

$\text{Now,}$

$\text{1+3+5......+x=400}$

$\implies\frac{n}{2}[a+l]=400$

$\implies\,n[a+l]=800$

$\implies\,\frac{x+1}{2}[1+x]=800$

$\implies\,(x+1)^2=1600$

$\implies\,x+1=40$

$\implies\,x=40-1$

$\implies\bf\,x=39$

$\therefore\textbf{The value of x is 39 }$

Find more:

Find the value of x if 2 + 7 + 12 +17+...... + x = 156.

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