Solve the equation

$ -4+(-1)+2+\ldots+x=437 $

AcademicMathematicsNCERTClass 10

Given:

$(-4) + (-1) + 2 + 5 + … + x = 437$.

To do:

We have to find the value of $x$.

Solution:

Given equation is \( (-4)+(-1)+2+5+\ldots+x=437 \quad \ldots(i) \)

LHS: \( (-4), (-1), 2, 5, \ldots x \)

This forms an A.P., where

First term \( a=-4, \) common difference \( =(-1)-(-4)=-1+4=3 \),

\( a_{n}=l=x \)

We know that,

\( n \) th term of an A.P., \( a_{n}=l=a+(n-1) d \)

\( \Rightarrow x=-4+(n-1) 3 \)

\( \Rightarrow \frac{x+4}{3}=n-1 \)

\( \Rightarrow n=\frac{x+7}{3} \)

We know that,

Sum of an A.P., $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{n}=\frac{x+7}{2 \times 3}[2(-4)+(\frac{x+4}{3})3]$

\( =\frac{x+7}{2 \times 3} (-8+x+4) \)

\( =\frac{(x+7)(x-4)}{2 \times 3} \)

\( \Rightarrow \frac{(x+7)(x-4)}{2 \times 3}=437 \)                (From (i))

\( \Rightarrow x^{2}+7 x-4 x-28=874 \times 3 \)

\( \Rightarrow x^{2}+3 x-2650=0 \)

\( x=\frac{-3 \pm \sqrt{(3)^{2}-4(-2650)}}{2} \)

\( x=\frac{-3 \pm \sqrt{9+10600}}{2} \)

\( x=\frac{-3 \pm \sqrt{10609}}{2} \)

\( x=\frac{-3 \pm 103}{2} \)

\( x=\frac{100}{2} \) or \( \frac{-106}{2} \)

\( x=50 \) or \( x=-53 \) which is not possible, for \( x=-53, n \) will be negative which is not possible.

Hence, the required value of \( x \) is 50.

raja
Updated on 10-Oct-2022 13:27:44

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