Question

Solve the equation

– 4 + (–1) + 2 + ... + x = 437

Answer 1

Given equation is,

– 4 + (–1) + 2 + ... + x = 437   ...(i)

Here, – 4 – 1 + 2 + ... + x forms an AP with first term = – 4,

Common difference = – 1 – (– 4) = 3,

a_n = l = x

∵ n^th term of an AP,

a_n = l = a + (n – 1)d

`\implies` x = – 4 + (n – 1)3   ...(ii)

`\implies` `(x + 4)/3` = n – 1

`\implies` n = `(x + 7)/3`

∴ Sum of an AP,

S_n = `n/2[2a + (n - 1)d]`

S_n = `(x + 7)/(2 xx 3)[2(-4) + ((x + 4)/3) * 3]`

= `(x + 7)/(2 xx 3)(-8 + x + 4)`

= `((x + 7)(x - 4))/(2 xx 3)`

From equation (i),

S_n = 437

`\implies ((x + 7)(x - 4))/(2 xx 3)` = 437

`\implies` x² + 7x – 4x – 28 = 874 × 3

`\implies` x² + 3x – 2650 = 0

x = `(-3 +- sqrt((3)^2 - 4(-2650)))/2`   ...[By quadratic formula]

= `(-3 +- sqrt(9 + 10600))/2`

= `(-3 +- sqrt(10609))/2`

= `(-3 +- 103)/2`

= `100/2, (-106)/2`

= 50, – 53

Here, x cannot be negative i.e., x ≠ – 53

Also for x = – 53, n will be negative which is not possible

Hence, the required value of x is 50.