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

⇒ x = – 4 + (n – 1)3   .......(ii)

⇒ `(x + 4)/3 = n - 1`

⇒ `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

⇒ `((x + 7)(x - 4))/(2 xx 3)` = 437

⇒ `x^2 + 7x - 4x - 28 = 874 xx 3`

⇒ `x^2 + 3x - 2650` = 0

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

= `(-3 +- sqrt(9 + 106000))/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.