# Solve the equation – 4 + (–1) + 2 + ... + x = 437 - Mathematics

Sum

Solve the equation – 4 + (–1) + 2 + ... + x = 437

#### Solution

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

an = l = x

∵ nth term of an AP, an = 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.

Concept: Sum of First n Terms of an A.P.
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.4 | Q 8 | Page 57
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