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प्रश्न
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
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उत्तर
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
`\implies` x = – 4 + (n – 1)3 ...(ii)
`\implies` `(x + 4)/3` = n – 1
`\implies` n = `(x + 7)/3`
∴ Sum of an AP,
Sn = `n/2[2a + (n - 1)d]`
Sn = `(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),
Sn = 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.
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