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प्रश्न
Solve for x: 1 + 4 + 7 + 10 + ... + x = 287.
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उत्तर
1 + 4 + 7 + 10 + ... + x = 287
Here, a = 1, d = 4 – 1 = 3, n = x
l = x = a = (n – 1)d = 1 + (n – 1) × 3
⇒ x – 1 = (n – 1)d
Sn = `n/(2)[2a + (n - 1)d]`
287 = `n/(2)[2 xx 1 + (n - 1)3]`
574 = n(2 – 3n – 3)
⇒ 3n2 – n – 574 = 0
⇒ 3n2 – 42n + 41n – 574 = 0
⇒ 3n(n – 14) + 41(n – 14) = 0
⇒ (n – 14)(3n + 41) = 0
Either n – 14 = 0
Then n = 14
or
3n + 41 = 0,
Then 3n = –41
⇒ n = `(-41)/(3)`
Which is not possible being negative.
∴ n = 14
Now, x = a + (n – 1)d
= 1 + (14 – 1) × 3
= 1 + 13 × 3
= 1 + 39
= 40
∴ x = 40
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