If 1 + 4 + 7 + 10 + … + x = 287, find the value of x.

1.	If 1 + 4 + 7 + 10 + … + x = 287, find the value of x.
Answer» Given: 1 + 4 + 7 + 10 + … + x = 287 Here a = 1, d = 3 and S_n = 287 Sum = S_n = n/2 [2a + (n-1)d] 287 = n/2 [2 + (n-1)3] 574 = 3n² – n Which is a quadratic equation. Solve 3n² – n – 574 = 0 3n² – 42n + 41n – 574 = 0 3n(n – 14) + 41(n-14) = 0 (3n + 41)(n-14) = 0 Either (3n + 41) = 0 or (n-14) = 0 n = -41/3 or n = 14 Since number of terms cannot be negative, so result is n = 14. => Total number of terms in AP are 14. Which shows, x = a₁₄ or x = a + 13d or x = 1 + 39 or x = 40 The value of x is 40.

Answer»

Given: 1 + 4 + 7 + 10 + … + x = 287

Here a = 1, d = 3 and S_n = 287

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

287 = n/2 [2 + (n-1)3]

574 = 3n² – n

Which is a quadratic equation.

Solve 3n² – n – 574 = 0

3n² – 42n + 41n – 574 = 0

3n(n – 14) + 41(n-14) = 0

(3n + 41)(n-14) = 0

Either (3n + 41) = 0 or (n-14) = 0

n = -41/3 or n = 14

Since number of terms cannot be negative, so result is n = 14.

=> Total number of terms in AP are 14.

Which shows, x = a₁₄

or x = a + 13d

or x = 1 + 39

or x = 40

The value of x is 40.

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