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

(a) 40

(b) 39

(c) 41

(d) 42

 

Solution:

Given 1 + 4 +7 +10….. + x = 287

The series is an AP with first term, a = 1

Common difference, d = 4-1 = 3

Sum of n terms, Sn = 287

We know Sn = (n/2)(2a + (n-1)d)

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

= (n/2)(2 + 3n-3)

= (n/2)(3n -1)

287 ×2 = 3n2 – n

574 = 3n2 – n

3n2 – n – 574 = 0

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

=> n = 14 ( -41/3 cannot be used)

nth term of AP = a + (n-1)d 

14th term = 1 + (13 × 3)

= 1+39

= 40

Hence option a is the answer.

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