Question

Find the 35th term of the sequence given: (3, 14, 25,...)

Find the 35th term of the sequence given: (3, 14, 25, ...)

Answer 1

377 = a(35)

Explanation:

This is an arithmetic sequence with first term 3 and common differernce 11. Notice how each new term is equal to the previous term, plus 11.

So we have a(n) = 3 + 11(n - 1), and here that becomes

a(35) = 3 + 11(35 - 1) = 3 + 11(34) = 3 + 374 = 377 = a(35)