How many terms of the given sequence: 1, 4, 7, … are needed to give the sum 715?

Approach: recognize this is an arithmetic sequence with first term 1 and common difference 3.

Use the sum formula for an arithmetic series:

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

Substitute a = 1, d = 3 and set S_n = 715:

715 = n/2 [2*1 + (n-1)*3] = n/2 (3n - 1).

Multiply both sides by 2:

1430 = n(3n - 1) → 3n^2 - n - 1430 = 0.

Solve the quadratic using the discriminant:

D = (-1)^2 - 4*3*(-1430) = 1 + 17160 = 17161 = 131^2.

n = (1 ± 131) / (2*3). The positive root is n = (1 + 131)/6 = 132/6 = 22.

Answer: 22 terms are needed to give the sum 715.