Question

How many terms of the series 2 + 5 + 8 + … must be taken if their sum is 155

Answer 1

Answer: 10 terms

Explanation:

`\displaystyle \ \Large \boldsymbol{} S_n=(2a_1+d(n-1))/(2) \cdot n =155 \\\\ (4+3(n-1))/(2) \cdot n =155 \\\\\\ 4n+3n^2-3n=310 \\\\ 3n^2+n-310=0 \\\\D=1+3720=3721=61^2\\\\n_1=(61-1)/(6) =\boxed{10} \\\\\\n_2=(-61-1)/(3) \ \ \o`

Answer 2

9514 1404 393

Answer:

10

Explanation:

The sum of terms of an arithmetic series is ...

Sn = (2a +d(n -1))·n/2 = (2an +dn^2 -dn)/2

For the series with first term 2 and common difference 3, the sum is 155 for n terms, where ...

155 = (3n^2 +n(2·2 -3))/2

Multiplying by 2, we have ...

3n^2 +n -310 = 0 . . . . . arranged in standard form

Using the quadratic formula, the positive solution is ...

n = (-1 +√(1 -4(3)(-310)))/(2(3)) = (-1 +√3721)/6 = (61 -1)/6 = 10

10 terms of the series will have a sum of 155.