Question

How many terms of the arithmetic sequence 5,7,9,.... must be added to get 140?​

Answer 1

First term a_1 = 5

Common difference, d = 2

S_n = 140

Using formula

$S_n = \frac{n}{2} (2a_1 + (n - 1)d) \\ 140 = \frac{n}{2} (2 \times 5+ (n - 1) \times 2) \\ 280 = 10n + 2 {n}^{2} - 2n \\ 140 = 5n + {n}^{2} - n \\ {n}^{2} + 4n - 140 = 0 \\ {n}^{2} + 14n - 10n - 140 = 0 \\ n(n + 14) - 10(n + 14) = 0 \\ (n + 14)(n - 10) = 0 \\ \\ n = - 14 \: and \: 10$

n cannot be negative Hence,

n = 10

10 terms of the arithmetic sequence 5,7,9,.... must be added to get 140

Answer 2

answer is10. 10 is term must be added