Question

How many term of the A.P 16,14,12,.....are needed to the sum 60?explain why do we get two answers.

Answer 1

Step-by-step explanation:

a = 16 d = -2 Sn = 60

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

$60 = \frac{n}{2} (2(16) + (n - 1)( - 2)) \\ 60 = \frac{n}{2} (32 - 2n + 2) \\ 60 = \frac{n}{2} (34 - 2n) \\ 60 = \frac{n}{2} \times 2(17 - n) \\ 60 = 17n - {n}^{2} \\ {n}^{2} - 17n + 60 = 0 \\ {n}^{2} - 12n - 5n + 60 = 0 \\ n(n - 12) - 5(n - 12) = 0 \\ (n - 12)(n - 5) = 0 \\ n - 12 = 0 \: \: \: \: n - 5 = 0 \\ n = 12 \: \: \: \: \: n = 5$

12 terms or 5 terms should be added to get 60

we get two answers as the series has negative numbers in it.

S5

$= \frac{5}{2} (2(16) + (5 - 1)( - 2)) \\ = \frac{5}{2} (32 + 4( - 2)) \\ = \frac{5}{2} (32 - 8) \\ = \frac{5}{2} \times 24 \\ = 5 \times 12 \\ = 60$

S12

$= \frac{12}{2} (2(16) + (12 - 1)( - 2)) \\ = 6(32 + (11 \times - 2)) \\ = 6(32 - 22) \\ = 6 \times 10 \\ = 60$

hope you get your answer