The first and second terms of an arithmetic sequence are 10 and 6 respectively.the sum of the first n terms of this sequence is -560. Calculate n
Let the common difference be d.
The nth term of an arithmetic sequence can be calculated using the formula:
a_n = a_1 + (n-1)d
Substitute the values given:
10 = 6 + (2-1)d
10 = 6 + d
d = 4
The sum of the first n terms of an arithmetic sequence can be calculated using the formula:
S_n = n/2 * (2a_1 + (n-1)d)
Substitute the values given:
-560 = n/2 * (2*10 + (n-1)*4)
-560 = n/2 * (20 + 4n - 4)
-560 = n/2 * (16 + 4n)
Solving for n:
-560 = 8n + 2n^2
2n^2 + 8n + 560 = 0
n^2 + 4n + 280 = 0
Using the quadratic formula, we get:
n = (-4 ± √(4^2 - 4*1*280)) / 2
n = (-4 ± √(16 - 1120)) / 2
n = (-4 ± √(-1104)) / 2
n = (-4 ± 33.245) / 2
Since n must be a positive integer, n = (-4 + 33.245) / 2 OR n = (-4 - 33.245) / 2
n = 29.245 / 2 OR n = -37.245 / 2
Thus, n = 14 (rounded to the nearest whole number).