Question

how many terms of the AP is -15,-13,-11......are needed to make the sum -55? Explain the reason for double answer​

Answer 1

Answer:

At 5 and 11 term it becomes -55

Answer 2

Answer:

5 terms or 11 terms

Step-by-step explanation:

Sum of the first "n" terms of an AP is

$S_{n} = \frac{n}{2}(2a_{1} + (n-1)d)$

~~~~~~~~

- 15, - 13, - 11, ...

$a_{1}$ = - 15

d = - 13 + 15 = 2

$S_{n} = -55$

- 55 = (n/2)[- 15×2 + 2(n - 1)]

n(2n - 2 - 30) = - 110

2n² - 32n + 110 = 0

n² - 16n + 55 = 0

$n_{1} = 5$

$n_{2} = 11$

- 15 + (- 13) + (- 11) + (- 9) + (- 7) = - 55

- 15 + (- 13) + (- 11) + (- 9) + (- 7) + [(- 5) + (- 3) + (- 1) + 1 + 3 + 5] = - 55 + 0 = -55