Question

please answer this question ​

Answer 1

Answer:

,The sum of the first n terms of an AP is

Sn = (n/2)[2a + (n-1)·d]

n = 10, S10 = -150

-150 = (10/2)[2a + (10 - 1)d]

-150 = 5[2a + 9d]

Dividing through by 5,

-30 = 2a + 9d ----------(1)

Given sum ofits next 10 terms is - 550.

n = 10, Sn = - 550

Here a = a + 10d

- 550 = (10/2)[2(a + 10d) + (10-1)d]

- 550 = 5[2a + 29d]

Dividing through by 10,

-110 = 2a + 29d ----------(2)

Solving (1) and (2) we get

d = -4 and a = 3.

Step-by-step explanation:

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