Question

What is the sum of the first ten terms of a_1 = -43, d=12 ?

Answer 1

`S_10 = 110`

Explanation:

`a_1 = -43`
`d = 12`
`n = 10`

The formula for first 10 terms is:
`S_n = 1/2n{2a +(n-1)d}`

`S_10 = 1/2(10){2(-43) +(10-1)12}`

`S_10 = (5){-86 +(9)12}`

`S_10 = (5){-86 +108}`

`S_10 = (5){22}`

`S_10 = 110`

Answer 2

110
(Assuming the question refers to an Arithmetic Progression)

Explanation:

If I'm understanding this right(the lack of math notation makes it ambiguous!), this is an Arithmetic Progression with its first term `a = -43` and common difference `d = 12`.

The formula for the sum of the first `n` terms of an A.P is `S = [n(2a + (n-1)d)]/2` .
Let's substitute `a = -43`, `d = 12` and `n = 10`
`S = [10(2(-43) + (10-1)12)]/2`
`S = 5(-86+ 9(12))`
`S = 5(108 - 86) = 5(22)`

Thus the answer is 110.

Answer 3

Sum of first `10` terms is `110`

Explanation:

Given first term of an arithmetic progression `a_1` and common difference `d`,

sum of first `n`terms is given by

`S_n=n/2(2a_1+(n-1)d)`

Here `a_1=-43` and `d=12`, hence

`S_10=10/2(2xx(-43)+(10-1)*12)`

= `5xx(-86+9xx12)`

= `5xx(-86+108)`

= `5xx22`

= `110`