Question

The first two terms in an arithmetic progression are 57 and 46. The last term is -207. Find the sum of all the terms in this progression. ​

Answer 1

-1875

Explanation:

57 , 46 , ......... -207

`First \ term = \ a_(1) = 57\\`

common difference = d = second term - first term = 46 - 57 = -11

`n^(th) \ term = -207\\\\a + (n-1)*d=t_(n)`

57 + (n-1)* (-11) = -207

57 - 11n + 11 = -207

-11n + 68 = -207

-11n = -207 - 68

-11n = -275

n = -275/-11

n = 25

`S_(n)=(n)/(2)(a_(1)+l)\\\\\\S_(25)=(25)/(2)*(57 + (-207) )\\\\\\ =(25)/(2)* (-150)\\\\\\= 25 *(-75)\\\\= -1875`

Answer 2

-1875

Explanation:

An arithmetic sequence has a common difference as a sequence. Here the common differnece is -11.

So our sequence so far looks like,

(57,46,35,24....). We know the last term of the sequence is -207 and we need to find the nth term of that series so we use arithmetic sequence

`a _(1) + (n - 1)d`

where a1 is the inital value,

d is the common differnece and n is the nth term.

We need to find the nth term so

`57 + (n - 1)( - 11) = - 207`

`(n - 1)( - 11) = - 264`

`n - 1 = 24`

`n = 25`

So the 25th term of a arithmetic sequence is last term, now we can use the sum of arithmetic sequence

which is

`(a _(1) + a _(n) )/(2) n`

`(57 + ( - 207))/(2) (25) =`

`( - 150)/(2) (25)`

`- 75(25) = - 1875`