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

Answer:

-1875

Step-by-step 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

$\frac{a _{1} + a _{n} }{2} n$

$\frac{57 + ( - 207)}{2} (25) =$

$\frac{ - 150}{2} (25)$

$- 75(25) = - 1875$

Answer 2

Answer:

-1875

Step-by-step 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

$S_{n}=\dfrac{n}{2}(a_{1}+l)\\\\\\S_{25}=\dfrac{25}{2}*(57 + (-207) )\\\\\\ =\dfrac{25}{2}* (-150)\\\\\\= 25 *(-75)\\\\= -1875$