Question

What is the sum of the first 53 terms of the sequence 140, 137, 134, 131, ...?

Answer 1

S =3,286

Explanation:

We are given the following sequence and we are to find the sum of the first 53 terms of this sequence:

`140, 137, 134, 131, ...`

Finding the common difference
`d` =
`137-140` =
`-3`

`a_1=140`

`a_n=?`

`a_n=a_1+(n-1)d`

`a_n = 140 + (53 - 1 ) -3`

`a _ n = -16`

Finding the sum using the formula
`S_n = (n)/(2)(a_1+a_n)`.

`S_n = (53)/(2)(140+(-16))`

S = 3,286

Answer 2

The sum of the first 53 terms of the sequence is 3286

Explanation:

* Lets talk about the arithmetic sequence

- There is a constant difference between each two consecutive numbers

Ex:

# 2 , 5 , 8 , 11 , ……………………….

# 5 , 10 , 15 , 20 , …………………………

# 12 , 10 , 8 , 6 , ……………………………

* General term (nth term) of an Arithmetic Progression:

- U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

between each two consecutive terms and n is the position of

the term in the sequence

* Sum of an Arithmetic Progression:

is calculate from Sn = n/2[2a + (n - 1)d]

* Lets solve the problem

- The sequence is 140 , 137 , 134 , 131 , .........

∵ 137 - 140 = -3 and 134 - 137 = -3

∴ The sequence is arithmetic

- The first term a = 140

- The common difference d = -3

- n = 53

∵ Sn = n/2[2a + (n - 1)d]

∴ S53 = 53/2[2 × 140 + (53 - 1)(-3)]

∴ S53 = 53/2[280 + 52(-3) = 53/2[280 + -156] = 53/2[124]

∴ S53 = 3286

* The sum of the first 53 terms of the sequence is 3286