Answer:
The sum of the first 53 terms of the sequence is 3286
Step-by-step 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
Answer:
S =3,286
Step-by-step explanation:
We are given the following sequence and we are to find the sum of the first 53 terms of this sequence:
[tex]140, 137, 134, 131, ...[/tex]
Finding the common difference [tex]d[/tex] = [tex]137-140[/tex] = [tex]-3[/tex]
[tex]a_1=140[/tex]
[tex]a_n=?[/tex]
[tex]a_n=a_1+(n-1)d[/tex]
[tex] a_n = 140 + (53 - 1 ) -3 [/tex]
[tex] a _ n = -16 [/tex]
Finding the sum using the formula [tex]S_n = \frac{n}{2}(a_1+a_n)[/tex].
[tex]S_n = \frac{53}{2}(140+(-16))[/tex]
S = 3,286
