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

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

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:

Finding the common difference = =

Finding the sum using the formula .

S = 3,286

answer: c

step-by-step explanation: because you don't have to draw a point of intersection. couldyou give me brainliest?

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