What is the sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms?

2 answers:

7 0

9 - 3 = 6, 15-9 = 6 the difference is 6, So d = 6

First term: a1 = 3

Sn = n*(a1 + an)/2

Sn = n*(a1 + a1 + (n-1)*d)/2

Sn = n*(2*a1 + (n-1)*d)/2

substitute 26 for n
 S26 = 26*(2*a1 + (26-1)*d)/2

substitute 3 for a1

S26 = 26*(2*3 + (26-1)*d)/2

substitute 6 for d

S26 = 26*(2*3 + (26-1)*6)/2 

S26 = 2,028

marissa [1.9K]4 years ago

3 0

Sum of an arithmetic series is:

S = n*(a_1+a_n)/2, that easy.

n is the number of terms, 26 in your case. a_n is a_26. We need to find the series!

a_n = a_1 + (n-1)*d, the difference, d, is 6.

So a_n = 3 + (n-1)*6 = 6*n-3 (check it works: 3, 9, 15, ...)

Now a_26 = 6*26-3 = 153, so finally:

S = 26*(3+153)/2 = 26*78 = 2028

2028 is the answer

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baherus [9]

Answer: Option B.

Step-by-step explanation:

For this exercise it is important to remember that, by definition, a relation is a function if and only if each input value has an unique output value.

The input values are the values of the variable "x" and the output values are the values of the variable "y".

You can observe in the table shown in the picture, that the input value -2 has two different output values:

$x=-2,y=3\\\\x=-2,y=-13$