Question

Find the sum of the first 40 terms of the arithmetic sequence 17, 24, 31, 38,...

Answer 1

Answer:

6120

Step-by-step explanation:

First let's calculate the 40th term.  Here a(n) = 17+7(n-1), so a(40) = 17 + 7(39).  This works out to 290.

To find the sum of the first 40 terms, add together the first and last terms and then divide the result by 2 (which gives us the average of the first and last terms), and then mult. this result by 40, the number of terms:

40(17 + 290)/2 = 40(306)/2 = 6120

Answer 2

The sum of the first 40 terms of 17, 24, 31, 38,..., with a common difference of 7 is: 6,140.

Recall:

• Formula for finding the sum of the first n terms of an arithmetic progression/sequence is given as:  $S_n = \frac{n}{2} [2a + (n - 1)d]$
• d = common difference between terms
• a = the first term of the sequence

Thus, the first term of the sequence, 17, 24, 31, 38,..., is: 17

• a = 17

Common difference (d) = 38 - 31 = 31 - 24 = 24 - 17 = 7

• d = 7

• n = 40 (sum of the first 40 terms is what is required

Plug in the values into the formula for finding the sum of the first n terms:

$S_{40} = \frac{40}{2} [2 \times 17 + (40 - 1)7] \\\\S_{40} = 20(34 + 273)\\\\S_{40} = 6,140$

Therefore, the sum of the first 40 terms of 17, 24, 31, 38,..., with a common difference of 7 is: 6,140.

Learn more here:

https://brainly.com/question/14311840