Calculate the sum of the first 36 terms of the arithmetic sequence defined in which a36=14 and the common difference is d=1/8

Since we are given 36th term as 14 and we know common difference is [tex]\frac{1}{8}[/tex], it means that from the first term, we add [tex]\frac{1}{8}[/tex] to each and get 14 on the 36th term. To figure out the first term, thus, we have to subtract [tex]\frac{1}{8}[/tex] 35 times from 14. Let's do it to get first term:

[tex]14-35(\frac{1}{8})=\frac{77}{8}=9.625[/tex]

The sum of arithmetic sequence formula is:

[tex]S_{n}=\frac{n}{2}[2a+(n-1)d][/tex]

Where,

[tex]S_{n}[/tex] is the sum of nth term (we want to figure this out for first 36 terms)
[tex]a[/tex] is the first term (we figured this out to be 9.625)
[tex]n[/tex] is the term number (36 for our case)
[tex]d[/tex] is the common difference (given as [tex]\frac{1}{8}[/tex])

Substituting all the values, we get:

[tex]S_{36}=\frac{36}{2}[2(9.625)+(36-1)(\frac{1}{8})]\\S_{36}=18[19.25+4.375]\\S_{36}=18[23.625]\\S_{36}=425.25[/tex]

First answer choice is right.

kudzordzifrancis kudzordzifrancis · Answer 2 · 2018-12-08T23:45:50+01:00

ANSWER

[tex]S_ {36} = 425.25[/tex]

EXPLANATION

Since we know the 36th term to be 14, we can find the first term using the formula,

[tex]u_n= a+(n-1)d[/tex]

This implies that,

[tex]14= a+(36-1) \times \frac{1}{8} [/tex]

[tex]14 = a + \frac{35}{8} [/tex]

[tex]a = 14 - \frac{35}{8} [/tex]

[tex] a =9.625[/tex]

We can now find the sum of the first 36 terms using the formula,

[tex]S_n= \frac{n}{2} (2a + (n - 1)d)[/tex]

[tex]S_ {36} = \frac{36}{2} (2 \times 9.625+ (36 - 1) \times \frac{1}{8} )[/tex]

[tex]S_ {36}= 18(2 \times 9.625+ (35) \times \frac{1}{8} )[/tex]

[tex]S_ {36}= 18(23.625 ) = 425.25[/tex]