Question

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

Answer 1

Answer:

425.25

Step-by-step explanation:

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

$14-35(\frac{1}{8})=\frac{77}{8}=9.625$

The sum of arithmetic sequence formula is:

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

Where,

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

Substituting all the values, we get:

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

First answer choice is right.

Answer 2

ANSWER

$S_ {36} = 425.25$


EXPLANATION

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

$u_n= a+(n-1)d$



This implies that,


$14= a+(36-1) \times \frac{1}{8}$




$14 = a + \frac{35}{8}$

$a = 14 - \frac{35}{8}$



$a =9.625$


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

$S_n= \frac{n}{2} (2a + (n - 1)d)$

$S_ {36} = \frac{36}{2} (2 \times 9.625+ (36 - 1) \times \frac{1}{8} )$


$S_ {36}= 18(2 \times 9.625+ (35) \times \frac{1}{8} )$



$S_ {36}= 18(23.625 ) = 425.25$