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zhenek [66]
3 years ago
13

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

Mathematics
2 answers:
user100 [1]3 years ago
7 0

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.


spin [16.1K]3 years ago
7 0
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



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