# How do find the first four terms of an arithmetic sequence when given the first term (A1), An, and Sn?example: Find the first four terms. A1=10 An=67 SN=680 ~ I have know idea how to do this and...

How do find the first four terms of an arithmetic sequence when given the first term (A1), An, and Sn?

example: Find the first four terms.

A1=10 An=67 SN=680

~ I have know idea how to do this and step by step answers would be best.

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### 3 Answers

The two formulas you need to know:

Find any term in an arithmetic sequence: An = A1 + (n-1)d

where A1 = is the first term, n is the number in the sequence, and d is the common difference.

Find the sum of an arithmetic sequence: Sn = 0.5n(A1+An).

To solve d, we need to know n for the An=67. To do this we can use the sum equation:

Sn = 0.5n(A1+An)

680 = 0.5n(10+67)

n = (2*680)/(10+67) = 17.66

This equation is not contingent on n being a whole number, and so it is irrelevant that it is not a whole number in this case. It is customary for it to be a whole number, but the equation does not fail if is not.

It is my opinion that rounding it makes your final answer less accruate. It is possible to have a value in between two values in an arthimetic series, and as such, insisting that the 3.2nd term in a sequence cannot exist is just bad math; values exist between numbers. If they did not, interpolating, and extrapolating would be moot concepts.

You can still use the data given to find the terms in the sequence, regardless of whether the value 67 is an nth term or a between nth term.

Now, to determine d we can use the first equation:

An = A1 + (n-1)d

67 = 10 + (17.66-1)d

d = (67-10)/(16.66) = 3.42

Now:

A1 = 10

A2 = 10 + (2-1)(3.42) = 13.42

A3 = 10 + (3-1)(3.42) = 16.84

A4 = 10 + (4-1)(3.42) = 20.26

A1 = 10

An = 67

Sn = 680

WE know that in arthimaticsal progression:

Sn = (n/2) (an + a1)

We have Sn =680

==> 680 = (n/2) (10 + 67)

==> 680 = (n/2) (77)

Multiply by 2/77

==> n = 680*2/77

==> n = 17.66

But n is supposed to be a natural number . let us suppose that it n = 17

Now We know that the progression has 17 terms

a1= 10

an = a17 = 67

==> a17 = a1 + 16r where r is the constant ratio between terms.

67 = 10 + 16r

==> 16 r = 57

==> r = 57/16 = 3.56

Then terms of progression are:

a1= 10

a2= 10 + 3.56 = 13.56

a3= 10 + 2(3.56) = 17.12

a4= 10 + 3(3.56) = 20.68

The terms of the arithmetic series is A1, A2, A3......An

The realtion between the A1 and An is An =A1+(n-1)d, where d is the common ration and n is the number of terms.

Sn =[ A1+A1+(n-1))d}n/2 = (A1+An)n/2 = (10+67)n/2 = 77n/2 which should be equal to 680 by hypothesis.

So 77n/2 =680

or 77n = 680*2 = 1360

n = 1360/77 = 117.6623... is not a positive integer.

In an arithmetic progression the number of terms is always a positive integer .

So there no such arithmetic progression that could be generated from the given hypothesis: A1 = 10, An =67 and Sn = 680.

However if you editthis something like: A1 = 10 , An = 58, and Sn = 680, then

(A1+A2)(n/2) = Sn

(10+58)(n/2) 680

68*n = 680*2

n = 680*2/68 = 20

d = (58-10)/(20-1) = 48/19.

Then the first 4 after A1 are terms are:

A2 = 10+48/19 = 12 10/19

A3 = 10+48*2/19 = 15 1/19

A4 = 10+48*3/19 = 17 11/19

A5 = 10+48*4/19 = 20 2/19