Find the number of terms in the arithmetic series 49 + 42 + 35 + … = 189.Arithmetic sequences and series

2 Answers

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lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

The formula for the sum of arithmetic series is:

`S_n = (a_1+a_n)n/2`

where `S_n` - sum of n terms

           `a_1`  - first term of the arithmetic series

          `a_n` - nth term or last term  and

            `n` - number of terms

Substitute `S_n=189` and `a_1=49` to the formula.


`378=49n+a_n n`        (Let this be EQ1.)

To determine the nth term, use the formula:


where d - is the common difference between terms.

Base on the given series, the common difference is

`d= 42-49= 35-42 = -7`

Then, substitute values of a1 and d to the formula of nth term.



`a_n= 56-7n`                 (Let this be EQ2.)

To solve for n, use substitution method of system of equations. Substitute EQ2 to EQ1.

`378=49n+a_n n`




Express the equation in a quadratic form `ax^2+bx+c=0` .


To simplify, divide both sides by 7.


Factor left side.


Set each factor to zero and solve for n.

`n-6=0 `                 and            `n-9=0`

     `n=6`                                        `n=9`       

To check, substitute the values of n to EQ2 and to the formula of the sum of arithmetic series.

`n=6` ,  `a_n=56-7(6)=14`      


             `189= 189`                (True)     

`n=9` ,  `a_n=56-7(9)=-7`


             `189=189`                  (True)

--------------------------------------------------------------------------------------------      The arithmetic series 49,42,35,.... will have a sum of 189 if there are 6 or 9 terms in the series.                   


User Comments

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ayush01 | Student, Grade 10 | (Level 1) Honors

Posted on



a (first term)=49

common diff. (d)=32-49=-7

We have,


or, 189=n/2(2*49+(n-1)*(-7))

or, 378=n(98-7n+7)

or, 378=105n-7n^2

or, 7n^2-105n+378=0


or, n^2-6n-9n+54=0

or, n(n-6)-9(n-6)=0


Thus, n=6 or 9.

Therefore the required terms to be added are 9 or 6.