The first term of an arithmetic series is 16 and the last is 60. The sum of an arithmetic series is 342. find the common difference.

2. Hello, greghunter!

You're expected to know this formula:

. . $\displaystyle S_n \;=\;n\left(\frac{a_1 + a_n}{2}\right)$

$\displaystyle \text{Sum} \;=\;n \times \text{(average of first and last terms)}$

The first term of an arithmetic series is 16 and the last is 60.
The sum of the arithmetic series is 342. .Find the common difference.
We have: .$\displaystyle a_1 = 16,\;a_n = 60,\;\text{Sum} = 342$

Then: .$\displaystyle 342 \:=\:n\left(\frac{16+60}{2}\right) \quad\Rightarrow\quad 342 \:=\:38n \quad\Rightarrow\quad\boxed{ n \:=\:9}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you didn't know that formula, you can still work it out.

First term: .$\displaystyle a_1 \:=\: 16$ .[1]

Last term: .$\displaystyle a_n \:=\:a_1 + (n-1)d \:=\:60 \quad\Rightarrow\quad 16 + (n-1)d \:=\:60 \quad\Rightarrow\quad (n-1)d \:=\:44$ .[2]

Sum of the first n terms: .$\displaystyle S_n \;=\;\frac{n}{2}\bigg[2a_1 + (n-1)d\bigg] \;=\;342$ .[3]

Substitute [1] and [2] into [3]: .$\displaystyle \frac{n}{2}\bigg[2(16) + 44\bigg] \;=\;342 \quad\Rightarrow\quad n \:=\:9$

. . . . . See?

3. ## Thanks!

Thanks Soroban.
I'm new to this, as you can see, and would just like to say its fantastic!

Your working out is very clear and easy to read thankyou very much.

Greghunter

4. Originally Posted by greghunter
The first term of an arithmetic series is 16 and the last is 60. The sum of an arithmetic series is 342. find the common difference.

Let first term=a & last term=l
sum=n(a+l)/2
342=n(16+60)/2
342=n(38)
342/38=n
n=9