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.
I can't work out n. Please help.
Hello, greghunter!
You're expected to know this formula:
. . $\displaystyle S_n \;=\;n\left(\frac{a_1 + a_n}{2}\right)$
$\displaystyle \text{Sum} \;=\;n \times \text{(average of first and last terms)}$
We have: .$\displaystyle a_1 = 16,\;a_n = 60,\;\text{Sum} = 342$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.
Then: .$\displaystyle 342 \:=\:n\left(\frac{16+60}{2}\right) \quad\Rightarrow\quad 342 \:=\:38n \quad\Rightarrow\quad\boxed{ n \:=\:9}$
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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?