+0

# Arithmetic Progression

+1
60
1

The nth term of an arithmetic progression is 1/2(3n-15).
Find the value of n for which the sum of the first n terms is 84.

ANYONE?

Apr 3, 2021

#1
+485
+1

Writing out your arithmetic progression in summation notation, we have $$\sum_{n=1}^{c}\frac{3n-15}{2}$$ where c is the number of terms, what we are trying to solve for.

Pulling the 1/2 out from the summation and breaking it up gives $1/2\sum_{n=1}^{c}3n+1/2\sum_{n=1}^{c}-15$.

Now, we first solve the summation on the left. Pulling out the three, and using the appropriate technique for a sum of $n$, we get $\frac{3}{2}\frac{c(c+1)}{2}$, and combining gives $\frac{3c^2+3c}{4}$.

Now, we solve the summation on the right. This is easy, for any constant c summed from 1 to n, the result is nc. So we have $1/2\sum_{n=1}^{c}-15=\frac{-15c}{2}$

Now, we have an equation for c!

$\frac{3c^2+3c}{4}-\frac{15c}{2}=84$

You should be able to find c(the number of terms) from here.

HINT: Expand all the terms and bring everything to the left side. Look familiar?

Apr 3, 2021