# Thread: Arithmetic progression

1. ## Arithmetic progression

Given an arithmetic progression -10,-5,0,........State three consecutive terms in this progression whose sum is 90.

2. Originally Posted by mastermin346
Given an arithmetic progression -10,-5,0,........State three consecutive terms in this progression whose sum is 90.
You know that the difference $d = 5$.

Your three terms are therefore $x, x + 5, x + 10$.

You know they add up to $90$.

So $x + x + 5 + x + 10 = 90$

$3x + 15 = 90$

$3x = 75$

$x = 25$.

So your three numbers are $25, 30, 35$.

3. Originally Posted by Prove It
You know that the difference $d = 5$.

Your three terms are therefore $x, x + 5, x + 10$.

You know they add up to $90$.

So $x + x + 5 + x + 10 = 90$

$3x + 15 = 90$

$3x = 75$

$x = 25$.

So your three numbers are $25, 30, 35$.
the three numbers are 25,30,35 or plus 25?=25,50,75?

4. ## aliter

sum of n terms of A.P is given by
(n/2)(2A+(n-1)d)
here n= number of terms
A= first term
d= common difference
in given problem
n=3,d=5 so putting in equation we get
(3/2)(2A+10)=90
2A+10=60
A=25
since first term is A=25 other terms will be 30 and 35
so three terms are 25,30,35