# Arithmetic progression

1. Dec 11, 2015

### REVIANNA

1. The problem statement, all variables and given/known data

the sum of four integers in A.P is 24 and their product is 945.find them

2. Relevant equations
$(a-d)+a+(a+d)+(a+2*d)=24$
$2a+d=12$

$(a+d)(a-d)(a)(a+2d)=945$
$(a^2-d^2)(a^2+2*a*d)=945$
3. The attempt at a solution

there are two equations and two unknowns a(one of the integers) and d(common diff of the A.P)
but I have trouble manipulating the second(product) equation so that the result of the 1st eq can be used.
help!!

2. Dec 11, 2015

### Staff: Mentor

Solve for one of the variables above in terms of the other, and then substitute for that variable in the equation below.

3. Dec 11, 2015

### REVIANNA

I could have done that but I found a more elegant solution(given at the back with hints which I previously ignored)

assume the integers
$(a-3*d)+(a-d)+(a+d)+(a+3*d)=4*a=24$

$(6-3*d)(6-d)(6+d)(6+3*d)=945$

this is much easier to solve.
and I think we can come up with such terms when 6 terms are asked for?

$(a-5*d)+(a-3*d)+(a-d)+(a+d)+(a+3*d)+(a+5*d)$

4. Dec 11, 2015

### REVIANNA

and for even number of terms we can have the common difference d instead of 2d

for 5 terms

$(a-2d)+(a-d)+a+(a+d)+(a+2d)$

5. Dec 11, 2015

### Staff: Mentor

In the above, you're assuming that the common difference between the integers in the AP is 2d, so you'll need to account for that in your answer.
Same here.