Solving Arithmetic Progression: Sum and Product of Four Integers

[REVIANNA]

Homework Statement

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

Homework Equations

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

$(a+d)(a-d)(a)(a+2d)=945$
$(a^2-d^2)(a^2+2*a*d)=945$

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!

 

[Mark44]

Homework Statement

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

Homework Equations

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

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

$(a+d)(a-d)(a)(a+2d)=945$
$(a^2-d^2)(a^2+2*a*d)=945$

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!

 

[REVIANNA]

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

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?

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

 

[REVIANNA]

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

for 5 terms

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

 

[Mark44]

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

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.

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

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

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

Same here.