1. Arithmatic Progression Word Problem

1) Find three numbers in A.P such that their sum is 21 and sum of their product is 280.

2) Also, what the general idea for assuming "any set of numbers" required by the condition?

For eg. if the problem demands four numbers, the assumed ones are (a-3d),(a-d),(a+d),(a+3d). Where "d" is the common difference.

How do you frame them?

2. Hello saberteeth
Originally Posted by saberteeth

1) Find three numbers in A.P such that their sum is 21 and sum of their product is 280.
Suppose that the first number is $a$, and that the common difference is $d$. Then the next two numbers are $(a+d)$ and $(a+2d)$. So we have:

Sum: $a+(a+d)+(a+2d) =21$

$\Rightarrow 3a+3d=21$

$\Rightarrow a+d=7$

$\Rightarrow d = 7-a$ (1)

Product: $a(a+d)(a+2d)=280$

So, substituting for $d$:

$7a(a+14-2a)=280$

$\Rightarrow a(14-a)=40$

$\Rightarrow 14a-a^2=40$

$\Rightarrow a^2-14a+40=0$

$\Rightarrow (a-4)(a-10)=0$

$\Rightarrow a=4,10$

Plugging these into (1) gives: $d =3,-3$

Each of these results gives the same set of numbers: $4, 7, 10$.

2) Also, what the general idea for assuming "any set of numbers" required by the condition?

For eg. if the problem demands four numbers, the assumed ones are (a-3d),(a-d),(a+d),(a+3d). Where "d" is the common difference.

How do you frame them?
You could use the ones you suggest here, since you have written 4 numbers with a common difference (in this case $2d$). Or, more staightforwardly, you could do as I have in question (1), and assume that the first is $a$, common difference $d$, and write them as $a, (a+d), (a+2d), (a+3d)$.

Either way, you'll have two unknowns, and you'll therefore need to be able set up two equations in order to solve a particular problem.