The sum of four consecutive numbers in an A.P is $28$. The product of the second and third numbers exceeds that of the first and last by $18$. Find the numbers.
I thought of this:
$$S_{4} = \frac{4}{2}[{2a+(4-1)d}] = 28$$
$$T_{2} * T_{3} > 18$$ where we know $$T_{n} = [a+(n-1)d]$$
And then I tried solving this simultaneously, is this right?
Sad part: My answers don't match :(
On
HINT:
As the number of terms the A.P. under consideration is even, the numbers can be chosen as $$a-3d,a-d,a+d,a+3d$$
So, we have $\displaystyle a-3d+a-d+a+d+a+3d=28\iff a=7$
and $$(a-d)(a+d)-(a-3d)(a+3d)=a^2-d^2-(a^2-9d^2)=8d^2$$
Had the number of terms been odd, we could choose them as $$a,a\pm d,a\pm 2d,\cdots$$
As you pointed out, you have $2a+3d=14$; you also have $a(a+3d)+18 = (a+d)(a+2d)$, or $a^2+3ad +18= a^2 + 3ad + 2d^2$. Thus $2d^2 = 18$ and $d=\pm 3$, so that $a = \frac{5}{2}$ and $d=3$ or $a=\frac{23}{2}$ and $d=-3$. These are the same sequences in opposite orders: $\frac{5}{2},\ \frac{11}{2},\ \frac{17}{2},\ \frac{23}{2}$.