Let a_1, a_2, a_3, a_4 be an A.P.

Question

Let $a_1, a_2, a_3, a_4$ be an A.P. of four terms such that each term and the common difference $d$ are integers. If $a_1+a_2+a_3+a_4 = 48$ and $a_1a_2a_3a_4 + d^4 = 361$, then the largest term of the A.P. is equal to:

Accepted Answer

Write the terms symmetrically as $a-\tfrac{3d}{2}, a-\tfrac{d}{2}, a+\tfrac{d}{2}, a+\tfrac{3d}{2}$ (spacing $d$). Their sum is $4a=48 \Rightarrow a=12$.

$$\left(144-\frac{d^2}{4}\right)\left(144-\frac{9d^2}{4}\right)+d^4 = 361 \implies \frac{25}{16}d^4 - 360d^2+20375=0.$$

Multiplying by $\tfrac{16}{5}$: $5d^4 - 1152d^2+65200=0$, giving (with $d^2$ real and the A.P. terms integers) $d^2 = 100$, so $d=\pm10$.

Taking $d=10$: terms are $-3, 7, 17, 27$ (or reversed for $d=-10$). The largest term is $\boxed{27}$.

Solutions · 1