In an A.P., the sixth term a_6=2.

Question

In an A.P., the sixth term $a_6=2$. If $a_1a_4a_5$ is the greatest possible value, then the common difference of the A.P. is equal to:

Accepted Answer

With $a+5d=2$, i.e. $a=2-5d$: $a_1=a$, $a_4=a+3d=2-2d$, $a_5=a+4d=2-d$.

$$P(d) = (2-5d)(2-2d)(2-d) = 8-32d+34d^2-10d^3.$$

$$P'(d) = -32+68d-30d^2 = 0 \implies 15d^2-34d+16=0 \implies d = \frac{34\pm14}{30} = \frac85 \text{ or } \frac23.$$

Checking $P''(d)=68-60d$: at $d=\tfrac85$, $P''<0$ (maximum). So $d = \dfrac85$.

Solutions · 1