Let a_1,a_2,,a_21 be an A.P.

Question

Let $a_1,a_2,\ldots,a_{21}$ be an A.P. such that $\displaystyle\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}=\frac49$. If the sum of this A.P. is $189$, then $a_6a_{16}$ is equal to:

Accepted Answer

$\displaystyle\sum\frac{1}{a_na_{n+1}} = \frac1d\left(\frac{1}{a_1}-\frac{1}{a_{21}}\right) = \frac49$; and $S_{21}=\dfrac{21}{2}(a_1+a_{21})=189 \Rightarrow a_1+a_{21}=18.$

Let $a_1=a$; then $a_{21}=18-a$, $d=\dfrac{18-2a}{20}$. Substituting into the sum condition:

$$\frac{20}{18-2a}\cdot\frac{18-2a}{a(18-a)} = \frac49 \implies a(18-a)=45 \implies a^2-18a+45=0 \implies a=3 \text{ or } 15.$$

Taking $a=3$, $d=0.6$: $a_6=3+3=6$? --- recomputing carefully: $a_6=a+5d=3+3=6$, $a_{16}=a+15d=3+9=12$, so $a_6a_{16}=72$ (the same product results by symmetry for $a=15$).

Solutions · 1