Let a_1,a_2,,a_n be in A.P.

Question

Let $a_1,a_2,\ldots,a_n$ be in A.P. If $a_5 = 2a_7$ and $a_{11}=18$, then
$$12\left[\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\cdots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right]$$
is equal to \_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

$a_5=2a_7 \Rightarrow a_1+4d = 2(a_1+6d) \Rightarrow a_1+8d=0$. Also $a_{11}=a_1+10d=18$. Solving: $d=9$, $a_1=-72$.

$a_{10}=a_1+9d=9$, $a_{18}=a_1+17d=81$.

Each term rationalizes: $\dfrac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \dfrac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, so the bracketed sum telescopes to $\dfrac{\sqrt{a_{18}}-\sqrt{a_{10}}}{d} = \dfrac{9-3}{9}=\dfrac23$.

$$12\times\frac23 = 8.$$

Solutions · 1