Let 1x_1,1x_2,1x_3, (x_i0) be in A.P.

Question

Let $\dfrac{1}{x_1},\dfrac{1}{x_2},\dfrac{1}{x_3},\ldots$ ($x_i
e0$) be in A.P. such that $x_1=4$ and $x_{21}=20$. If $n$ is the least positive integer for which $x_n>50$, then $\displaystyle\sum_{i=1}^{n}\frac{1}{x_i}$ is equal to:

Accepted Answer

Common difference of $\{1/x_i\}$: $\dfrac{1}{x_{21}}-\dfrac{1}{x_1} = 20d \Rightarrow d = -\dfrac{1}{100}$. $x_n>50 \Rightarrow \dfrac{1}{x_n} < \dfrac{1}{50}$; solving $\dfrac14+(n-1)\left(-\dfrac{1}{100}\right)<\dfrac{1}{50}$ gives $n>24$, so $n=25$. $$\sum_{i=1}^{25}\frac{1}{x_i} = \frac{25}{2}\left[\frac12+24\left(-\frac{1}{100}\right)\right] = \frac{13}{4}.$$

Solutions · 1