Sequences and Series 5 Question 10
1. If $a _1, a _2, a _3, \ldots$ are in a harmonic progression with $a _1=5$ and $a _{20}=25$. Then, the least positive integer $n$ for which $a _n<0$, is
(2012)
(a) 22
(b) 23
(c) 24
(d) 25
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Answer:
Correct Answer: 1. (d)
Solution:
- PLAN $n$th term of HP, $t _n=\frac{1}{a+(n-1) n}$
Here, $\quad a _1=5, a _{20}=25$ for HP
$$ \begin{aligned} & \therefore & \frac{1}{a} & =5 \text { and } \frac{1}{a+19 d}=25 \\ & \Rightarrow & \frac{1}{5}+19 d & =\frac{1}{25} \Rightarrow 19 d=\frac{1}{25}-\frac{1}{5}=-\frac{4}{25} \\ & \therefore & d & =\frac{-4}{19 \times 25} \end{aligned} $$
Since, $\quad a _n<0$
$\Rightarrow \quad \frac{1}{5}+(n-1) d<0$
$\Rightarrow \quad \frac{1}{5}-\frac{4}{19 \times 25}(n-1)<0 \Rightarrow(n-1)>\frac{95}{4}$
$\Rightarrow \quad n>1+\frac{95}{4}$ or $n>24.75$
$\therefore \quad$ Least positive value of $n=25$
