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Question: Let \({a_1},{a_2},{a_3},..........\) be in harmonic progression with \({a_1} = 5\) and, \({a_{20}} =...

Let a1,a2,a3,..........{a_1},{a_2},{a_3},.......... be in harmonic progression with a1=5{a_1} = 5 and, a20=25{a_{20}} = 25. The least positive integer n for which an<0{a_n} < 0 is:
A) 22
B) 23
C) 24
D) 25

Explanation

Solution

According to given in the question we have to determine the least positive integer n for which an<0{a_n} < 0 and we have to let a1,a2,a3,..........{a_1},{a_2},{a_3},.......... be in harmonic progression with a1=5{a_1} = 5 and, a20=25{a_{20}} = 25. So, first of all we have to convert the given harmonic progression a1,a2,a3,..........{a_1},{a_2},{a_3},.......... in to arithmetic progression but before that we have to understand about the harmonic progression which is explained below:
Harmonic progression: Harmonic progression is defined as a sequence of real numbers which are formed by reciprocal of the given arithmetic progression.
Now, we have to find the common difference (d) of the obtained arithmetic progression which can be obtained by finding the term a20{a_{20}} with the help of the formula as mentioned below:

Formula used: 1an=1a1+(n1)d...............(A) \Rightarrow \dfrac{1}{{{a_n}}} = \dfrac{1}{{{a_1}}} + (n - 1)d...............(A)
Now, with the help of the formula can determine the value of common difference by substituting all the given values.
Now, with the help of the condition in the question which is an<0{a_n} < 0 we can determine the value of n.

Complete step-by-step answer:
Given,
a1=5\Rightarrow {a_1} = 5, and
a20=25\Rightarrow {a_{20}} = 25
Step 1: First of all we have to convert the given harmonic progression a1,a2,a3,..........{a_1},{a_2},{a_3},.......... into arithmetic progression as mentioned in the solution hint. Hence,
1a1,1a2,1a3.............\Rightarrow \dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_3}}}............. are in arithmetic progression.
Step 2: Now, we have to determine the common difference (d) with the help of the formula (A) as mentioned in the solution hint. Hence, on substituting all the values in the formula (A),
125=15+(201)d 125=15+19d \Rightarrow \dfrac{1}{{25}} = \dfrac{1}{5} + (20 - 1)d \\\ \Rightarrow \dfrac{1}{{25}} = \dfrac{1}{5} + 19d
On solving the expression as obtained just above,
125=1+19×5d5 125=1+95d5 \Rightarrow \dfrac{1}{{25}} = \dfrac{{1 + 19 \times 5d}}{5} \\\ \Rightarrow \dfrac{1}{{25}} = \dfrac{{1 + 95d}}{5}
Now, on applying cross-multiplication in the expression as obtained just above,

525=1+95d 15=1+95d 5+475d=1 d=4775 \Rightarrow \dfrac{5}{{25}} = 1 + 95d \\\ \Rightarrow \dfrac{1}{5} = 1 + 95d \\\ \Rightarrow 5 + 475d = 1 \\\ \Rightarrow d = \dfrac{{ - 4}}{{775}}

Step 3: Now, we have to put all the values and with the help of the condition an<0{a_n} < 0 we can determine the value of n. Hence,
=1a1+(n1)(4475)<0= \dfrac{1}{{{a_1}}} + (n - 1)\left( {\dfrac{{ - 4}}{{475}}} \right) < 0
As we know that a1=5{a_1} = 5 and substituting the value in the expression obtained just above,
=2(n1)19×5>1 =(n1)>19×54 =n>19×54+1 =n24.7525 =n25 = \dfrac{{2(n - 1)}}{{19 \times 5}} > 1 \\\ = (n - 1) > \dfrac{{19 \times 5}}{4} \\\ = n > \dfrac{{19 \times 5}}{4} + 1 \\\ = n \geqslant 24.75 \approx 25 \\\ = n \geqslant 25
Final solution: Hence, with the help of the formula (A) we have obtained n25n \geqslant 25.

Therefore our correct option is (D).

Note: In arithmetic progression there is a fixed difference between every pair of two terms and the common difference can be calculated by subtracting the second term by first term and the same as we can obtain in by subtracting third term by second term.
If the difference between second term and first term and third and second term are not equal then the given sequence is not an arithmetic sequence.