Question

If \[{a_1},{a_2},{a_3}, \cdots \] are in a harmonic progression with \[{a_1} = 5\] and \[{a_{20}} = 25\]. Find the least positive integer \[n\] for which \[{a_n} < 0\].
A. 22
B. 23
C. 24
D. 25

Answer 1

Hint: By using formula \[{n^{th}}\] term of a harmonic progression \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\], we will find \[{20^{th}}\] term of the harmonic progression. Then put the value of \[a\] and \[{a_{20}}\]to calculate the value of \[d\]. If the \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}} < 0\], then \[a + \left( {n - 1} \right)d\] must be less than zero. By putting the value of \[a\] and \[d\] in the inequality \[a + \left( {n - 1} \right)d < 0\] and solving the inequality, we get the desired result.

Formula Used:
\[{n^{th}}\] term of a harmonic progression is \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\], where \[a\] is the first term of the AP and \[d\] is common difference.

Complete step by step solution:
Given that, \[{a_1},{a_2},{a_3}, \cdots \] are in a harmonic progression.
So, \[\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_2}}}, \cdots \] are in AP.
Given that \[{a_1} = 5\] and \[{a_{20}} = 25\]
The first term of the Ap is \[a = \dfrac{1}{{{a_1}}} = \dfrac{1}{5}\].
Apply the formula of \[{n^{th}}\] term of a harmonic progression is \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\] to calculate the \[{20^{th}}\] term of HP:
\[{a_{20}} = \dfrac{1}{{a + \left( {20 - 1} \right)d}}\]
Now putting the value of \[a\] and \[{a_{20}}\]
\[ \Rightarrow 25 = \dfrac{1}{{\dfrac{1}{5} + 19d}}\]
Solving the equation to calculate the value of \[d\]
\[ \Rightarrow 25 = \dfrac{5}{{1 + 95d}}\]
Apply cross multiplication
\[ \Rightarrow 25\left( {1 + 95d} \right) = 5\]
Divide both sides by 25
\[ \Rightarrow \left( {1 + 95d} \right) = \dfrac{5}{{25}}\]
Subtract 1 from both sides
\[ \Rightarrow 95d = \dfrac{1}{5} - 1\]
\[ \Rightarrow 95d = - \dfrac{4}{5}\]
Divide both sides by 95
\[ \Rightarrow d = - \dfrac{4}{{5 \times 95}}\]
 We have to calculate the least positive integer \[n\] for which \[{a_n} < 0\]. The \[{n^{th}}\] term of a harmonic progression is \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\]. If \[{a_n} < 0\], then \[a + \left( {n - 1} \right)d\] must be less than zero.
So, \[a + \left( {n - 1} \right)d < 0\]
Putting the value of \[a\] and \[d\]
\[ \Rightarrow \dfrac{1}{5} + \left( {n - 1} \right)\left( { - \dfrac{4}{{5 \times 95}}} \right) < 0\]
Multiply \[5 \times 95\] on both sides of inequality
\[ \Rightarrow 95 - 4\left( {n - 1} \right) < 0\]
\[ \Rightarrow 95 < 4\left( {n - 1} \right)\]
Apply distributive property
\[ \Rightarrow 95 < 4n - 4\]
Add 4 on both sides
\[ \Rightarrow 95 + 4 < 4n - 4 + 4\]
\[ \Rightarrow 99 < 4n\]
Divide both sides by 4
\[ \Rightarrow \dfrac{{99}}{4} < n\]
\[ \Rightarrow 24.75 < n\]
\[n\] must be an integer. So, the least positive value of \[n\] is 25.

Hence option D is correct.

Note: Sometimes students plug a wrong value of \[a\]. They put \[a = 5\] in \[{a_n} = \dfrac{1}{{a + \left( {n - 1} \right)d}}\]. But the value of \[a\] is \[\dfrac{1}{{{a_1}}}\]. The correct value of \[a\] is \[\dfrac{1}{5}\].