Question

If the $7^{th}$ term of HP is \[\dfrac{1}{{10}}\] $12^{th}$ term is \[\dfrac{1}{{25}}\], then the $20^{th}$term is
A.\[\dfrac{1}{{41}}\]
B. \[\dfrac{1}{{45}}\]
C. \[\dfrac{1}{{49}}\]
D. \[\dfrac{1}{{37}}\]

Answer 1

Hints: If \[\dfrac{1}{a}\] is $n^{th}$ term of HP then a is $n^{th}$ term of AP. Now, apply this concept to obtain the $7^{th}$ term and $12^{th}$ term of AP, then find the values of a and d by the $n^{th}$ term formula of AP. Apply the $n^{th}$ term formula of AP to find the $20^{th}$ term. Substitute 3 for d and -8 for a in the expression \[a + 19d\] and take the reciprocal of 49 to obtain the required value.

Formula used
The $n^{th}$ term of AP is \[{t_n} = a + (n - 1)d\] , where a is the first term and d is the common difference.

Complete step by step solution
It is given that; $7^{th}$ term of HP is \[\dfrac{1}{{10}}\] $12^{th}$ term is \[\dfrac{1}{{25}}\].
Therefore, $7^{th}$ term of AP is 10 and $12^{th}$ term is 25.
Now, the $n^{th}$ term of AP is \[{t_n} = a + (n - 1)d\].
So, \[a + 6d = 10 - - - (1)\]
And
\[a + 11d = 25 - - - (2)\]
Subtract (1) from (2) to obtain the value of d.
\[a + 11d - a - 6d = 25 - 10\]
\[5d = 15\]
\[d = 3\]
Substitute 3 for d in (1) to obtain the value of a.
\[\begin{array}{l}a + 6 \times 3 = 10\\a = 10 - 18\\a = - 8\end{array}\]
The $20^{th}$ term is \[{t_{20}} = a + 19d\]
Substitute 3 for d and -8 for a in the equation \[{t_{20}} = a + 19d\] to obtain the $20^{th}$ term of AP.
\[\begin{array}{I}{t_{20}} = - 8 + 19 \times 3\\ = - 8 + 57\\ = 49\end{array}\]
Therefore, the $20^{th}$ term of HP is \[\dfrac{1}{{49}}.\]

The correct option is C.

Note: Recall the concept that “if \[\dfrac{1}{a}\] is $n^{th}$ term of HP then a is the $n^{th}$ term of AP”- by using this concept students can easily solve this type of problems. Sometimes students solve the problem directly with HP which leads to a critical calculation. Other than that, convert the terms of HP into AP and then solve the problem easily.