
If \[{{\text{m}}^{{\text{th}}}}\] term of HP is n while \[{{\text{n}}^{{\text{th}}}}\] term is m find its \[{{\text{(m + n)}}^{{\text{th}}}}\] term.
Answer
606k+ views
Hint – In order to solve this question we need to know that nth term of HP is inverse of nth term of AP. After using this concept as per the conditions given we will get the answer.
Complete step-by-step answer:
As we know nth term of HP is inverse of nth term of AP.
It is given that \[{{\text{m}}^{{\text{th}}}}\] the term of HP is n.
So, ${{\text{T}}_{\text{m}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (m - 1)d}}}}{\text{ = n}}$ ……(1)
And ${{\text{T}}_{\text{n}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (n - 1)d}}}}{\text{ = m}}$ ……(2)
Equation (1) can be written as:
${\text{a + (m - 1)d = }}\dfrac{{\text{1}}}{{\text{n}}}$= a + md – d ……(3)
Equation (2) can be written as:
${\text{a + (n - 1)d = }}\dfrac{{\text{1}}}{{\text{m}}}$= a + nd – d ……(4)
On subtracting equation (4) from (3) we get the new equation as:
a – a +md – nd – d + d = $\dfrac{{\text{1}}}{{\text{n}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{m}}}$
(m - n)d = $\dfrac{{\text{1}}}{{\text{n}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{m}}}$
(m - n)d = $\dfrac{{{\text{m - n}}}}{{{\text{mn}}}}$
Then we get, d = $\dfrac{{\text{1}}}{{{\text{mn}}}}$
On putting the value of d in equation (3) we get the new equation as:
$
{\text{a + (n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{m}}} \\
{\text{a = }}\dfrac{{\text{1}}}{{\text{m}}}{\text{ - (n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}} \\
{\text{a = }}\dfrac{{{\text{n - n + 1}}}}{{{\text{mn}}}} \\
{\text{a = }}\dfrac{{\text{1}}}{{{\text{mn}}}} \\
$
Now we have first term and common difference so now we can find the
(m + n)th term of HP.
${{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (m + n - 1)d}}}}$
On putting the value of a and d in above equation we get,
$
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{\dfrac{{\text{1}}}{{{\text{mn}}}}{\text{ + (m + n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}}}} \\
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{\dfrac{{{\text{1 + m + n - 1}}}}{{{\text{mn}}}}}} \\
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{{\text{mn}}}}{{{\text{m + n}}}} \\
$
So, we get the \[{\left( {{\text{m + n}}} \right)^{{\text{th}}}}\] term of the HP.
Note – Whenever you face such types of problems you have used the concept that nth term of HP is inverse of nth term of AP. Here in this question we have made equations of AP with the help of given HP then solved it to find the first term and common difference then you can find any of the terms with the help of first term and common difference. Proceeding like this will take you to the right solution of the question asked.
Complete step-by-step answer:
As we know nth term of HP is inverse of nth term of AP.
It is given that \[{{\text{m}}^{{\text{th}}}}\] the term of HP is n.
So, ${{\text{T}}_{\text{m}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (m - 1)d}}}}{\text{ = n}}$ ……(1)
And ${{\text{T}}_{\text{n}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (n - 1)d}}}}{\text{ = m}}$ ……(2)
Equation (1) can be written as:
${\text{a + (m - 1)d = }}\dfrac{{\text{1}}}{{\text{n}}}$= a + md – d ……(3)
Equation (2) can be written as:
${\text{a + (n - 1)d = }}\dfrac{{\text{1}}}{{\text{m}}}$= a + nd – d ……(4)
On subtracting equation (4) from (3) we get the new equation as:
a – a +md – nd – d + d = $\dfrac{{\text{1}}}{{\text{n}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{m}}}$
(m - n)d = $\dfrac{{\text{1}}}{{\text{n}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{m}}}$
(m - n)d = $\dfrac{{{\text{m - n}}}}{{{\text{mn}}}}$
Then we get, d = $\dfrac{{\text{1}}}{{{\text{mn}}}}$
On putting the value of d in equation (3) we get the new equation as:
$
{\text{a + (n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{m}}} \\
{\text{a = }}\dfrac{{\text{1}}}{{\text{m}}}{\text{ - (n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}} \\
{\text{a = }}\dfrac{{{\text{n - n + 1}}}}{{{\text{mn}}}} \\
{\text{a = }}\dfrac{{\text{1}}}{{{\text{mn}}}} \\
$
Now we have first term and common difference so now we can find the
(m + n)th term of HP.
${{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{a + (m + n - 1)d}}}}$
On putting the value of a and d in above equation we get,
$
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{\dfrac{{\text{1}}}{{{\text{mn}}}}{\text{ + (m + n - 1)}}\dfrac{{\text{1}}}{{{\text{mn}}}}}} \\
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{\text{1}}}{{\dfrac{{{\text{1 + m + n - 1}}}}{{{\text{mn}}}}}} \\
{{\text{T}}_{{\text{m + n}}}}{\text{ = }}\dfrac{{{\text{mn}}}}{{{\text{m + n}}}} \\
$
So, we get the \[{\left( {{\text{m + n}}} \right)^{{\text{th}}}}\] term of the HP.
Note – Whenever you face such types of problems you have used the concept that nth term of HP is inverse of nth term of AP. Here in this question we have made equations of AP with the help of given HP then solved it to find the first term and common difference then you can find any of the terms with the help of first term and common difference. Proceeding like this will take you to the right solution of the question asked.
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 6 Maths: Engaging Questions & Answers for Success

Class 6 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

