
The ${7}^{\text{th}}$ term of an H.P is $\dfrac{1}{10}$ and the ${12}^{\text{th}}$ term is $\dfrac{1}{25}$ , find the ${20}^{\text{th}}$ term.
Answer
578.4k+ views
Hint:
Here we have to calculate the ${20}^{\text{th}}$ term of an H.P. For that, we will write the term H.P series. Then we will first equate the ${7}^{\text{th}}$ term of an H.P with the value of the term given in the question and then again we will equate the ${12}^{\text{th}}$ term of an H.P series with the value of the ${12}^{\text{th}}$ term given in the question.
From there we will get the value of the two unknowns, ‘a’ and ‘d’ where is actually the inverse of first term of an H.P. and inverse of all the term of an H.P forms an A.P, so ‘d’ here is the common difference between inverse of the terms of an H.P series.
Complete step by step solution:
First we will write the H.P series.
$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+........$
It is given that the ${7}^{\text{th}}$ term of an H.P is $\dfrac{1}{10}$ and the ${12}^{\text{th}}$ term is $\dfrac{1}{25}$
We know, ${7}^{\text{th}}$ term of an H.P $=\dfrac{1}{a+6d}$
Now, we will put the value of the ${7}^{\text{th}}$ term of an H.P here.
$\dfrac{1}{10}=\dfrac{1}{a+6d}$
By cross multiplying the numerators and denominators, we get.
$\Rightarrow a+6d=10$…………….$(1)$
We know, ${12}^{\text{th}}$ term of an H.P $=\dfrac{1}{a+11d}$
Now, we will put the value of the ${12}^{\text{th}}$ term of an H.P here.
$\Rightarrow \dfrac{1}{25}=\dfrac{1}{a+11d}$
By cross multiplying the numerators and denominators, we get.
$\Rightarrow a+11d=25$…………….$(2)$
We will subtract equation $(1)$ from equation $(2)$.
$\Rightarrow a+11d-a-6d=25-10$
Simplifying the terms further, we get
$\Rightarrow 5d=15$
Now, we divide 15 by 5.
$\Rightarrow d=\dfrac{15}{5}=3$
$\therefore d=3$
Now, we will put the value of ‘d’ in equation $(1)$
$\Rightarrow a+6\times 3=10$
We will multiply 6 by 3 now.
$\Rightarrow a+18=10$
We will subtract 18 from 10 now.
$\Rightarrow a=10-18=-8$
$\therefore a=-8$
Now, we have to calculate ${20}^{\text{th}}$ term of an H.P which is equal to $\Rightarrow \dfrac{1}{a+19d}$
We will put the value of ‘a’ and ‘d’ here.
$\Rightarrow {20}^{\text{th}}$ term of an H.P $=\dfrac{1}{-8+19\times 3}$
Simplifying further, we get
$\Rightarrow {20}^{\text{th}}$ term of an H.P $=\dfrac{1}{49}$
Therefore, the ${20}^{\text{th}}$ term of an H.P is equal to $\dfrac{1}{49}$.
Note:
Since we have applied a cross multiplication method here, let’s understand it deeply.
A cross multiplication method is defined as a method in which we multiply the numerator of the first fraction with the denominator of the second fraction and multiply the numerator of the second fraction with the denominator of the first fraction.
A cross multiplication method is also used in the addition and subtraction of unlike fractions.
Here we have to calculate the ${20}^{\text{th}}$ term of an H.P. For that, we will write the term H.P series. Then we will first equate the ${7}^{\text{th}}$ term of an H.P with the value of the term given in the question and then again we will equate the ${12}^{\text{th}}$ term of an H.P series with the value of the ${12}^{\text{th}}$ term given in the question.
From there we will get the value of the two unknowns, ‘a’ and ‘d’ where is actually the inverse of first term of an H.P. and inverse of all the term of an H.P forms an A.P, so ‘d’ here is the common difference between inverse of the terms of an H.P series.
Complete step by step solution:
First we will write the H.P series.
$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+........$
It is given that the ${7}^{\text{th}}$ term of an H.P is $\dfrac{1}{10}$ and the ${12}^{\text{th}}$ term is $\dfrac{1}{25}$
We know, ${7}^{\text{th}}$ term of an H.P $=\dfrac{1}{a+6d}$
Now, we will put the value of the ${7}^{\text{th}}$ term of an H.P here.
$\dfrac{1}{10}=\dfrac{1}{a+6d}$
By cross multiplying the numerators and denominators, we get.
$\Rightarrow a+6d=10$…………….$(1)$
We know, ${12}^{\text{th}}$ term of an H.P $=\dfrac{1}{a+11d}$
Now, we will put the value of the ${12}^{\text{th}}$ term of an H.P here.
$\Rightarrow \dfrac{1}{25}=\dfrac{1}{a+11d}$
By cross multiplying the numerators and denominators, we get.
$\Rightarrow a+11d=25$…………….$(2)$
We will subtract equation $(1)$ from equation $(2)$.
$\Rightarrow a+11d-a-6d=25-10$
Simplifying the terms further, we get
$\Rightarrow 5d=15$
Now, we divide 15 by 5.
$\Rightarrow d=\dfrac{15}{5}=3$
$\therefore d=3$
Now, we will put the value of ‘d’ in equation $(1)$
$\Rightarrow a+6\times 3=10$
We will multiply 6 by 3 now.
$\Rightarrow a+18=10$
We will subtract 18 from 10 now.
$\Rightarrow a=10-18=-8$
$\therefore a=-8$
Now, we have to calculate ${20}^{\text{th}}$ term of an H.P which is equal to $\Rightarrow \dfrac{1}{a+19d}$
We will put the value of ‘a’ and ‘d’ here.
$\Rightarrow {20}^{\text{th}}$ term of an H.P $=\dfrac{1}{-8+19\times 3}$
Simplifying further, we get
$\Rightarrow {20}^{\text{th}}$ term of an H.P $=\dfrac{1}{49}$
Therefore, the ${20}^{\text{th}}$ term of an H.P is equal to $\dfrac{1}{49}$.
Note:
Since we have applied a cross multiplication method here, let’s understand it deeply.
A cross multiplication method is defined as a method in which we multiply the numerator of the first fraction with the denominator of the second fraction and multiply the numerator of the second fraction with the denominator of the first fraction.
A cross multiplication method is also used in the addition and subtraction of unlike fractions.
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