Question

The ratio of ${{7}^{th}}$ to the ${{3}^{rd}}$ term of an AP is 12:5. Find the ratio ${{13}^{th}}$ to the ${{4}^{th}}$ term.

Answer 1

Hint: The formula for finding the n-th term of an AP is ${{t}_{n}}=a+\left( n-1 \right)d$, where “a” is the first term “d” is the common difference of the progression. We have to use these formulae to find the place value of ${{7}^{th}}$,${{3}^{rd}}$, ${{13}^{th}}$, ${{4}^{th}}$ by replacing the value of “n” by 7, 3, 13, 4 in the general formulae. Then we have to calculate accordingly to find the suitable result.

Complete step-by-step answer:
We know the formula for finding the n-th term of an AP is ${{t}_{n}}=a+\left( n-1 \right)d$.
Now, replacing the value of “n” by 7 and 3 we get ${{t}_{7}}=a+\left( 7-1 \right)d$ and ${{t}_{3}}=a+\left( 3-1 \right)d$ respectively.
According to the question we also know that the ratio of ${{7}^{th}}$ to the ${{3}^{rd}}$ term is given is 12:5.

Now, let us substitute the values as
$\dfrac{{{t}_{7}}}{{{t}_{3}}}=\dfrac{12}{5}$
$\Rightarrow$ $\dfrac{a+\left( 7-1 \right)d}{a+\left( 3-1 \right)d}=\dfrac{12}{5}$
$\Rightarrow$ $\dfrac{a+6d}{a+2d}=\dfrac{12}{5}$
Now cross multiplying the numbers,
$\Rightarrow$ $5\left( a+6d \right)=12\left( a+2d \right)$
By multiplying and removing the brackets we get,
$\Rightarrow$ $5a+30d=12a+24d$
Taking “a” variant to the left hand side and “d” variant to the right hand side,
$\Rightarrow$ $12a-5a=30d-24d$
$\Rightarrow$ $7a=6d$
$\Rightarrow$ $\dfrac{a}{d}=\dfrac{6}{7}$
$\Rightarrow$ $\dfrac{d}{a}=\dfrac{7}{6}$

We get the value of $\dfrac{d}{a}=\dfrac{7}{6}$
Now we have to find the ratio of ${{13}^{th}}$ term to the ${{4}^{th}}$ term, i.e $\dfrac{{{t}_{13}}}{{{t}_{4}}}$.
Replacing the value of “n” by 13 and 4 we get,
$\Rightarrow$ $\dfrac{a+\left( 13-1 \right)d}{a+\left( 4-1 \right)d}$
$\Rightarrow$ $\dfrac{a+12d}{a+3d}$
Now we have to divide “a” with numerator and denominator.
$\Rightarrow$ $\dfrac{1+12\dfrac{d}{a}}{1+3\dfrac{d}{a}}$
Now putting the value of $\dfrac{d}{a}=\dfrac{7}{6}$
$\Rightarrow$ $\dfrac{1+12\times \dfrac{7}{6}}{1+3\times \dfrac{7}{6}}$
\Rightarrow $$$$\dfrac{1+14}{1+3.5}
$\Rightarrow$ $\dfrac{15}{4.5}$
$\Rightarrow$ $\dfrac{150}{45}$
\Rightarrow $$$$\dfrac{10}{3}

The ratio ${{13}^{th}}$ to the ${{4}^{th}}$ term is 10:3.

Note: We have to remember the formulae for the nth term of an AP is ${{t}_{n}}=a+\left( n-1 \right)d$ and evaluate calculation accordingly. Student also get confused with the formulae and write ${{t}_{n}}=a+\left( n+1 \right)d$ instead of ${{t}_{n}}=a+\left( n-1 \right)d$. Sometimes student don’t understand what is told in the question and wrote the ratio of ${{7}^{th}}$ to the ${{3}^{rd}}$ term of an AP is $\dfrac{{{t}_{7}}}{{{t}_{3}}}=\dfrac{5}{12}$ instead of $\dfrac{{{t}_{7}}}{{{t}_{3}}}=\dfrac{12}{5}$. We have to look out for this silly mistake and solve the problem properly.