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.