Question

If the ratio of sum of \[n\] terms in two AP is \[2n:n + 1\], then the ratio of \[{8^{th}}\] terms is
A. \[15:8\]
B. \[8:133\]
C. \[5:17\]
D. none

Answer 1

Hint:
Here we use the formulas of AP (arithmetic progression)
* The sum of \[n\] terms of an AP is given by \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]where \[a\]is the first terms and \[d\] is the common difference.
* The \[{n^{th}}\] terms of an AP is given by \[{T_n} = a + \left( {n - 1} \right)d\]where \[a\]is the first terms and \[d\]is the common difference.
* Ratio of two numbers is the same when divided or multiplied by the same number.
i.e. ratio \[a:b = m \times a:m \times b\]

Complete step by step solution:
Here we have to find the ratio between the \[{8^{th}}\] term of two AP’s.
Given, the ratio of \[n\] terms in two AP is \[2n:n + 1\].
Let \[T\]and \[T'\]be the two arithmetic series with first terms as \[a\]and \[a'\]having common differences as \[d\]and \[d'\].
Sum of \[n\] terms of the first AP is given by \[S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Sum of \[n\] terms of an second AP is given by \[S' = \dfrac{n}{2}\left[ {2a' + \left( {n - 1} \right)d'} \right]\].
Find the ratio of the two APs.
\[\dfrac{S}{{S'}} = \dfrac{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\dfrac{n}{2}\left[ {2a' + \left( {n - 1} \right)d'} \right]}}\]
Cancel out the common factor \[\dfrac{n}{2}\] on the right side.
\[\dfrac{S}{{S'}} = \dfrac{{\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\left[ {2a' + \left( {n - 1} \right)d'} \right]}}\]
Equate the obtained ratio to the given ratio, that is \[2n:n + 1\]
\[\dfrac{{\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\left[ {2a' + \left( {n - 1} \right)d'} \right]}} = \dfrac{{2n}}{{n + 1}}\]
Divide the numerator and denominator of the left side by 2 to simplify.
\[\dfrac{{a + \dfrac{{\left( {n - 1} \right)}}{2}d}}{{a' + \dfrac{{\left( {n - 1} \right)}}{2}d'}} = \dfrac{{2n}}{{n + 1}}\]
It can be observed that left side of the equation is similar to the ratio of \[{(\dfrac{{n - 1}}{2} + 1)^{th}}\] terms that is \[\dfrac{{{T_{\dfrac{{n - 1}}{2} + 1}}}}{{{{T'}_{\dfrac{{n - 1}}{2} + 1}}}} = \dfrac{{a + \dfrac{{\left( {n - 1} \right)}}{2}d}}{{a' + \dfrac{{\left( {n - 1} \right)}}{2}d'}}\].
Therefore \[\dfrac{{{T_{\dfrac{{n - 1}}{2} + 1}}}}{{{{T'}_{\dfrac{{n - 1}}{2} + 1}}}} = \dfrac{{2n}}{{n + 1}}\]
To find the 8th terms equate \[\dfrac{{n - 1}}{2}\] to 7 and solve to find the value of \[n\].
\[
  \dfrac{{n - 1}}{2} = 7 \\
  n = 15 \\
 \]
Substitute 15 for \[n\] into \[\dfrac{{{T_{\dfrac{{n - 1}}{2} + 1}}}}{{{{T'}_{\dfrac{{n - 1}}{2} + 1}}}} = \dfrac{{2n}}{{n + 1}}\] and solve to obtain the ratio of 8th term.
\[
  \dfrac{{{T_{\dfrac{{15 - 1}}{2} + 1}}}}{{{{T'}_{\dfrac{{15 - 1}}{2} + 1}}}} = \dfrac{{2\left( {15} \right)}}{{15 + 1}} \\
  \dfrac{{{T_8}}}{{{{T'}_8}}} = \dfrac{{30}}{{16}} \\
   = \dfrac{{15}}{8} \\
 \]

Therefore, Option A is correct.

Note:
When solving these types of questions, care should be taken for \[n\]in the nth term formula and the sum formula, as \[n - 1\] is used in the formula. If \[n\]is used , the wrong result will be produced. Also, students should keep in mind that the present term is dependent on the previous term, therefore finding the previous term is the main key to find the required term.