Question

Let \[{a_1},{a_2},{a_3},...\] be terms of an arithmetic progression. If \[\dfrac{{\left( {{a_1} + {a_2} + ... + {a_p}} \right)}}{{\left( {{a_1} + {a_2} + ... + {a_q}} \right)}} = \dfrac{{{p^2}}}{{{q^2}}}\], where \[p \ne q\]. Then what is the value of \[\dfrac{{{a_6}}}{{{a_{21}}}}\]?
A. \[\dfrac{7}{2}\]
B. \[\dfrac{2}{7}\]
C. \[\dfrac{{11}}{{41}}\]
D. \[\dfrac{{41}}{{11}}\]

Answer 1

Hint:
First, simplify the given equation using the formula of the sum of the first \[n\] terms of an arithmetic progression. Solve the equation and calculate the common difference of the given arithmetic progression. After that, rewrite the required ratio using the formula of the \[{n^{th}}\] term of an arithmetic progression. In the end, substitute the value of the common difference in the ratio and solve it to reach the required answer.



Formula Used:
The sum of the first \[n\] terms of an arithmetic progression: \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
The \[{n^{th}}\] term of an arithmetic progression: \[{a_n} = a + \left( {n - 1} \right)d\]
Where, \[a\] is the first term and \[d\] is the common difference of an arithmetic progression.


Complete step-by-step answer:
Given:
\[{a_1},{a_2},{a_3},...\] be terms of an arithmetic progression.
\[\dfrac{{\left( {{a_1} + {a_2} + ... + {a_p}} \right)}}{{\left( {{a_1} + {a_2} + ... + {a_q}} \right)}} = \dfrac{{{p^2}}}{{{q^2}}}\] where \[p \ne q\]

Let’s solve the above equation.
Apply the formula of the sum of the first \[n\] terms of an arithmetic progression.
Let \[d\] be the common difference of the given arithmetic progression.
\[\Rightarrow \dfrac{{\dfrac{p}{2}\left[ {2{a_1} + \left( {p - 1} \right)d} \right]}}{{\dfrac{q}{2}\left[ {2{a_1} + \left( {q - 1} \right)d} \right]}} = \dfrac{{{p^2}}}{{{q^2}}}\]
Cancel out the common terms from both sides.
\[\Rightarrow \dfrac{{2{a_1} + \left( {p - 1} \right)d}}{{2{a_1} + \left( {q - 1} \right)d}} = \dfrac{p}{q}\]
By cross multiplication
\[\Rightarrow 2{a_1}q + \left( {p - 1} \right)qd = 2{a_1}p + \left( {q - 1} \right)pd\]
\[ \Rightarrow 2{a_1}q + pqd - qd = 2{a_1}p + pqd - pd\]
Cancel out the common terms from both sides.
\[\Rightarrow 2{a_1}q - qd = 2{a_1}p - pd\]
\[ \Rightarrow d\left( {p - q} \right) = 2{a_1}\left( {p - q} \right)\]
\[ \Rightarrow d = 2{a_1}\]
Thus, the common difference of the given arithmetic progression is \[2{a_1}\].

We need to find \[\dfrac{{{a_6}}}{{{a_{21}}}} \]
Apply the formula of the \[{n^{th}}\] term of an arithmetic progression.
\[\dfrac{{{a_6}}}{{{a_{21}}}} = \dfrac{{{a_1} + (6-1)d}}{{{a_1} + (21-1)d}}\]
Substitute the value of the common difference.
\[\Rightarrow \dfrac{{{a_6}}}{{{a_{21}}}} = \dfrac{{{a_1} + 5\left( {2{a_1}} \right)}}{{{a_1} + 20\left( {2{a_1}} \right)}}\]
\[ \Rightarrow \dfrac{{{a_6}}}{{{a_{21}}}} = \dfrac{{{a_1} + 10{a_1}}}{{{a_1} + 40{a_1}}}\]
\[ \Rightarrow \dfrac{{{a_6}}}{{{a_{21}}}} = \dfrac{{11{a_1}}}{{41{a_1}}}\]
\[ \Rightarrow \dfrac{{{a_6}}}{{{a_{21}}}} = \dfrac{{11}}{{41}}\]
Hence the correct option is C.



Note:
Students often make a common mistake that is they are using the formula for \[{n^{th}}\] term as \[{a_n} = a + nd\] which is a wrong formula. The correct formula is \[{a_n} = a + \left( {n - 1} \right)d\].