Question

If $\dfrac{{\left( {5 + 9 + 13 + ....n{\text{ terms}}} \right)}}{{\left( {7 + 9 + 11 + ...\left( {n + 1} \right){\text{ terms}}} \right)}} = \dfrac{{17}}{{16}}$ then find the value of $n$.
A. $7$
B. $6$
C. $9$
D. None of these

Answer 1

Hint: In this question, for determining the value of $n$, we have to use the concept of arithmetic progression. We can find the sum of all the terms for both the sequences and simplify the equation.

Formula Used: The sum of all the terms of an Arithmetic progression sequence is given by ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Here, $n$ is total number of terms of a sequence
$a$ is the first term of a sequence
$d$ is the common difference of a sequence
${S_n}$ is the sum of all the terms of a sequence

Complete step by step solution:
We know that $\dfrac{{\left( {5 + 9 + 13 + ....n{\text{ terms}}} \right)}}{{\left( {7 + 9 + 11 + ...\left( {n + 1} \right){\text{ terms}}} \right)}} = \dfrac{{17}}{{16}}$
Let ${S_1} = 5 + 9 + 13 + ....n{\text{ terms}}$
The sum of all the terms of an AP sequence is given by
${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Where, $a$ is the first term
                $d$ is the common difference of the sequence.
So, the common difference is given by
 $ d = 9 - 5 \\ \Rightarrow d = 4 $
Here, $a = 5,d = 4$
By applying the concept of Arithmetic progression, we get
$ {S_1} = \left( {\dfrac{n}{2}} \right)\left( {2\left( 5 \right) + \left( {n - 1} \right)4} \right) \\ \Rightarrow {S_1} = \left( {\dfrac{n}{2}} \right)\left( {10 + 4n - 4} \right) \\ \Rightarrow {S_1} = \left( {\dfrac{n}{2}} \right)\left( {6 + 4n} \right) $
Let ${S_2} = 7 + 9 + 11 + ...\left( {n + 1} \right){\text{ terms}}$
Here, the common difference is given by
 $ d = 9 - 7 \\ \Rightarrow d = 2 $
So, $a = 7,d = 2$
By applying the concept of Arithmetic progression, we get
$ {S_2} = \left( {\dfrac{{n + 1}}{2}} \right)\left( {2\left( 7 \right) + \left( {\left( {n + 1} \right) - 1} \right)2} \right) \\ \Rightarrow {S_2} = \left( {\dfrac{{n + 1}}{2}} \right)\left( {14 + \left( n \right)2} \right) \\ \Rightarrow {S_2} = \left( {\dfrac{{n + 1}}{2}} \right)\left( {14 + 2n} \right) \\ \Rightarrow {S_2} = \left( {n + 1} \right)\left( {7 + n} \right) $
Thus, we get
$ \dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{\left( {\dfrac{n}{2}} \right)\left( {6 + 4n} \right)}}{{\left( {n + 1} \right)\left( {7 + n} \right)}} = \dfrac{{17}}{{16}} \\ \Rightarrow \dfrac{{n\left( {3 + 2n} \right)}}{{\left( {n + 1} \right)\left( {7 + n} \right)}} = \dfrac{{17}}{{16}} \\ \Rightarrow 16n\left( {3 + 2n} \right) = 17\left( {n + 1} \right)\left( {7 + n} \right) \\ \Rightarrow 48n + 32{n^2} = 17\left( {7n + {n^2} + 7 + n} \right) $
By simplifying further we get
$48n + 32{n^2} = \left( {17{n^2} + 136n + 119} \right) \\ \Rightarrow 48n - 136n + 32{n^2} - 17{n^2} - 119 = 0 \\ \Rightarrow 15{n^2} - 88n - 119 = 0 $
Let us factorize the above quadratic equation.
$ \Rightarrow 15{n^2} - 105n + 17n - 119 = 0 \\ \Rightarrow 15n\left( {n - 7} \right) + 17\left( {n - 7} \right) = 0 \\ \Rightarrow \left( {n - 7} \right)\left( {15n + 17} \right) = 0 $
Thus, we get
$n = 7,\dfrac{{ - 17}}{{15}}$
Here, neglect the negative value as $n$ cannot be negative.
So, consider $n = 7$i since it is a whole number.

Option ‘A’ is correct

Additional Information: A set of data is an arithmetic progression (AP) or an arithmetic sequence in mathematics in which the difference between subsequent terms remains constant. That means an arithmetic progression is a progression in which the difference between any two consecutive terms is always constant. Also, an arithmetic progression is a series that finds each phrase by adding a specified amount to the preceding term.

Note: Many students make mistakes in calculating the sum of all the terms of an arithmetic progression and further part. This is the only way, through which we can solve the example in the simplest way.