Question

If sequence \[\dfrac{(5+9+13+.......\,n\,\,terms\,)}{(7+9+11+.......\,(n+1)\,terms\,)}=\dfrac{17}{16}\], then \[n\] is equal to
1. 7
2. 6
3. 9
4. None of these

Answer 1

Hint: If you carefully look at the pattern of the series, you will find that the series which is present in both numerator as well as denominator are in A.P. (Arithmetic progression) with first term as \[(a=5)\]and common difference as \[(d=4)\] in numerator and in denominator first term as \[(a=7)\]and common difference as\[(d=2)\]. We know that the general sum of \[n\] terms in A.P. is as follows:\[S=\dfrac{n}{2}(2a+(n-1)d)\] so, using this general formula, we can find the value of “n”.

Complete step by step answer:
The series given in the above problem is as follows:
\[\dfrac{(5+9+13+.......\,n\,\,terms\,)}{(7+9+11+.......\,(n+1)\,terms\,)}=\dfrac{17}{16}---(1)\]
As you can see here in above series in numerator is in A.P
\[(5+9+13+.......\,n\,\,terms\,)\]
A.P with \[a=5\] and \[d=4\](in numerator)
Common difference is calculated by taking any term and then subtracting it to its successive term.
Calculation of common difference in the above problem by taking any term for example 9 and the successive term from 9 is 5 so subtracting 9 and 5 we get:
\[d=9-5\]
\[d=4\] (In numerator)
Now we are we are going to use general sum of \[n\] terms which is the following form
\[{{S}_{1}}=\dfrac{n}{2}(2a+(n-1)d)---(2)\]
Substitute the value of \[(a=5)\] and \[(d=4)\] in equation\[(2)\] we get:
\[{{S}_{1}}=\dfrac{n}{2}((2\times 5)+(n-1)\times 4)\]
By simplifying further we get:
\[{{S}_{1}}=\dfrac{n}{2}(10+(n-1)\times 4)---(3)\]
\[{{S}_{1}}=\dfrac{n}{2}(6+4n)---(3)\]
Now, we are solving the given series in denominator which is also in A.P. that is
\[(7+9+11+.......\,(n+1)\,terms\,)\]
A.P with \[a=7\] and \[d=2\] (in denominator)
Here, in these series common difference in the above problem by taking any term for example 9 and the successive term from 9 is 7 so subtracting 9 and 7 we get:
\[d=9-7\]
\[d=2\]
Now we are we are going to use general sum of \[\,(n+1)\,\] terms formula becomes
\[{{S}_{2}}=\dfrac{(n+1)}{2}(2a+((n+1)-1)\times d)\]
After simplifying we get:
\[{{S}_{2}}=\dfrac{(n+1)}{2}(2a+(n\times d))---(4)\]
By substituting the value of \[a=7\] and \[d=2\] on equation \[(4)\] we get:
\[{{S}_{2}}=\dfrac{(n+1)}{2}((2\times 7)+(n-1)\times 2)\]
By simplifying further we get:
\[{{S}_{2}}=\dfrac{(n+1)}{2}((2\times 7)+(n\times 2))\]
\[{{S}_{2}}=\dfrac{(n+1)}{2}(14+2n)---(5)\]
By dividing the equation \[(3)\] and equation \[(5)\]
\[\dfrac{{{S}_{1}}}{{{S}_{2}}}=\dfrac{17}{16}--(6)\]
By substituting the value of equation \[(3)\] and equation \[(5)\] in equation \[(6)\]
\[\dfrac{\left( \dfrac{n}{2} \right)\left( 6+4n \right)}{\left( \dfrac{\left( n+1 \right)}{2} \right)\left( 14+2n \right)}=\text{ }17/16\]
After cross multiply and simplify we get:
\[8n(6+4n)=\dfrac{17\times \left( n+1 \right)}{2}\left( 14+2n \right)\]
2 multiply on both sides
\[16n(6+4n)=17\times \left( n+1 \right)\left( 14+2n \right)\]
Further simplifying we get:
\[96n+6{{n}^{2}}=17\times \left( 14n+2{{n}^{_{_{2}}}}+14+2n \right)\]
\[96n+6{{n}^{2}}=17\times \left( 16n+2{{n}^{_{_{2}}}}+14 \right)\]
After multiplying 17 inside the bracket we get:
\[96n+6{{n}^{2}}=272n+34{{n}^{_{_{2}}}}+238\]
After rearranging the term we get:
\[30n^2=176n+238\]
\[30n^2-176n-238=0\]
Multiply 2 on both the sides of this equation we get:
\[15n^2-88n-119=0\]
By solving this quadratic equation we get:
\[15n(n-7)+17(n-7)=0\]
By taking \[(n-7)\]common we get:
\[(15n+17)(n-7)=0\]
We get two roots
\[n=7\] Or \[n=\dfrac{-17}{15}\]
\[n\]Cannot be negative
\[\therefore n\text{ }=\text{ }7\text{ }\]

So, the correct answer is “Option 1”.

Note: The mistake that could be possible in the above problem is that while selecting the value of\[n\]
Because there are two values of \[n\]if we choose the negative value of \[n\]then you get the wrong answer because series which is given in problem that is A.P. Number in series is increasing that means value of \[n\] cannot be negative and\[n\]will be positive and also be careful while subtracting the larger number with smaller number for finding the value of common difference. Don’t subtract smaller with larger value because we have increasing A.P. so make sure you won’t make this mistake.