Question

Answer the following question in one word or one sentence or as per the exact requirement of the question.
If $\dfrac{3+5+7+......+\; \text{upto n terms}}{5+8+11+...... +\; \text{upto 10 terms}}=7$, then find value of n.

Answer 1

Hint: First observe the sequences on numerator and denominator. By using definition, you can identify them as arithmetic progression. Now you can use all the formulae of an arithmetic progression in numerator and denominator separately. By this you get an equation of n. Now try to bring everything on to the left-hand side and then find the roots possible for n. if we take an arithmetic progression with first term a, common difference d, we can say the formula of sum of n terms.
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$

Complete step-by-step solution -
A sequence of numbers such that the difference of any two consecutive numbers is a constant, called an Arithmetic Progression. For example the sequence 1, 2, 3, 4….. is an arithmetic progression with common difference = 2-1 =1. Now, we need to find ${{n}^{th}}$ term of such sequence. Let us have a sequence with the first term as ‘a’ and common difference d.
We know difference between successive terms is d. so, second term = first term+d = a+d
Third term = second term+d= a+2d.
And so on calculating, we get:
${{n}^{th}}={{n-1}^{th}}term+d=a+\left( n-2 \right)d+d=a+\left( n-1 \right)d$
Given equation in the question which we need to solve:
$\dfrac{3+5+7+.......n\,terms}{5+8+11+......10\,terms}=7.......(i)$
By taking numerator of above fraction separately and assume it as N, we get
N=3+5+7+……..n terms.
We know sum of n terms of an arithmetic progression of first term a and common difference d is given by:
\[S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)...........(ii)\]
Common difference of N
$\Rightarrow d=5-3=2$
By substituting a=3, d=2, we get the above sequence
$N=\dfrac{n}{2}\left( 2\times 3 + \left( n-1 \right)2 \right)$
By simplifying above equation, we get value of N as:
$N=\dfrac{n}{2}\left( 6+2n+2 \right)=\dfrac{n}{2}\left( 2n+4 \right)$
By simplifying above equation, we get final value of N as:
$N={{n}^{2}}+2n......(iii)$
Now, by taking denominator of equation (i), assume it as M, we get,
$M=5+8+11.......10\,terms$
Common difference of M is $d=8-5=3$
So, by substituting a=5, d=3, n=10 in equation, we get:
$M=\dfrac{10}{2}\left( 2\times 5+\left( 10-1 \right)3 \right)$
By simplifying above equation, we get value of M as:
$M=5\times 37$
By substituting values of M, N into equation (i), we get:
$\dfrac{{{n}^{2}}+2n}{5\times 37}=7$
By cross multiplying we can say the equation as:
${{n}^{2}}+2n=35\times 37=1295$
${{n}^{2}}+2n-1295=0$
By breaking $2n$ as $37n-35n$, we can get the equation as:
${{n}^{2}}+37n-35n-1295=0$
By factorization we can write it as $(n+37) (n-35)$ = 0.
So roots are 35, -37. As n is the number of terms $n > 0$.
So, $n=35$. Therefore, the value of n in the given question is 35.

Note: Be careful at substituting into sum formula as it forms the main base to the solution. Alternate method is to write ${{n}^{2}}+2n\,$ as $n\left( n+2 \right)=35\times 37$ by comparing you can see that the solution is 35 directly without factorising. But we need factorization to prove that the other root of the quadratic which we neglected does not satisfy the given question.