If 5+9+13+... to n terms7+9+11+...to (n+1) terms=1716, then n =

Question:

If $\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$, then $n=$

(a) 8

(b) 7

(c) 10

(d) 11

Solution:

Here, we are given,

$\frac{5+9+13+\ldots \text { to } n \text { terms }}{7+9+11+\ldots \text { to }(n+1) \text { terms }}=\frac{17}{16}$ .......(1)

We need to find n.

So, first let us find out the sum of n terms of the A.P. given in the numerator. Here we use the following formula for the sum of n terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

Here,

Common difference of the A.P. (d) = 

$=9-5$

$=4$

Number of terms (n) = n

First term for the given A.P. (a) = 

So, using the formula we get,

$S_{n}=\frac{n}{2}[2(5)+(n-1)(4)]$

$=\left(\frac{n}{2}\right)[10+(4 n-4)]$

$=\left(\frac{n}{2}\right)(6+4 n)$

 

$=n(3+2 n)$ ...........(2)

Similarly, we find out the sum of $(n+1)$ terms of the A.P. given in the denominator $(7+9+11+\ldots)$.

Here,

 

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=9-7$

 

$=2$

Number of terms (n) = n

First term for the given A.P. (a) = 

So, using the formula we get,

$S_{n+1}=\frac{n+1}{2}[2(7)+[(n+1)-1](2)]$

$=\left(\frac{n+1}{2}\right)[14+(n)(2)]$

$=(n+1)(7+n)$

$=7 n+7+n^{2}+n$

$=n^{2}+8 n+7$$\ldots(3)$

Now substituting the values of (2) and (3) in equation (1), we get,

$\frac{2 n^{2}+3 n}{n^{2}+8 n+7}=\frac{17}{16}$

$16\left(2 n^{2}+3 n\right)=17\left(n^{2}+8 n+7\right)$

$32 n^{2}+48 n=17 n^{2}+136 n+119$

$32 n^{2}-17 n^{2}+48 n-136 n-119=0$

$15 n^{2}-88 n-119=0$

Further solving the quadratic equation for n by splitting the middle term, we get,

$15 n^{2}-88 n-119=0$

$15 n^{2}-105 n+17 n-119=0$

$15 n(n-7)+17(n-7)=0$

 

$(15 n+17)(n-7)=0$

So, we get

$15 n+17=0$

$15 n=-17$

$n=\frac{-17}{15}$

Or

$n-7=0$

$n=7$

Since is a whole number, it cannot be a fraction. So, 

Therefore, the correct option is (b).

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