Question

please solve this question.​

Answer 1

$\Huge\text{(A) 7}$

$\Large\text{\underline{\underline{Explanation}}}$

The A.P sum of the numerator has the following properties.

• First term$\large(a)$: 5
• Common difference$\large(d)$: 4
• Up to $\largen$ terms

We know that the sum of A.P is -

$\text{ \cdots\longrightarrow\boxed{S=\dfrac{n}{2}\{2a+(n-1)d\}.} }$

So, the sum is -

$\cdots\longrightarrow S_{a}=\dfrac{n}{2}\{10+4(n-1)\}$

$\cdots\longrightarrow S_{a}=n\{5+2(n-1)\}$

$\cdots\longrightarrow S_{a}=n(2n+3)$

$\large\therefore S_{a}=n(2n+3).\ \cdots[1]$

The A.P sum of the denominator has the following properties.

• First term$\large(a)$: 7
• Common difference$\large(d)$: 2
• Up to $\largen+1$ terms

We know that the sum of A.P is -

$\text{ \cdots\longrightarrow\boxed{S=\dfrac{n}{2}\{2a+(n-1)d\}.} }$

So, the sum is -

$\cdots\longrightarrow S_{a}=\dfrac{n+1}{2}(14+2n)$

$\large\therefore S_{a}=(n+1)(n+7).\ \cdots[2]$

As we know from [1] and [2], -

$\cdots\longrightarrow\dfrac{n(2n+3)}{(n+1)(n+7)}=\dfrac{17}{16}.$

Now, let's solve this.

$\cdots\longrightarrow16(2n+3)=17(n+1)(n+7)$

$\cdots\longrightarrow32n^{2}+48=17n^{2}+136n+119$

$\cdots\longrightarrow15n^{2}-88n-119=0$

$\cdots\longrightarrow(15n+17)(n-7)=0$

$\large\therefore n=7.$

So, the correct option is choice (a).

Answer 2

${ \huge{ \boxed{ \bf{\underline{ \red{Answer}}}}}} : -$

Option A Is Correct Answer

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