Question

Guys!!!!........ May I have yr attention please??

Answer 1

We know that the sum of 'n' terms of an A.P is given by :

Sn = n/2 × { 2a + (n - 1)d }

Let us consider Series in the Numerator :

5 + 9 + 13 + . . .to n terms

In Numerator series : 'a' = 5 and 'd' = 9 - 5 = 4

Sum of 'n' terms of Numerator series = n/2 × {(2 × 5) + (n - 1)4}

Sum of 'n' terms of Numerator series  = n/2 × {4n + 6} = n(2n + 3)

= 2n² + 3n

Let us consider Series in the Denominator :

7 + 9 + 11 + . . .to (n + 1) terms

In Denominator series : 'a' = 7 and 'd' = 9 - 7 = 2

Sum of 'n + 1' terms of Denominator series = (n + 1)/2 × {(2 × 7) + 2n}

⇒ Sum of 'n + 1' terms of Denominator series  = (n + 1)/2 × {14 + 2n}

⇒ Sum of 'n + 1' terms of Denominator series = (n + 1)(n + 7)

= n² + 8n + 7

The Question reduces to :

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

⇒ 16(2n² + 3n) = 17(n² + 8n + 7)

⇒ 32n² + 48n = 17n² + 136n + 119

⇒ 15n² - 88n - 119 = 0

⇒ 15n² - 105n + 17n - 119 = 0

⇒ 15n(n - 7) + 17(n - 7) = 0

⇒ (n - 7)(15n + 17) = 0

⇒ n = 7 or n = -17/15

n cannot be -17/15, because n is positive number.

⇒ n = 7