First slide

Arithmetic progression

Question

$If \frac{5 + 9 + 13 + \dots upto n terms}{7 + 9 + 11 + \dots . upto (n + 1) terms} = \frac{17}{16}, then n is equal to$

Moderate

A

7
B

6
C

9
D

10

Solution

$Given that \frac{5 + 9 + 13 + \dots upto n terms}{7 + 9 + 11 + \dots . upto (n + 1) terms} = \frac{17}{16}$
$By the given condition \frac{\frac{n}{2} [2.5 + (n - 1) 4]}{\frac{n + 1}{2} [2.7 + (n + 1 - 1) 2]} = \frac{17}{16}$
$\Rightarrow \frac{n}{n + 1} [\frac{10 + 4 n - 4}{14 + 2 n}] = \frac{17}{16}$
$\Rightarrow \frac{n (4 n + 6)}{(n + 1) (2 n + 14)} = \frac{17}{16}$
$\Rightarrow \frac{n (2 n + 3)}{(n + 1) (n + 7)} = \frac{17}{16}$
$\Rightarrow 32 n^{2} + 48 n = 17 n^{2} + 136 n + 119$
$\Rightarrow 15 n^{2} - 88 n - 119 = 0$
$\Rightarrow (n - 7) (15 n + 17) = 0 \Rightarrow n = 7. [∵ n = - \frac{17}{15} is not possible]$

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