Let Sn = an2+ bn be sum of n terms of a series of real numbers with Tn as its nth term. If the ratio T4nTn is independent of n, then the value of S8S9 is

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Let S_n = an²+ bn be sum of n terms of a series of real numbers with T_n as its n^th term. If the ratio $\frac{T_{4 n}}{T_{n}}$ is independent of n, then the value of $\frac{S_{8}}{S_{9}}$ is

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$\frac{9}{8}$

$\frac{8}{9}$

$\frac{5}{4}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

Using

T_n = S_n – S_n–1 = 2an + b – a
$\Rightarrow \frac{T_{4 n}}{T_{n}} = \frac{8 an + b - a}{2 an + b - a}$
will be independent of n $\Rightarrow$ a = b
$\Rightarrow S_{n} = an (n + 1) \Rightarrow \frac{S_{8}}{S_{9}} = \frac{4}{5}$