Question

If 5+9+13+.... to n terms/7+9+11+... to (n+1) terms=17/16, Find the value of n.

Answer 1

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Answer 2

Answer:

Value of n is 7.

Step-by-step explanation:

We are given APs in both numerator and denominator,

AP in numerator,

5 + 9 + 13 + ... + n

a = 5 and d = 4

So, $S_n=\frac{n}{2}(2a+(n-1)d)$

$S_n=n(2\times5+(n-1)2)$

$S_n=n(5+(n-1)2)$

$S_n=n(3+2n)$

AP in denominator,

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

a = 7 and d = 2

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

$S_{n+1}=\frac{n}{2}(2\times7+(n)2)$

$S_{n+1}=(n+1)(7+n)$

$S_{n+1}=(n+1)(7+n)$

Now we get,

$\frac{S_n}{S_{n+1}}=frac{17}{16}$

$\frac{n(3+2n)}{(n+1)(7+n)}=frac{17}{16}$

$16n(3+2n)=(17n+17)(7+n)$

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

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

On solving we get,

$n=7$

Therefore, Value of n is 7.