Question

if the ratio of the sum of the 1st n terms of the APs is (7n+1):(4n+27),tjen find the ratio of their 9th term.

Answer 1

Solution:

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Given:

   The ratio of the sum of the 1st n terms of the APs is (7n+1):(4n+27)

 $P_n : Q_n = 7n+1: 4n+27$

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To find :

Ratio of their 9th term,.

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By converting the ratios into fraction , we get,

=> $\frac{P_n}{Q_{n}} = \frac{7n+1}{4n+27}$

As, we know that,

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

So,

=> $\frac{ \frac{n}{2}(2a_1+(n-1)d_1) }{ \frac{n}{2}(2a_2 + (n-1)d_2)} = \frac{7n+1}{4n+27}$

=> $\frac{ (2a_1+(n-1)d_1) }{ (2a_2 + (n-1)d_2)} = \frac{7n+1}{4n+27}$

=> $\frac{ 2a_1+nd_1-d_1 }{ 2a_2 + nd_2-d_2} = \frac{7n+1}{4n+27}$....(i)

By replacing value of n = 2l - 1 in (i),

We get,

=> $\frac{2a_1+(2l-1-1)d_1}{2a_2 + (2l-1-1)d_2} = \frac{7(2l-1)+1}{4(2l-1)+27}$

=>   $\frac{2a_1+(2l-2)d_1}{2a_2 + (2l-2)d_2} = \frac{7(2l-1)+1}{4(2l-1)+27}$

=> $\frac{2(a_1+(l-1)d_1)}{2(a_2 + (l-1)d_2)} = \frac{14l-7+1}{8l-4+27}$

=> $\frac{a_1+(l-1)d_1}{a_2 + (l-1)d_2} = \frac{14l-6}{8l+23}$

=> $a_{l1} : a_{l2} = 14l - 6 : 8l + 23$


∴ $a_9 : a_9 = 14(9) -6 : 8(9)+23$

=> $a_9 : a_9 = 126 - 6 : 72 + 23$

=> $a_9 : a_9= 120: 95$

=> $a_9 : a_9= 24 : 19$

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                            Hope it Helps !!

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