Question

The sum of first n terms of two aps are in the ratio ( 3n+ 8) (7n+15) find the ratio of their 12th terms​

Answer 1

Answer :-

• Required ratio is 7 : 16

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Step-by-step explanation :-

Given : Ratio of sum of nth terms of two AP's is (3n+8) : (7n+15)

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‎To find : Ratio of 12th terms of APs

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‎Solution :

We have formula for sum of nth term of AP as follows:

➝ $\sf Sn = \dfrac{n}{2}\bigg[2a + (n-1)d\bigg]$

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‎Also the sum of nth term of AP is given by:

➝ $\sf an = a + (n-1)d$

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‎Given that the ratio of sum is (3n + 8) : (7n + 15)

$\sf{ \implies\dfrac{Sn_1}{Sn_2}=\dfrac{(3n + 8)}{ (7n + 15)}}$

${\implies \sf \dfrac{ \left \{\dfrac{ n}{2}[2a + (n-1) d ] \right \} }{ \left \{\dfrac{n}{2} \left[2A + (n-1) D \right] \right \}} = \dfrac{(3n + 8) }{(7n + 15)}}$

${\implies \sf \dfrac{ \cancel\dfrac{ n}{2}[2a + (n-1) d ] }{ \cancel\dfrac{n}{2} \left[2A + (n-1) D \right] } = \dfrac{(3n + 8) }{(7n + 15)}}$

${\implies \sf \dfrac{ 2a + (n-1) d }{ 2A + (n-1) D } = \dfrac{3n + 8}{7n + 15}}$

${\implies \sf \dfrac{ 2 \left(a + \dfrac{ (n-1)}{2} d \right)}{ 2 \left(A + \dfrac{(n-1)}{2} D \right) } = \dfrac{3n + 8}{7n + 15}}$

${ {\implies \sf \dfrac{ a + \dfrac{ (n-1)}{2} d }{ A + \dfrac{(n-1)}{2} D } = \dfrac{3n + 8}{7n + 15}} - - - (1.)}$

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We have to find ratio of 12th terms, i.e.

${\sf \implies \dfrac{a + 11d}{A + 11 D}---(2.)}$

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By comparing eq(1) and (2), we get:

$\sf\implies\dfrac{n-1}{2}=11$

$\sf\implies{n-1} = 2\times 11$

$\sf\implies{n-1} = 22$

$\sf\implies{n} = 22+1$

$\boxed{\sf\implies{n} = 23}$

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‎Now put n=23 in equation (1.)

${ {\implies \sf \dfrac{ a + \dfrac{ (23-1)}{2} d }{ A + \dfrac{(23-1)}{2} D } = \dfrac{3(23) + 8}{7(23) + 15}} }$

${ {\implies \sf \dfrac{ a + \dfrac{ (22)}{2} d }{ A + \dfrac{(22)}{2} D } = \dfrac{69 + 8}{161 + 15}} }$

${ {\implies \sf \dfrac{ a + 11d }{ A + 11D } = \dfrac{69 + 8}{161 + 15}} }$

${ {\implies \sf \dfrac{ a12 }{ A12 } = \dfrac{77}{176}} }$

${ {\implies \sf \dfrac{ a12 }{ A12 } = \dfrac{7}{16}} }$

Hence ratio of 12th term of the given AP's is 7 : 16.

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