Question

Please reply , I'm stuck in this question no. 27​

Answer 1

Answer: 120:95

__________________________________________________

Nice question.

 

A SIMPLE METHOD!

   

We know that,

 

$\bold{S_n=n \times [\frac{n+1}{2}]^{th}\ term}$

 

   

Let sum of n terms of first AP be $\bold{n \times [a_{\frac{n+1}{2}}]}$.

 

Let sum of n terms of second AP be $\bold{n \times [b_{\frac{n+1}{2}}]}$

   

$\bold{n \times [a_{\frac{n+1}{2}}]}:\bold{n \times [b_{\frac{n+1}{2}}]}\ \bold{=(7n+1)(4n+27)} \\ \\ \bold{a_{\frac{n+1}{2}}:b_{\frac{n+1}{2}}=(7n+1):(4n+27)}$

 

This means the ratio of sum of n terms of the two APs is equal to the ratio of the average of n terms.

 

So let n = 17.

 

$\bold{17 \times [a_{\frac{17+1}{2}}]}:\bold{17 \times [b_{\frac{17+1}{2}}]}\ \bold{=(7 \times 17+1)(4 \times 17+27)} \\ \\ \bold{a_{\frac{17+1}{2}}:b_{\frac{17+1}{2}}=(119+1):(68+27)} \\ \\ \bold{a_{\frac{18}{2}}:b_{\frac{18}{2}}=120:95} \\ \\ \bold{a_9:b_9=120:95}$

 

So the answer is 120:95.

 

Please ask me if you have ANY doubts on my answer. The answer is on my own and not from any sources.

 

Hope this helps you.

 

Please mark it as the brainliest if it helps.

 

Thank you. :-))

Answer 2

S

n

=n×[

2

n+1

]

th

term

Let sum of n terms of first AP be \bold{n \times [a_{\frac{n+1}{2}}]}n×[a

2

n+1

] .

Let sum of n terms of second AP be \bold{n \times [b_{\frac{n+1}{2}}]}n×[b

2

n+1

]

\begin{lgathered}\bold{n \times [a_{\frac{n+1}{2}}]}:\bold{n \times [b_{\frac{n+1}{2}}]}\ \bold{=(7n+1)(4n+27)} \\ \\ \bold{a_{\frac{n+1}{2}}:b_{\frac{n+1}{2}}=(7n+1):(4n+27)}\end{lgathered}

n×[a

2

n+1

]:n×[b

2

n+1

] =(7n+1)(4n+27)

a

2

n+1

:b

2

n+1

=(7n+1):(4n+27)

This means the ratio of sum of n terms of the two APs is equal to the ratio of the average of n terms.

So let n = 17.

\begin{lgathered}\bold{17 \times [a_{\frac{17+1}{2}}]}:\bold{17 \times [b_{\frac{17+1}{2}}]}\ \bold{=(7 \times 17+1)(4 \times 17+27)} \\ \\ \bold{a_{\frac{17+1}{2}}:b_{\frac{17+1}{2}}=(119+1):(68+27)} \\ \\ \bold{a_{\frac{18}{2}}:b_{\frac{18}{2}}=120:95} \\ \\ \bold{a_9:b_9=120:95}\end{lgathered}

17×[a

2

17+1

]:17×[b

2

17+1

] =(7×17+1)(4×17+27)

a

2

17+1

:b

2

17+1

=(119+1):(68+27)

a

2

18

:b

2

18

=120:95

a

9

:b

9

=120:95

So the answer is 120:95.