Question

If the ratio of the sum of first n terms of two A.P.s is (7n+1):(4n+27), then find the ratio of their 9th terms.

Answer 1

Answer:

Step-by-step explanation:

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)

We can consider the 9th term as the m th term.

Let’s consider the ratio these two AP’s m th terms as am : a’m →(2)

Recall the nth term of AP formula, an = a + (n – 1)d

Hence equation (2) becomes,

am : a’m = a + (m – 1)d : a’ + (m – 1)d’

On multiplying by 2, we get

am : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]

= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]

= S2m – 1 : S’2m – 1 

= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]

= [14m – 7 +1] : [8m – 4 + 27]

= [14m – 6] : [8m + 23]

Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].

now substitute the value of m as 9

so the answer becomes

120/95

Answer 2

             

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. :-))

Read more on Brainly.in - https://brainly.in/question/8391638#readmore