If the sum of 'n' terms of two different AP are in

1.	If the sum of 'n' terms of two different AP are in ratio (2n+4) : (5n+5),then find the respective ratio of their 12th terms.
Answer» {n/2(2a1+(n-1)d1}/{n/2(2a2+ (n-1)d2} =(2n+4)/(5n+5) Where a1,a2 are first terms of two APs and d1 and d2 are common difference. We have {(2a1+(n-1)d1}/{2a2+(n-1)d2}=(2n+4)/(5n+5) Putting n=25 we have (A1)12/(A2)12 = 54/130=27/65 Given, (2n + 4) / (5n + 5) = { 2a1 + (n-1) d1 } / { 2a2 + (n-1) d2 } Putting n = 1, we get a2 = 5/3 a1 .....(1) Putting n = 2 and n = 3, we get, d1 = 2/5 d2 .....(2) 3/5 d2 = a1 ......(3) So, we have the relations b/w a1, a2, d1 and d2. Hence, the ratio comes as 5/12

If the sum of 'n' terms of two different AP are in ratio (2n+4) : (5n+5),then find the respective ratio of their 12th terms.

Answer» {n/2(2a1+(n-1)d1}/{n/2(2a2+ (n-1)d2} =(2n+4)/(5n+5)

Where a1,a2 are first terms of two APs and d1 and d2 are common difference.

We have

{(2a1+(n-1)d1}/{2a2+(n-1)d2}=(2n+4)/(5n+5)

Putting n=25 we have

(A1)12/(A2)12 = 54/130=27/65

Given, (2n + 4) / (5n + 5) = { 2a1 + (n-1) d1 } / { 2a2 + (n-1) d2 }

Putting n = 1, we get a2 = 5/3 a1 .....(1)

Putting n = 2 and n = 3, we get,

d1 = 2/5 d2 .....(2)

3/5 d2 = a1 ......(3)

So, we have the relations b/w a1, a2, d1 and d2. Hence, the ratio comes as 5/12

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