Question

maths problem
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Answer 1

Solution:-

Let the first AP be in the form a1,a2,a3...with common difference d

Let the second eqaution be A1,A2,A3,.....with common difference D

sum of n terms of first AP=n/2{2a1+(n-1)d}

sum of n terms of second AP=n/2{2A1+(n-1)D}

when we take ratio n/2 will be cancelled out from both the terms

now according to question

{2a1+(n-1)d}/{2a2+(n-1)D}=(7n+1)/(4n+27)

this is the ratio of sum of n terms

If we have to take out the ratio of (2m+1)terms then simply we will put this value in place of n

so,

{2a1+(2m+1-1)d}/{2a2+(2m+1-1)D}={7(2m+1)+1}/{4(2m+1)+27}

=>{2(a1+md)}/{2(a2+mD)=(14m+8)/(8m+31)

cancelling out 2 we get

(a1+md)/(a2+mD)=(14m+8)/(8m+31)

a1+md denotes the (m+1)th term of AP

a2+mD denote th (m+1)th term of AP

so if we put m =16,,we will get (16+1)=17th term

a17/A17=(14×16+8)/(8×16+31)

a17/A17=232/159