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27th sum
Asked by lovemaan5500 | 28 Jan, 2019, 06:35: AM
Let a and d are first term and common difference of first A.P.
Sum S1n of n terms :-  S1n = 2a+(n-1)d = (7n+1)k   ......................(1)
where k is constant.   (S1n /2 ) = a+[ (n-1)/2 ]d = [ (7n+1)k ]/2 ..............(2)

let us substitute (n-1)/2 = (m-1) in eqn.(2).  This substitution gives n = 2m-1

hence Eqn.(2) is written as:  (S1n /2 ) = a+(m-1)d = (7m-3)k ...............(3)
if we put m =11, eqn(3) becomes 11th term of A.P :  a+10d = 74k ......................(4)

Let b and e are first term and common difference of second A.P.
Sum S2n of n terms :-  S2n = 2b+(n-1)e = (4n+27)k   ......................(5)
where k is constant.   (S2n /2 ) = b+[ (n-1)/2 ]e = [ (4n+27)k ]/2 ..............(6)

let us substitute (n-1)/2 = (m-1) in eqn.(6).  This substitution gives n = 2m-1

hence Eqn.(6) is written as:  (S2n /2 ) = b+(m-1)e = (4m + 23/2 )k ...............(7)
if we put m =11, eqn(7) becomes 11th term of A.P :  b+10e = (101/2)k ......................(8)

From eqn.(4) and eqn.(8), we get the required ratio =
Answered by Thiyagarajan K | 28 Jan, 2019, 09:10: AM

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