The sum of n terms of two ap are in ratio (5n+4) : (9n+6) Find the ratio of their 18th term?
plzz reply with solution
Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
Substituting n = 35 in (1), we obtain
From (2) and (3), we obtain
Thus, the ratio of 18th term of both the A.P.s is 179: 321.
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Hey!
this is the answer to why n=35?
Suppose we want to get a19 th terms ?
then a19=a+(19-1)=a+18d
now notice 18 .(n-1) multiply it with 2 . we get 36.
now ques. urself how to get T36th .
ans. You need to subtract 1 from 37 (substituting this value=37 in eqn.), then only u'll get a+18d form .,(since we take 2 common and what remains inside is a1+18d)
thanks! Hope it helps. any mistake pl reply me .!
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