Question

The sum of n terms of two A.P.'s are in the ratio(5n+4):(9n+6).Find the ratio of their eighteenth term.

Answer 1

let the two AP's be a1, a2,a3,,,,,,,,,,,,andb1,b2,b3,,,,,,,,,,,,,,,,
Sna/Snb=(5n+4)/(9n+6)
by multiplying numerator and denominator bu n we get,
Sna/Snb=(5n^2+4n)/(9n^2+6n)
We know that,
an=Sn-S(n-1)
Then we get,
an/bn=[5n^2+4n-{5(n-1)^2+4(n-1)}]/[9n^2+6n-{9(n-1)^2+6n-1}]
an/bn=[5n^2+4n-{5n^2+5-10n+4n-4}]/[9n^2+6n-{9n^2+9-18n+6n-6}
after solving it we get,
an/bn=(-5+10n+4)/-9+18n+6)
an/bn=(10n-1)/18n-3)
Now by replacing n=18 we get
a18/b18=(10×18-1)/(18×18-3)
a18/b18=179/325
Hence ratio of their 18th term=179/325

Answer 2

I hope it's help you...