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algebra
by rudrendra Kashyap - Wednesday, 27 August 2014, 11:01 PM
  y(1),y(2)......y(n) is a sequence of positive integers
n^2(y(n)-y(n-1))-n(y(n)+y(n-1))+6(y(n-1)-y(n))=0 where n>3
y(1)=y(2)=y(3)=3
then y(50) is
a. 2544
b. 424
c. 1272
d. 1296
Re: algebra
by TG Team - Thursday, 28 August 2014, 03:39 PM
  Hi Rudrendra smile

You can rearrange the given equation to find the ratio (yn/yn-1) = (n2 + n - 6)/(n2 - n - 6) = (n + 3)(n - 2)/(n - 3)(n + 2) = {(n + 3)/(n + 2)}{(n - 2)/(n - 3)}

So y50 = [{53/52}{48/47}][{52/51}{47/46}].....[{7/6}{2/1}]*y3 = {(53*48)/(6*1)}*3 = 1272. smile

Kamal Lohia