Q5 Easy AP
An arithmetical progression all 2 positive terms. The ratio of the difference of the 4th and 8th term to the 15th term is and the square of the difference of the 4th and the 1st term is 225. Which term of the series is 2015?
This problem is a part of my set NMTC 2015
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1 solution
\(The first term is just a1 Let the 4th term be given by: a4 = a1 + d(4 -1) = a1 + 3d And the difference between the 4th and 1st terms is just 3d And the square of this difference = 225 So (3d)^2 = 225 → 9d^2 = 225 → d^2 = 225/9 → d = 15/3 so d =5 [since all the terms are positive]
And we're told that [ (a1 + d(7) ) - (a1 + d(3)) ] / [a1 + d (14) ] = 4/15 simplify
[d(7) - d(3)] /[ a1 + d(14)] = 4/15 and sustituting 5 for d we have [5(7) - 5(3)] /[ a1 + 5(14)] = 4/15 5*4 = [4/15] (a1 + (5)*14) multiply through by 15 5*4*15 = 4[a1 + 70] 300 = 4a1 + 280 20 = 4(a1) So a1 = 5
So 2015 = 5 + 5(n -1) 2010 = 5(n -1) 402 = n - 1 add 1 to both sides which gives 403 as answer\(\(LaTeX\)))