Problematic Progressions

Algebra Level 4

Two A.P's have the same number of terms. The ratio of the last term of the first progression to the first term of the second progressions equals the ratio of the last term of the second progression to the first term of the first progression both of which are numerically equal to 4 4 . The ratio of the sum of the first k k terms of the first to the second progression is equal to 2 2 . Let α \alpha be the ratio of the common difference of the first and the second progressions. Let λ \lambda be the ratio between their m m th terms. Compute α + 2 λ \alpha + 2\lambda


The answer is 33.

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1 solution

Let 1st series be 2/7,4 Let second be 1, 8/7 alpha=26 and lambda= 7/2 So ans is 33 (Sorry for bad formatting)

You haven't explained how you arrived at this.

Krishna Ar - 6 years, 1 month ago

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Let no. of terms be 2 (for the sake of simplicity)

let A.P'S be a1,a2 and b1,b2 respectively

let a1 = p and a2= q It's given that a2/b1 = b2/a1 = 4

Therefore, new series becomes p,4q and q,4p

If we take k=2 & q=1, we get p+4/1+4p = 2 i.e. p= \boxed {2/7}

So final series is 2/7, 4 & 1,8/ 7

So \alpha = 26 and \lambda = 7/2

Finally \alpha + 2 \lambda = 33 :-)

Harvindersingh Chavan - 6 years, 1 month ago

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Too many assumptions LOL. This is not a proper solution though.

Krishna Ar - 6 years, 1 month ago

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@Krishna Ar I know, but i atleast for the answer though. After preparing 2 years for JEE i got habitual of this type of solutions...

And this is MCQ not theory question :-p

Harvindersingh Chavan - 6 years, 1 month ago

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@Harvindersingh Chavan True. Exam prep has longstanding effects on thought

Krishna Ar - 6 years, 1 month ago

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