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In an increasing sequence of four positive integer, the first three terms are in AP, the last three terms are in GP and the fourth term exceed the first term by $30$, then the common difference of AP lying in the intervsl $[1,9]$ is:

MY ATTEMPT:

Let the series is $W,X,Y,Z$.

The first three terms are $a-d,a,a+d$ and the last three terms are $b/r$,$b$,$br$.

So we get $b=a+d$ and $$r=\frac{30 + a-d}{a+d}$$

Then I replaced $b$ and $r$ in terms of $a$ and $d$.

I obtained a quadratic equation. After this I could not do anything as I used all the data given.

lioness99a
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Priyanka Vithani
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1 Answers1

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Let $a,b,c,d$ be the four positive integers. The first three are in AP and the last three are in GP so we have $$b-a=x,c-b=x,\dfrac{d}{c}=\dfrac{c}{b}\rightarrow bd=c^2$$ furthermore The 4th is the 1st plus 30, that is $d=a+30$

Let substitute the last information into the system $$b-a=x;\;c-b=x;\;b(a+30)=c^2$$ we get $$c=b+x;\;a=b+x\rightarrow b(b+x+30)=(b+x)^2\rightarrow b^2+bx+30=b^2+2bx+x^2\rightarrow $$ $$b=\frac{x^2}{30-3x}$$ as the numbers are positive integers the only value for $x$ is $x=9$ and the four terms are $a= 18,\;b= 27,\;c= 36,\;d= 48$

Raffaele
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