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I came across a question today

Given a sequence of four numbers such that the first three are in G.P. and the last three are in A.P. with common difference $6$. If the first and the fourth number are equal, then common ratio of the G.P. is?


Now correct way of doing this is given as:

Let the last three numbers be $a, a+6, a+12$, so that the first three numbers are $a+12, a, a+6$. As these are in G.P. $$a^2 = (a+12)(a+6) \Rightarrow a=-4$$ Common ratio is $-2$


But how I did this is

I took first term of G.P. as $a_1$ and the first term of A.P. as $a_2$. So the series is $$a_1, (a_1r) \text{ or }(a_2), (a_1r^2) \text{ or } (a_2+6), a_2 +12$$

So, $a_1 = a_2+12$ and $r= \dfrac{a_2+6}{a_2}$ and $a_1r^2 - a_1r=6$

Solving these three equations I get Common ratio (r) = $-0.5$


What is wrong in my solution? I know that my solution is bit more time consuming but why is it wrong?

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3 Answers 3

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Let the four terms be: $$\begin{align} &a\\ &ar=&&a-12\\ &ar^2=&&a-6\\ & &&a\end{align}$$ From the above, we have $$6=a(1-r^2)\\ 12=a(1-r)$$ Dividing (allowed as $r\neq 1$) and solving gives $$r=-\frac12\quad\blacksquare$$


Solving gives $a=8$, hence the four numbers are $$\lbrace 8,-4,\;2,\;8\rbrace$$

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The common ratio is indeed $-\frac{1}{2}$. You can check that the answer $a=-4$ of the first solution produces common ratio $-\frac{1}{2}$, and not $-2$.

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<>let the last three numbers of the set which are in A.P be b,b+6,b+12 and the first number be a.

→thus ,the four numbers are a,b,b+6,b+12

given,a=b+12-----(1)

also, given a,b,b+6 are in G.P

by equation (1)

→b+12,b,b+6 are in G.P

→b/(b+12)=(b+6)/b

→b²=(b+6)(b+12)

→b²=b²+18b+72

→18b=-72

→b=-4

→a=-4+12=8

hence the four numbers are 8,-4,2,8

I hope this will help u ;)

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