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Please help with my past year paper question, answer not given to us. The question tell us that it is aritmetic, however it says the three numbers are in geo progession. I do not understand the question. Kindly help to solve the quesiton. Thank you

The sum of three number in an arithmetic progession is $24$. If the first term is decreased by $1$ and the second term is decreased by $2$, the three numbers are in a geometric progression.

a) Let $a$ be the first term and $d$ be the common difference of the arithmetic progession. Use this information to write down the second and third terms in terms of $a$ and $d$.

b) Solve for the values of $a$ and $d$, by first formulating two equations.

c) List down the values of the three numbers.

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  • $\begingroup$ Consider three numbers, say $4,8,12$. Obviously, there are in arithmetic progression, where $d=4$. Now if the first number is decreased by $1$ and the second by $2$ then we get a geometric progression. That is, $3,6,12$ is a geometric progression with the common ratio of $2$. $\endgroup$
    – Math Lover
    Nov 12, 2017 at 14:36

3 Answers 3

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I think the best way to solve similar problems is to use following statements (prove their!).

Numbers $a$, $b$ and $c$ they are three successive members of the arithmetic progression iff $$2b=a+c.$$

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Non-zero numbers $a$, $b$ and $c$ they are three successive members of the geometric progression iff $$b^2=ac.$$

Now, solve the following system: $$a+b+c=24,$$ $$2b=a+c$$ and $$(b+2)^2=c(a-1).$$

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The arithmetic progression is $a, a+d, a+2d$. You are told that $$a+(a+d)+(a+2d)=24 \tag 1$$

The geometric progression is $a-1, a+d-2, a+2d$. Let $r$ be the common ratio of the geometric progression. Then $$r=\frac{a+d-2}{a-1}=\frac{a+2d}{a+d-2} \tag 2$$

Now solve the equations $(1)$ and $(2)$ for $a$ and $d$.

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  • $\begingroup$ Just a tip for enumerating equations. Use \tag 1 instead of \quad (1) etc. to number an equation. $\endgroup$
    – Math Lover
    Nov 12, 2017 at 14:53
  • $\begingroup$ @MathLover thanks $\endgroup$
    – A. Goodier
    Nov 12, 2017 at 14:55
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Three numbers in AP can be written as $a, a+d, a+2d$. Their sum is $3a+3d = 24$, from which you can deduce that $a+d = 8$ so the middle term is known. Since the common difference (difference between second and first term) is $8-a$, the third term can be rewritten as $8 + 8-a = 16-a$.

So the original AP is $a, 8, 16-a$ (you've managed to eliminate one unknown, the common difference, so that's progress).

Now use the second clue. $a-1, 6, 16-a$ are in GP.

That means that $6$ is the geometric mean of the first and third terms, i.e.

$6^2 = (a-1)(16-a)$

Rearranging, we get the quadratic: $a^2-17a+52 = 0 \implies (a-4)(a-13) = 0$

That gives $a=4$ or $a=13$, so the original AP can be either $4, 8, 12$ or $13, 8, 3$.

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