Three different numbers, whose product is 125, are 3 consecutive terms in a geometric sequence. At the same time they are the first, third and sixth terms of an arithmetic sequence. Find the numbers.

Question

anonymous · Answer

$$
xyz=125\
xn^2=yn=z\
x+5m=y+3m=z
$$Right?

apoorvk · Answer

okay, let the middle term be 'a', and the common ratio be 'r'.
then, the 3 terms would be = a/r , a, ar

so,

$$\frac a {\cancel{r}} \times a \times a \cancel{r} = 125$$
so, a= 5

apoorvk · Answer

let 'p' being the first term of the AP series
now, this 'a' is the third term of an AP = 5 = p + 2d   ...{i} 
first term = p = a/r

or, p = 5/r  ...{ii}

sixth term of the AP = ar = 5r = p +5d ...{iii}


so, 3 equations, and 3 variables 'p', 'd' and 'r' - justified. can you solve them?

apoorvk · Answer

'd' is the common difference in the AP series.

anonymous · Answer

It makes sense if r = 5, and the Geometric sequence would be 1,5,25...

But the arithmetic sequence would have first term as 1, third term as 5, and sixth term as 25. This doesn't make sense. Since first term is 1 and third is 5, common difference must be 2. Then the sixth term must be 11 not 25. How does that work then?

apoorvk · Answer

Why are you trying hit 'n' trial - 'r' can be any number, not necessarily the '5' you assumed - think it out.. And what you're trying doesn't always yield right results.

just solve these 3 equations, and you'll get exact values:
 5 = p + 2d   ...{i}
p = 5/r  ...{ii}
5r = p +5d  ...{iii}

anonymous · Answer

i don't seem to be able to solve these equations =/

apoorvk · Answer

From 'i',
p = 5 -2d ...{iv}

Substituting this 'p' in 'ii',
5-2d = 5/r

              5
or, r = ------    ...(v)
           5 - 2d

Substitute values of 'p' from 'iv' and and 'r' from 'v' in equation 'iii' in the prev. post, and get an equation in 'd'. can you solve that?

anonymous · Answer

If I am right, then according to you apoor, d = 5

anonymous · Answer

Let the  three numbers are: $ a, ar, ar^2$

From the given condition,

$a = a$

$ar = a + 2d$

$ar^2=a+5d$

and,

$a(ar)(ar^2) = 125 \implies a =\frac 5 r \tag{1}$

Now,

$ar = a + 2d \implies  a = \frac{2d}{ (r - 1)}$

$ ar^2 = a + 5d \implies  a = \frac{5d }{ (r^2 - 1)} $

So,
$\frac{2d}{ (r - 1)} = \frac{5d }{ (r - 1)(r+1)} \implies  2dr=3d\implies r = \frac 3 2 $


substituting in $(1)$ you will get $a$ and then ...