1. Sequence

The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric.P.Find the common ratio of G.P and obtain an expression for nth term of the G.P. (question)
my first step is to find the common ratio, and i get r=2.....but i cant obtain others constant to find the expression...anyone know how to find that?

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2. Re: Sequence

Originally Posted by goldsomecriss
my first step is to find the common ratio, and i get r=2
HOW did you get that?

3. Re: Sequence

Set the starting point in the arithmetic sequence asa2=a1+k

Then we have:

a2=a1+k=br0 ..................Equation(1)
a6=a1+5k=br1 ................Equation(2)
a8=a1+7k=br2 ................Equation(3)

Eqiation(2)−Equation(1) → 4k=br−b

→ 4k=b(r−1)..Eqn(4)

Equation(3)−Equation(2) → 2k=br2−br

→ 2k=br(r−1)..Eqn(5)

Eqn(5)÷Eqn(4) → 4k2k=br(r−1)b(r−1)

→ 2.= r

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4. Re: Sequence

Originally Posted by DenisB
HOW did you get that?
do u know the next step?

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5. Re: Sequence

Arithmetic: 1st term = 1, common difference = 0
1,1,1,1,1,1,1,1,1,.........

Geometric: 1st term = 1, common ratio = 1
1,1,1,1,1,1,1,1,1,.........

YES, I'm lazy

6. Re: Sequence

Originally Posted by DenisB
Arithmetic: 1st term = 1, common difference = 0
1,1,1,1,1,1,1,1,1,.........

Geometric: 1st term = 1, common ratio = 1
1,1,1,1,1,1,1,1,1,.........

YES, I'm lazy
Lol, all zeros would work, as well. But, the common ratio may be any real number.

7. Re: Sequence

The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric.P.Find the common ratio of G.P and obtain an expression for nth term of the G.P. (question)
$a_2 = a_1 + d = b_1$

$a_6 = a_1 + 5d = b_1 \cdot r$

$a_8 = a_1 + 7d = b_1 \cdot r^2$

from the first two equations, $d = \dfrac{b_1(r-1)}{4}$

from the first and third equation $d = \dfrac{b_1(r^2-1)}{6}$

$\dfrac{b_1(r-1)}{4} = \dfrac{b_1(r-1)(r+1)}{6}$

for $b_1 \ne 0$ and $r \ne 1$ ...

$\dfrac{1}{4} = \dfrac{r+1}{6} \implies r = \dfrac{1}{2}$

$b_n = b_1 \cdot \left(\dfrac{1}{2}\right)^{n-1}$