How can I find the first term of this geometric progression?

Question

All the terms in this geometric progression are positives. Given the fifth term is four times that of the third term and the second term is $\frac 18$.

So, we have $T_2 = \frac18$

And let say that $T_3 = x$ so $T_5 = 4x$

So the common ratio would be,

$r = \dfrac {T_3}{T_2}$

$r = 8x$

But why is the common ration bigger than the fifth term? I mean $T_4$ would be $8x^2$ if I use this.

I'm confused and can't go any further. Can anyone help point out any mistakes here? Thanks.

Answer 1

The first terms of a geometric progression with first term $a$ and common ratio $r$ are $T_1=a, T_2=ar, T_3=ar^2, T_4=ar^3, T_5=ar^4$

You are given three pieces of information. First that all the terms are positive, which means (check you can show this) that both $a$ and $r$ are positive.

Next you are told $T_5=4T_3$ which becomes $ar^4=4ar^2$

Finally you are told the piece of information you jumped to first, without properly considering all the other information, that $T_2=\frac 18$ or $ar=\frac 18$

You need all three pieces of information to solve the problem, and rather than speculating on the nature of the solution before you have completed the computation, it is best to work systematically through what you have been told. Then you can check at the end whether your solution does in fact meet all the constraints in the problem.

Answer 2

It means $T_4=8x^2\Rightarrow T_5=64x^3=4x\Rightarrow x=\frac{1}{4}$ and $r=8x=2$> Therefore $T_1=T_2/r=\frac{1}{16}$

Answer 3

Let Terms of Geometric Progression be $a,ar,ar^2,ar^3,ar^4,......\;,$ Where $a$ is first term

and $r$ is common ratio.

Now Given $ar^4 = 4\cdot ar^2\Rightarrow r^2=4\Rightarrow r=\pm 2$

and $\displaystyle ar = \frac{1}{8}$

So we get $\displaystyle a = \pm \frac{1}{16}$

But above Given all terms are positive.

So we get $\displaystyle a= \frac{1}{16}$ and $r=2$

Answer 4

Third term: $x$

fifth term: $4x$

second term: $\frac{1}{8}$

$$ ar^4 = 4x$$

$$ ar^2 = x$$

This implies $r^2 = 4$, $r = \pm2$

Assuming $r>0$ , the series looks like the following:

$\frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1$

Answer 5

Since $T_5=rT_4=r^2T_3=r^2\cdot \frac{T_5}{4}$, we have $r^2=4$, i.e. $r=2$. ($r\not=-2$ because all terms have to be positive)

So, $T_1=\frac{T_2}{r}=\frac{1}{16}$.

why is the common ration bigger than the fifth term?

Indeed, we have $2=r\gt T_5=1$, but I don't think there is something wrong. (You might have missed that terms can be less than $1$.)

Answer 6

You're overthinking the problem. Let $T_1 = a$ (first term) and the common ratio be $r$.

You're given:

$T_5 = 4T_3 \implies ar^4 = 4ar^2 \implies r = 2$.

Take only the positive value as you're told all terms are positive.

Also given, $T_2 = \frac 18 \implies ar = \frac 18 \implies a = \frac {1}{16}$.

So the sequence to $5$ terms is:

$\frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1$

Answer 7

Let the first term be $a$ & common ratio be $r$ then we have $$T_2=ar=\frac{1}{8}\tag 1$$ $$\frac{T_5}{T_3}=4=\frac{ar^4}{ar^2}=r^2$$ $$r=2$$ Substituting $r=2$ in (1), we get $$2a=\frac{1}{8}\iff a=\frac{1}{16}$$

Hence, $$T_5=\frac{1}{16}\left(2\right)^4=1<r$$ Fifth term is smaller than the common ratio $r=2$ as the term $T_1=\frac{1}{16}$ is much smaller. G.P. can be expressed as $$\frac{1}{16}, \frac{1}{8}, \frac{1}{4}, \frac{1}{2}, 1, 2, 4\ldots $$

Answer 8

Part of solving math problems is knowing when to stop and try something else. The other part is knowing that you did everything right and, even if things look weird, going on and seeing what happens. You did this much

$T_2 = \frac 18$
$T_3 = x$
$T_5 = 4x$

$r = \dfrac{T_3}{T_2} = 8x$

So the first term must be
$T_1 = \dfrac{T_2}{r} = \dfrac{1}{64x}$. Also
$T_4 = T_3 \cdot r =8x^2$
$T_5 = T_4 \cdot r = 64x^3$

So we must have $64x^3 = 4x$
$16x^2 = 1$
$x = \frac 14$

So
$r = 2$
$T_1 = \frac{1}{16}$
$T_2 = \frac 18$
$T_3 = \frac 14$
$T_4 = \frac 12$
$T_5 = 1$