# Sequences and Series - Arithmetic and Geometry

If the 3rd , 5th and 8th terms of an arithmetic progression with a common difference of 3 are three consecutive terms of a geometric progression , then what is common ratio? Help me with step-by-step.

Regards!

• This is an interesting problem, although initially it may seem like any ordinary high-school textbook AP/GP problem. The solution is $r=\frac 32=\frac{8-5}{5-3}$. It can also be shown that if the GP terms are the $p-n, p, p+m$-th terms of the AP, the solution is $r=\frac mn$. See my solution below. Apr 15 '16 at 23:51

$$\frac{a+2d}{a}=\frac{a+5d}{a+2d}\implies a^2+4ad+4d^2=a^2+5ad\implies4d^2=ad$$ Thus, $a=4d$ or $d=0$. Since we are given $d=3$, we must have $a=12$. Thus, the common ratio is $$\frac{a+2d}{a}=\frac{a+5d}{a+2d}=\frac32$$

You can try to express the 3rd, 5th, 8th terms as following:

$T(3)=a+2d$

$T(5)=a+4d$

$T(8)=a+7d$

where $d=3$

Then try to find out the ratio by dividing terms

If we write $a_n$ for the $n$th term of the arithmetic sequence and $g_n$ for the $n$th term of the geometric sequence, we have (we can assume $g$ starts with these terms, otherwise it's just a shift): $$g_1 = a_3 \quad , \quad g_2 = a_5 = a_3+2v=a_3+6 \quad , \quad g_3 = a_8 = a_3+5v=a_3+15$$ where $v=3$ is the common difference. But you also know that: $$g_2 = qg_1 \quad \mbox{and} \quad g_3 = q^2g_1$$ Can you take it from there?

Or more visual: $$g_1 \overset{\cdot q}{\longrightarrow} g_2 \overset{\cdot q}{\longrightarrow} g_3$$ With the info above, this becomes: $$a_3 \overset{\cdot q}{\longrightarrow} a_3 + 6 \overset{\cdot q}{\longrightarrow} a_3 + 15$$ which means that: $$q = \frac{a_3 + 15}{a_3 + 6} = \frac{a_3 + 6}{a_3}$$

A slightly non-conventional approach:

Assume, WLOG, that

• $d=\mu a$
• the AP terms are $\lbrace a-2\mu a, a, a+3\mu a\rbrace$
• the GP terms are $\bigg\lbrace \dfrac ar, a, ar\bigg\rbrace$.

Equating terms and dividing by $a$ gives

$$\begin{cases} \dfrac 1r=1-2\mu\\ r=1+3\mu\\ \end{cases}\\\\ \frac {3\mu}{2\mu}=\frac{r-1}{1-\frac 1r}\\ \qquad r=\frac 32\qquad\blacksquare$$

This means the GP terms (and also the 3rd, 5th and 8th terms of the AP) are: $\lbrace\frac 23 a, a, \frac 32 a\rbrace$. As it is given that $d(=\mu a)=3$, this means $a=18$, hence the terms are $\lbrace 12, 18, 27 \rbrace$. (This part is not required in the question.)

It can be seen from the method above that the common ratio of the GP is the ratio of the term-distances of the AP taken from the centre term i.e. replacing $2,3$ with $n,m$ respectively gives $r=\dfrac mn$ . However this is only valid for $\mu=\dfrac 1n-\dfrac 1m$.