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The exercise reads as follows:

The sum of the first 5 terms in a geometric progression is 62. The 5th, 8th and 11th term of this geometric sequence are also the 1st, 2nd and 10th term of an arithmetic sequence. Find the first number of the geometric sequence.

This was a rough translation so my attempt at explaining the exercise-question better is:

GP $a,b,c,d,e,f,g,h,i,j,k$

$a+b+c+d+e=62$

AP $e,h,x_3,x_4,x_5,x_6,x_7,x_8,x_9,k$

Find the number $a$.

GP is geometric progression, AP is arithmetic progression. $x_3,x_4,x_5,x_6,x_7,x_8,x_9$ are the other terms of the arithmetic progression which we know nothing of (and they probably do not matter, not sure).

Any hints? I thought about using the ratio from the geometric progression in two different ways: either starting from $e$ (therefore $d=e/r$ and $h=e*r^3$) or starting from $a$ (and we use $e=a*r^4$). The information about the sum is probably valuable but I might not know how to use it properly.

Thanks to the suggestions, I've made it a bit further:

$a=a$
$b=a*q$
$c=a*q^2$
$d=a*q^3$
$e=a*q^4$
$h=a*q^7$
$k=a*q^{10}$

I've wrote the important terms using $a$ and the geometric progression ratio. We have the following equations:

$a*(1+q+q^2+q^3+q^4)=62$ (sum)
$a*q^7=a*q^4+r$ (h)
$a*q^{10}=a*q^4+9r$ (k)

$q$ is the ratio from the geometric progression, $r$ is the ratio from the arithmetic progression

EDIT: I've managed to find some answers with the help of Ross. The arithmetic progression has the ratio equal to 224, the geometric progression has the ratio 2. The first term is equal to 2. Sadly the book has no answers and I can't double-check, they were discovered by giving values and seeing if it works, assuming all the numbers are integers (which they probably are but the exercise doesn't specify).

If anyone is still willing to help with a solution that doesn't require "trial and error" so to say, feel free.

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  • $\begingroup$ It's likely easier to use $ a, ar, ar^2, \ldots $ and $ b, b+d, b+2d, \ldots$ for the progressions. Can you write down what equations you have from that? $\endgroup$
    – Calvin Lin
    Jan 7, 2021 at 14:31
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    $\begingroup$ I think writing terms of geometric progression in the form $aq^n$ would make it easier to solve the problem. $\endgroup$
    – Etemon
    Jan 7, 2021 at 14:32
  • $\begingroup$ thanks for the help :) $\endgroup$
    – questionasker22
    Jan 7, 2021 at 14:33
  • $\begingroup$ I have edited the post with the "new" equations $\endgroup$
    – questionasker22
    Jan 7, 2021 at 14:44
  • $\begingroup$ You have three equations in three unknowns. If you assume the variables are integers, there are few choices for factors of $62$, so just try them $\endgroup$
    – Ross Millikan
    Jan 7, 2021 at 14:46

2 Answers 2

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We have sum of the first $5$ terms in a geometric progression. sum of first $n$ terms can be calculated by $S_n=\frac{a(1-q^n)}{1-q}$. for $n=5$:

$$S_5=\frac{a(1-q^5)}{1-q}=62\tag1$$

The $5$th, $8$th and $11$th term of this geometric sequence are also the $1$st, $2$nd and $10$th term of an arithmetic sequence:$$\underbrace{{aq^4}}_{m},\underbrace{aq^7}_{m+d},\cdots,\underbrace{aq^{10}}_{m+9d}$$

By extracting $d$ we get:

$$d=a(q^7-q^4)=\frac{a(q^{10}-q^{7})}{8}$$ $$8(q^7-q^4)=q^3(q^7-q^4)$$

Therefor $q=2$. substitute this in the equation $(1)$ to find $a$.

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    $\begingroup$ Thank you! To be fair I hadn't thought of finding the ratio of the geometric progression by using the ratio from the arithmetic progression. Nice move $\endgroup$
    – questionasker22
    Jan 7, 2021 at 15:08
  • $\begingroup$ You're welcome. $\endgroup$
    – Etemon
    Jan 7, 2021 at 15:08
  • $\begingroup$ I'm too slow again +1. $\endgroup$
    – Oscar Lanzi
    Jan 7, 2021 at 15:17
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In the arithmetic progression you need one step to go from $e=aq^4$ to $h=aq^7$, but then eight steps to go from $h$ to $k=aq^{10}$. So we must have

$aq^7-aq^4=d,aq^{10}-aq^7=8d$

$\color{blue}{8}d=aq^{10}-aq^7=q^3(aq^7-aq^4)=\color{blue}{q^3}d$

Can you find $q$ (and verify that there is only one real root for that variable) and continue?

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  • $\begingroup$ $q^3=8 => q=2$ is what I'd say, if q were negative then the cube would be negative. Odd powers are negative if the number is negative, whereas even powers are always positive. $\endgroup$
    – questionasker22
    Jan 7, 2021 at 15:13
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    $\begingroup$ Yup, that is correct. Go on and solve the problem. $\endgroup$
    – Oscar Lanzi
    Jan 7, 2021 at 15:15
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    $\begingroup$ don't worry mate, the more people that have the same answer the better. You both had similar ways of solving so that eliminates possible confusion and helps me understand the exercise better $\endgroup$
    – questionasker22
    Jan 7, 2021 at 15:18

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