Sum of the series of numbers consisting of AP and GP both.

Question

Find the sum of all the terms, if the first $3$ terms among $4$ positive $2$ digit integers are in AP and the last $3$ terms are in GP. Moreover the difference between the first and last term is 40.

I have assumed four numbers to be $a,b,c\text{ & }d$ where $a,b\text{ & }c$ are in AP and $b,c\text{ & }d$ are in GP.
Let $b= x$, then $a=x-k$ and $c=x+k$ where $k$ is the common difference of the AP.
Also as per the question, $|a-d|=40$. This means that $d=a-40$ or $d=40+a$. One place that I am getting confused is here as to which relation should I take. How can we decide this?

Now if I take $d=40+a$ i.e. $d=40+x-k$, then I can say that ;
$(x+k)^2=x(40+x-k)$
$\Rightarrow x = \frac{k^2}{40-3k} \gt 0$

Now solving the above inequality, we get the range for $k$ to be $(-\infty,-8)\cup(5,\infty)$ but I don't know if this is helpful as I have to plug in the values of $k$ into the relation $x = \frac{k^2}{40-3k}$ and find out that value which can give a positive $2$ digit number.

Is my approach correct? Is there a better way to solve this? Also how to decide which relation to consider for $|a-d|=40$?
Please help !!!

Thanks in advance !!!

Answer 1

Let the first three terms be $x$, and $x+d$, and $x+2d$. Then the fourth term is $(x+2d)^2/(x+d)$. If we first assume $d$ is positive, then we have

$${(x+2d)^2\over x+d}-x=40$$

which reduces to

$$x={4d(d-10)\over40-3d}$$

Now the only positive integer values of $d$ that give positive integer values of $x$ are $d=12$ and $13$, which give $x=24$ and $156$, respectively. Of these, only $x=24$ is a two-digit number. So if $d$ is positive, then we see that the four numbers must be $24$, $36$, $48$, and $48^2/36=64$, the sum of which is

$$24+36+48+64=172$$

On the other hand, if we assume that $d$ is negative, then we have

$$x-{(x+2d)^2\over x+d}=40$$

which reduces to

$$x=-{4d(d+10)\over40+3d}$$

or, writing $D=40+3d$, so that $d=(D-40)/3$,

$$x={(40-D)(D-10)\over9D}$$

This expression gives an integer value of $x$ if and only if $D$ is a divisor of $400$ congruent to $1$ mod $3$. Since $400=2^4\cdot5^2$, this limits $D$ to $-200$, $-80$, $-50$, $-20$, $-8$, $-5$, $-2$, $1$, $4$, $10$, $16$, $25$, $40$, $100$, and $400$. From $D=40+3d\lt40$ (since $d$ is assumed to be negative), we can drop the last three possibilities, and from $x\gt0$ we can drop the possibilities $D=1$, $4$, and $10$. From $0\lt x+2d\lt100+2d$ we can drop $D=-200$, $-80$, and $-50$. This leaves six values to check: $D=-20$, $-8$, $-5$, $-2$, $16$, and $25$, which give $x=10$, $12$, $15$, $28$, $1$, and $1$ (again), respectively. But none of these is consistent with the requirement that $x\gt40$. So there are no solutions with $d$ negative.

Remark: The argument here ruling out a descending sequence (i.e., the case with $d$ negative) feels a bit clumsy, but I don't see any easy way to streamline it. Any comments (or alternate answer) improving it would be most welcome.

Answer 2

As an alternative, we could approach this using the properties of the numbers involved. The requirements for the sequence $ \ a_1 \ , \ a_2 \ , \ a_3 \ , \ a_4 \ = \ a_1 + 40 \ \ $ that they must be integers with $ \ 10 \ \le \ a_{k} \ \le \ 99 \ $ tells us that $ \ a_4 \ \ge \ 50 \ \ . $ For the ascending sequence, that the last three terms are in geometric progression means that $ \ a_3 \ $ is the geometric mean of the other members of the sub-series, that is, $ \ a_3 \ = \ \sqrt{a_2 · a_4} \ \ . $ This means that our sequence behaves schematically like this: $$ a_1 \underbrace{\quad \quad}_{d} a_2 \ = \ s·p^2 \underbrace{\quad \quad}_{d \ , \ r = \frac{q}{p}} a_3 \ = \ s·p·q \underbrace{\quad \quad}_{ r = \frac{q}{p}} a_4 \ = \ s·q^2 \ \ \ , $$ where $ \ p \ , \ q \ , \ s \ $ are positive integers and the ratio $ \ r \ = \ \frac{p}{q} > 1 \ $ being a rational number. This is actually fairly limiting, since the last three terms must all be multiples of the same integer (so $ \ s \ge 1 \ $ cannot be too large) and $ \ a_2 \ $ and $ \ a_4 \ $ must be multiples of perfect squares. The value for $ \ d \ $ will be set by the difference between $ \ a_2 \ $ and $ \ a_3 \ \ , $ and of course we will need to be sure that $ \ a_4 - a_1 \ = \ 40 \ \ . $

The candidates for $ \ a_4 \ $ above $ \ 50 \ $ are

$ 64 \ = \ 8^2 \ \ , \ \ 81 \ = \ 9^2 \ = \ 3^2·3^2 \ \ ; \ \ 50 \ = \ 2·5^2 \ \ , \ \ 72 \ = \ 2·6^2 \ = \ 2·2^2·3^2 \ \ , \ \ 98 \ = \ 2·7^2 \ \ ; $

$ 75 \ = \ 3·5^2 \ \ ; \ \ 64 \ = \ 4 · 4^2 \ \ , \ \ 80 \ = \ 5·4^2 \ \ , \ \ 96 \ = \ 6·4^2 \ \ ; $

and, naturally, we could include other appropriate multiples of $ \ 9 \ = \ 3^2 \ $ and $ \ 4 \ = \ 2^2 \ \ . $ But the requirement that the geometric mean for $ \ a_3 \ $ and any perfect-square multiple $ \ a_2 \ $ must be multiples of the same integer reduces our "options" for the geometric sub-series quite a bit:

$ 36 \ = \ 4·3^2 \ \ , \ \ 48 \ = \ 4·3·4 \ \ , \ \ 64 \ = \ 4·4^2 \ \ \ [ \ r \ = \ \frac43 \ \ , \ \ d \ = \ 12 \ ] \ \ ; \ \ \ \mathbf{[1]} $

$ 64 \ = \ 1·8^2 \ \ , \ \ 72 \ = \ 1·8·9 \ \ , \ \ 81 \ = \ 1·9^2 \ \ \ [ \ r \ = \ \frac98 \ \ , \ \ d \ = \ 8 \ ] \ \ ; \ \ \ \mathbf{[2]} $

$ 50 \ = \ 2·5^2 \ \ , \ \ 60 \ = \ 2·5·6 \ \ , \ \ 72 \ = \ 2·6^2 \ \ \ [ \ r \ = \ \frac65 \ \ , \ \ d \ = \ 10 \ ] \ \ ; \ \ \ \mathbf{[3]} $

$ 72 \ = \ 2·6^2 \ \ , \ \ 84 \ = \ 2·6·7 \ \ , \ \ 98 \ = \ 2·7^2 \ \ \ [ \ r \ = \ \frac76 \ \ , \ \ d \ = \ 12 \ ] \ \ ; \ \ \ \mathbf{[4]} $

We begin to see in this that we cannot permit $ \ d \ $ to become too large, as we must maintain $ \ a_1 \ \ge \ 10 \ \ $ and $ \ a_4 - a_1 \ = \ 40 \ \ $ ; so we could consider

$ 32 \ = \ 2·4^2 \ \ , \ \ 40 \ = \ 2·4·5 \ \ , \ \ 50 \ = \ 2·5^2 \ \ \ [ \ r \ = \ \frac54 \ \ , \ \ d \ = \ 8 \ ] \ \ ; \ \ \ \mathbf{[5]} $

but not

$ 18 \ = \ 2·3^2 \ \ , \ \ 30 \ = \ 2·3·5 \ \ , \ \ 50 \ = \ 2·5^2 \ \ \ [ \ d \ = \ 12 \ ] \ \ \ $ or $ 50 \ = \ 2·5^2 \ \ , \ \ 70 \ = \ 2·5·7 \ \ , \ \ 98 \ = \ 2·7^2 \ \ \ [ \ d \ = \ 20 \ ] \ \ ; $

similarly, we cannot have $ \ s \ $ become too large, or we will not be able to keep the difference $ \ a_4 - a_1 \ $ small enough, leading us to "reject", for instance,

$ 54 \ = \ 6·3^2 \ \ , \ \ 72 \ = \ 6·3·4 \ \ , \ \ 96 \ = \ 6·4^2 \ \ \ [ \ d \ = \ 18 \ ] \ \ . $

A search "by hand" does not take all that long under these restrictions (computational assistance could complete this rather quickly). The two other sub-series we can produce are

$ 45 \ = \ 5·3^2 \ \ , \ \ 60 \ = \ 5·3·4 \ \ , \ \ 80 \ = \ 5·4^2 \ \ \ [ \ r \ = \ \frac43 \ \ , \ \ d \ = \ 15 \ ] \ \ ; \ \ \ \mathbf{[6]} $

$ 48 \ = \ 3·4^2 \ \ , \ \ 60 \ = \ 3·4·5 \ \ , \ \ 75 \ = \ 3·5^2 \ \ \ [ \ r \ = \ \frac54 \ \ , \ \ d \ = \ 12 \ ] \ \ . \ \ \ \mathbf{[7]} $

When we complete these sequences by calculating the implied values for $ \ a_1 \ \ , $ however, we find only one that satisfies the specified conditions:

$ [1] \ : \ \ a_1 \ = \ 24 \ \ , \ \ a_4 - a_1 \ = \ \mathbf{40} \ \ ; \ \ [2] \ : \ \ a_1 \ = \ 56 \ \ , \ \ a_4 - a_1 \ = \ 25 \ \ ; $ $ [3] \ : \ \ a_1 \ = \ 40 \ \ , \ \ a_4 - a_1 \ = \ 32 \ \ ; \ \ [4] \ : \ \ a_1 \ = \ 60 \ \ , \ \ a_4 - a_1 \ = \ 38 \ \ $ (close); $ [5] \ : \ \ a_1 \ = \ 24 \ \ , \ \ a_4 - a_1 \ = \ 26 \ \ ; \ \ [6] \ : \ \ a_1 \ = \ 30 \ \ , \ \ a_4 - a_1 \ = \ 50 \ \ ; $ $ [7] \ : \ \ a_1 \ = \ 36 \ \ , \ \ a_4 - a_1 \ = \ 39 \ \ $ (closer!) .

So the unique increasing sequence is $ \ \mathbf{24 \ , \ 36 \ , \ 48 \ , \ 64} \ \ $ with sum $ \ 172 \ \ $ (as Barry Cipra has already shown).

For a descending sequence, the scheme is the same as above, with the behavior altered slightly; we have $ \ a_1 \ \ge \ 50 \ $ with $ \ a_1 - a_4 \ = \ 40 \ $ and $ \ r < 1 \ \ . $ This means, so to say, that the game is played the same way, but there is "less room" to play it in, since there will be fewer multiples of perfect squares as we “get closer” to $ \ 10 \ \ . $ The square-multiples for $ \ a_4 \ $ below $ \ 50 \ $ are

$ 16 \ = \ 4^2 \ = \ 2^2·2^2 \ \ , \ \ 25 \ = \ 5^2 \ \ , \ \ 36 \ = \ 6^2 \ = \ 2^2·3^2 \ \ , \ \ 49 \ = \ 7^2 \ \ ; $

$ 18 \ = \ 2·3^2 \ \ , \ \ 32 \ = \ 2·4^2 \ \ ; \ \ 12 \ = \ 3·2^2 \ \ , \ \ 27 \ = \ 3 · 3^2 \ \ , \ \ 48 \ = \ 3·4^2 \ \ ; $

$ 16 \ = \ 4·2^2 \ \ , \ \ 36 \ = \ 4·3^2 \ \ ; \ \ 20 \ = \ 5·2^2 \ \ , \ \ 45 \ = \ 5·3^2 \ \ $

and a few other multiples of $ \ 2^2 \ $ and $ \ 3^2 \ \ . $ While this appears to be more options for the "end" of our series, it is actually harder to construct suitable candidates:

$ 49 \ = \ 7^2 \ \ , \ \ 35 \ = \ 5·7 \ \ , \ \ 25 \ = \ 5^2 \ \ \ [ \ r \ = \ \frac57 \ \ , \ \ d \ = \ -14 \ ] \ \ ; \ \ \ \mathbf{[1']} $

$ 64 \ = \ 8^2 \ \ , \ \ 56 \ = \ 7·8 \ \ , \ \ 49 \ = \ 7^2 \ \ \ [ \ r \ = \ \frac78 \ \ , \ \ d \ = \ -8 \ ] \ \ ; \ \ \ \mathbf{[2']} $

$ 50 \ = \ 2·5^2 \ \ , \ \ 40 \ = \ 2·4·5 \ \ , \ \ 32 \ = \ 2·4^2 \ \ \ [ \ r \ = \ \frac45 \ \ , \ \ d \ = \ -10 \ ] \ \ ; \ \ \ \mathbf{[3']} $

$ 45 \ = \ 5·3^2 \ \ , \ \ 30 \ = \ 5·2·3 \ \ , \ \ 20 \ = \ 5·2^2 \ \ \ [ \ r \ = \ \frac23 \ \ , \ \ d \ = \ -15 \ ] \ \ . \ \ \ \mathbf{[4']} $

It is rather more difficult to produce sequences for which the difference $ \ a_1 - a_4 \ $ is not obviously too large or too small. For the sequences above, we find

$ [1'] \ : \ \ a_1 \ = \ 63 \ \ , \ \ a_1 - a_4 \ = \ 38 \ \ $ (close) ; $ \ \ [2'] \ : \ \ a_1 \ = \ 72 \ \ , \ \ a_1 - a_4 \ = \ 23 \ \ ; $ $ [3'] \ : \ \ a_1 \ = \ 60 \ \ , \ \ a_1 - a_4 \ = \ 28 \ \ ; \ \ [4'] \ : \ \ a_1 \ = \ 60 \ \ , \ \ a_1 - a_4 \ = \ \mathbf{40} \ \ $ (!) .

So we find there is also a unique descending sequence, $ \ \mathbf{60 \ , \ 45 \ , \ 30 \ , \ 20} \ \ $ with sum $ \ 155 \ \ . $

We observe from all of this that it is not especially challenging to construct these hybrid arithmetic-geometric sequences. The genuinely severe restriction for answering the question posed is in obtaining a specified difference between the first and last terms.

ADDENDUM [2/3] --

Using the method of the earlier posts (from June 2021), if we write the descending sequence as $ \ a_1 \ , \ a_2 \ = \ a_1 - d \ , \ a_3 \ = \ a_1 - 2d \ , \ a_4 \ = \ a_1 - 40 \ \ , \ \ d \ > \ 0 \ \ , $ we are led to $$ \frac{a_1 \ - \ 2d}{a_1 \ - \ 40} \ \ = \ \ \frac{a_1 \ - \ d}{a_1 \ - \ 2d} \ \ \Rightarrow \ \ 40·d^2 \ - \ 40·d \ \ = \ \ -40·a_1 \ + \ 3·a_1·d $$ $$ \Rightarrow \ \ a_1 \ = \ \frac{4·d·(d \ - \ 10)}{3·d \ - \ 40} \ \ . $$

We require $ \ a_1 \ \ge \ 50 \ \ $ (since $ \ a_4 \ \ge \ 10 \ ) \ $ and this can met by the "branch" of the function curve for $ \ d \ > \ \frac{40}{3} \ \ . $ There is a minimum at $ \ d \ = \ 20 \ \ , \ \ a_1 \ = \ 40 \ \ $ and there are two possible values for $ \ a_1 \ $ for other values of $ \ d \ \ . $ However, the portion of the branch for $ \ d \ \ge \ 20 \ \ $ does not yield admissible values for $ \ a_4 \ \ , $ so we must look at $ \ d \ < \ 20 \ \ . $ There, we only find integer values for $ \ a_1 \ $ with

$ \mathbf{d \ = \ 16 \ \ : } \ \ a_1 \ = \ 48 \ \ , \ \ $ series -- $ \ 48 \ , \ 32 \ , \ 16 \ , \ 8 \ \ $ [inadmissible] ;

$ \mathbf{d \ = \ 15 \ \ : } \ \ a_1 \ = \ 60 \ \ , \ \ $ series -- $ \ \mathbf{60 \ , \ 45 \ , \ 30 \ , \ 20} \ \ ; $

$ \mathbf{d \ = \ 14 \ \ : } \ \ a_1 \ = \ 112 \ \ , \ \ $ series -- $ \ 112 \ , \ 98 \ , \ 84 \ , \ 72 \ \ $ [inadmissible] .

So the descending series can be found by this technique.