Sequences and series(Arithmetic and Geometric progression)

Question

Anyone can help me solve this question？

The first three of four integers are in an a.p. and the last three are in g.p. Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.

Thanks!

Answer 1

Let the second digit be $n$, and the common difference of the AP be $d$, the common ratio of the GP be $r$:

$(n-d), n, nr, nr^2$

So as the middle two integers are also in AP we have that

$d=nr-n=n(r-1)$

so we can eliminate $d$ and have the four integers as

$(n-n(r-1)), n, nr, nr^2$

or more tidily

$n(2-r), n, nr, nr^2$

Now we can make a few equations based on the sums of the first and last etc that you state:

$n(2-r)+nr^2=37$

factorising out the $n$ gives

$n(r^2-r+2)=37$ ............... Equation 1

and the middle number sum:

$n+nr=36$

which tidies to

$n(1+r)=36$ ............... Equation 2

combining equations 1 and 2 to eliminate $n$ gives:

$\frac{36}{1+r}=\frac{37}{r^2-r+2}$

which, I assure you, tidies to

$36r^2-73r+35=0$

solving the quadratic gives $r=\frac{5}{4}$. The other value it gives won't give integer solutions (try it yourself and see). Putting this value of $r$ into equation 2 gives us $n=16$. So we have

12, 16, 20, 25

Answer 2

Call the three numbers $a,b,c,d$. Then you know that $a+c=2b$ and $bd=c^2$. Next you know that $a+d=37$ and $b+c=36$.

Leave aside the degree two equation. The linear ones can be solved for $b$ by $d=37-a$, $c=36-b$ and so $$ a-2b+36-b=0 $$ that yields $a=3b-36$ and therefore $d=37-3b+36=73-3b$. Finally $$ b(73-3b)=(36-b)^2 $$ Can you finish?