Geometric to Arithmetic Progression

The elements of our sequence are, as defined:

1. $12$

2. $12r = \frac{45}{2} - 2d$

3. $12r^{2} = \frac{45}{2} - d$

4. $\frac{45}{2}$

Now we can solve for $d$ in the third element:

$12r^{2} = \frac{45}{2} - d \Leftrightarrow d = \frac{45}{2} - 12r^{2}$

Now we plug $d$ back into the second element:

$2d = 45 - 24r^2 \Leftrightarrow 12r = \frac{45}{2} - 45 + 24r^2 \Leftrightarrow 24r^2 - 12r - \frac{45}{2} = 0$

And then we solve for $r$ :

$r = \frac{12 \pm \sqrt{2304}}{48} \Leftrightarrow r = -\frac{3}{4} \wedge \frac{5}{4}$

We are looking for a positive $r$ because $12$ is smaller than $\frac{45}{2}$ , thus $r = \frac{5}{4}$ . Lastly, we can evaluate elements two and three, sum them and add the numerator and denominator:

$12 \times \frac{5}{4} + 12 \times (\frac{5}{4})^2 = \frac{135}{4}$

$135 + 4 = 139$

let the numbers are 12 b c $\frac{45}{2}$

$12c = b^{2}$

$b + \frac{45}{2} = \frac{b^{2}}{6}$ i.e

$b^{2} - 6b - 135 = 0$ which gives b = 15

ans so c = $\frac{75}{4}$

We call the sequence : 12,x,y,45/2 We create these equations : 12 . d = a; 12 . d^2 = b; (b-a) = 12d(d-1) ; 12d(d-1) + 12d^2 =45/2 ( d is a real number) The first 2 equations are based on the thesis as the first three are in geometrix progression. The next 2 are based on arithmetic progression. Solve these equation, we can find "d" and find x,y, then a,b.

the sequence can be written 12, x, y, $\frac{45}{2}$

the first three are in geometric progression so

$\frac{x}{12}$ = $\frac{y}{x}$

$x^{2}$ = 12y

y = $\frac{x^{2}}{12}$

the last three are in arithmetic progression so x + $\frac{45}{2}$ = 2y

x + $\frac{45}{2}$ = 2 $\frac{x^{2}}{12}$

6x + 135 = $x^{2}$

0 = $x^{2}$ -6x - 135

0 = (x-15) (x+9)

x1 = 15 . x2 = -9

y1 = 15. $\frac{15}{12}$ ........ y2 = -9 . $\frac{-9}{12}$

y1 = $\frac{75}{4}$ ....... y2 = $\frac{27}{4}$

x1+y1 = $\frac{135}{4}$ ............ x2+y2 = $\frac{-9}{4}$

x2+y2 is negative . so we use x1+y2 = $\frac{135}{4}$

a= 135 .. b = 4

a+b = 139

Sequence 12, a, b, 22,5

b/a=a/12 22,5-b=b-a

a^2=12b 2b=a+22,5

12b=6a+135

a^2-6a-135=0

delta = 36-4.1.(+135) delta = 36+540 delta = 576 sqrt delta = 24

a= 3 + 12 a=15

a+b=33,75

33,75 = 135/4

135+4=139

(I'm a foreigner from Vietnam so maybe i'm not good at describing) we sort 4 numbers as: 12,v,t,45/2. then we have: t-v=(45/2)-t so t=(45/4)+(v/2). the 4 numbers will be 12,v,(45/4+v/2),45/2. then we have v/12=(45/4+v/2)/v so v=15 or -9( discarded as it's positive ). then we find out t=18.75 . 15+18.75=33.75=135/4 so the answer is 139

Let $f$ and $g$ represent the two middle numbers, respectively in order. Let $k$ represent the constant ratio between the first three numbers, and let $x$ represent the common difference between the last three numbers. Then $f = 12k$ and $g = 12k^2$ . Additionally, $f + x = g$ , and $g + x = \frac{45}{2}$ . Through a system of equations, $f + \frac{45}{2} = 2g$ . Substituting $12k$ for $f$ , and $12k^2$ for $g$ , we now have a quadratic, namely $16k^2-8k-15=0$ . This can be factored into $(4k+3)(4k-5) = 0$ , and thus $k = -\frac{3}{4}, \frac{5}{4}$ . Now the sum of the two middle numbers $f + g = 12k^2 + 12k = 12k(k + 1)$ , and since $\frac{a}{b}$ must be positive, we substitute for $k = \frac{5}{4}$ . This gives us $f + g = 15 \times \frac{9}{4} = \frac{135}{4}$ . And $a + b = 135 + 4 = 139$

knowing that the first three in the sequence are geometric and the last three are arithmetic.. the first number is 12 and the last is $\frac{45}{2}$ , we can conclude that the sequence is

12, 12x, $12x^{2}$ , $\frac{45}{2}$ where x is the common ratio

and knowing the fact that the last three are in arithmetic progression, we can conclude that

$\frac{45}{2}$ - $\frac{45}{2}$ = $\frac{45}{2}$ - 12x lets equate this to zero;

$24x^{2}$ - 12x - $\frac{45}{2}$ = 0 using the quadratic formula, we can get the roots which are $\frac{45}{2}$ and - $\frac{3}{4}$ but the progression is ascending so we will get the positive value of x.. thus,

12, 12 $\frac{5}{4}$ , 12 $\frac{5}{4}^{2}$ , $\frac{45}{2}$

the middle terms will be 15 and $\frac{75}{4}$

add them and get $\frac{135}{4}$ thus a + b = 135 + 4 = 139

Since the first three are in geometric progression we can write b/a = c/b and so b^2 = a * c. Also, since the three last are in arithmetic progression we can write c - b = d - c. We know that a = 12 and d = 45/2. Plugging this into the ecuations we get that: b^2 = 12c and c - b = 45/2 - c. Solving this we get b = 15 and c = 75/4. So c + b = 135/4 and the anser is 135 + 4 = 139

12,12r,12r^2,45/2 and 12r^2-12r=45/2-12r^2

48r^2-24r-45=0 solve for r=5/4.

then 12r+12r^2=12r(1+r)=12(5/4)(1+5/4)=135/4=a/b

and a+b=135+4=139

Let the sequence : 12 , x , y , \frac {45}{2}

The first three are in geometric progression, then y - x = x - 12 ...........(1). The last three are in arithmetic progression, then \frac {\frac {45}{2}}{y} = \frac {y}{x} ......(2)

Based on (2), we have y = \frac {x^2}{12}. Substitute y = \frac {x^2}{12} on (1). Then we have

\frac {x^2}{6} - 6x - \frac {45}{2} = 0

x^2 - 6x - 135 = 0

So, x = -6 or x = 15, we take x = 15 because based on the sequence we have 12 \leq x \leq \frac {45}{2}

and y = \frac {15^2}{12} = \frac {75}{4}

Then the sum of the two middle numbers is x + y = 15 + \frac {75}{4} = \frac {135}{4} 134 and 4 are coprime, then a + b =135 + 4 = 139

Consider the same sequence ${a,b,c,d}$ written in the two different ways: $12,12x,12x^{2},22.5$ and $12,22.5-2k,22.5-k,22.5$
Then: $12x+12x^{2}=22.5-2k+22.5-k$ and $22.5-k=x(22.5-2k)$
With x the geometric factor and k the arithmetic one.
Substituting the second equality in the first yields: $12x+12x^{2}=45-67.5(\frac{x-1}{2x-1})$
Quartic equation has three real roots: $x=-1 or - \frac{3}{4} or \frac{5}{4}$
As the sequence is increasing (from 12 to 22.5) and the terms are positive (no alternating series), then the geometric factor must be positive greater than 1, thus: $x=\frac{5}{4}$
we get: Desired fraction= $12x+12x^{2}=12*1.25+12*1.25^{2}=\frac{135}{4}$
So $a+b=135+4=139$

the G. P. is 12, 12r, 12r^2. the A. P. is 45/2-2d, 45/2-d, 45/2. 12r=45/2-2d ------------eq 1. 12r^2=45/2-d-------------eq 2. Solving simultaneously, r=-3/4 or r=5/4. Substituting the middle terms are 15 and 75/4. adding, we get 135/4. a+b=139

Geometric series hold the relationship of $\frac{x}{y}$ between values. Arithmetic series hold the relationship of x + z between values. Given the restraint that there are 4 values, 2 of which are known, allows for 3 relationships to be created.

12 x y 22.5 (equivalent to $\frac{45}{2}$ )

Geometric series have a constant multiple linking them. This allows us to create two relationships.

1. The relationship between 12 and x which is $\frac{x}{12}$
2. The relationship between x and y which is $\frac{y}{x}$

We also know these must be equivalent, which gives us the equation: $\frac{x}{12} = \frac{y}{x}$

We can then rewrite this equation into a two variable quadratic function: $x^{2} = 12y$

Arithmetic series have a constant additive. We will call this additive ${z}$ This also allows us to create two relationships.

1. The relationship between x and y which is $x + z = y$
2. The relationship between y and 22.5 which is $y + z = 22.5$

A three way system of equations can be developed from these relationships:

• $x^{2} = 12y$
• $x + z = y$
• $y + z = 22.5$

$( x + z = y) - (y + z = 22.5) = x - y = y - 22.5 = (x + 22.5 = 2y)$

$(x^{2} = 12y) = (x^{2} = 6(2y))$

We can now plug in $x + 22.5$ for $2y$ in the equation, which gives us:

$(x^{2} = 6(x + 22.5)) = (x^{2} = 6x + 135) = (x^{2} - 6x -135 = 0)$

This quadratic expression can then be solved for $x$ using whatever method (quadratic formula, etc.). This gives solutions of $-9$ and $15$ . We know our answer must be positive due to the sequence we are given already, so $x = 15$ .

Plugging this into the geometric relationship we created earlier, we can solve for $y$ .

• $15^{2} = 12y$
• $y = 18.75$

We are instructed to add the middle numbers together and write this as a fraction, so:

$x + y = 15 + 18.75 = 33.75 = \frac{135}{4}$

Finally we must add our numerator and denominator together, giving us: $135 + 4 = 139$

According to the question,

sq(b) = a.c ; 2.c = b + d

a = 12 and d = 45/2.

so,the equations become:

sq(b)=12.c .....(1)

2.c = b + 45/2 => 4.c = 2.b + 45 ......(2)

Putting equation (1) in equation (2), we get

sq(b) - 6.b - 135 = 0

which gives b = 15, -9

Since the question considers only positive numbers, therefore b = 15.

Putting b = 15 in equation (1), we get

c = 75/4.

Now, b + c =135/4.

The required answer = 135 + 4 = 139.

12, 12q, 12q² são os 3 primeiros números. 12q, 12q+r, 12q²+r (ou 12q+2r ou ainda 45/2) são os 3 últimos números. 45/2=12q+2r :. 24q+4r=45 :. r=(45-24q)/4 12q²=12q+r :. 12q²=12q+(45-24q)/4 :. 48q²=48q+45-24q :. 48q²=24q+45 :. 48q²-24q-45=0 :. 16q²-8q-15=0 :. Δ=64-4.(-15).16=64+960=1024 :. √Δ=2^5=32 :. q=(8+_32)/32:. q'=40/32=5/4 e q"=-24/32=-3/4 :. 12q'=12.5/4=15 e 12q'²=12.25/16=75/4 e 15+75/4=(60+75)/4=135/4 e 135+4=139

12r=a; 12r^2=a+d; a+2d=22.5; 12r^2=12r+d => d=12r^2-12r
12r + 2( 12r^2-12r )= 22.5 => 48r^2-24r-45=0 r= 1.25 and -0.75 since all numbers are positive r = 1.25.
a= 12 * 1.25 = 15; 15 * 1.25 = 18.75. 15+18.75= 33.75=3375/100=135/4 (on division by 25). 135+4= 139

a/r a a*r b be the numbers

a+b=2 a r as they are in AP => a/r + b/r=2 a =>12+45/2 r=2 12 r By solving the above equation we get a quadratic expression as

16 r^2 -8 r -15=0

roots are 5/4 and -3/4

required r=5/4 =>a=15 and a r=75/4 and a+a r=15+75/4=135/4

required result is 135+4=139 Hence the result

Let the two middle terms be x,y. Therefore the 4 +ve integers are 12,x,y,45/2 . Since first 3 are in G.P, so x^{2}=12y or (x^{2})/6=2y....(i)

As x,y,45/2 are in A.P, so 2y=x+45/2 or (x^{2})/6=x+45/2.....(ii) Solving equation (ii) we get the value of x as 15 or -9, we take the +ve value i.e 15. Similarly we get y, as 75/4. so, x+y=135/4.

here a=135 and b=4, so a+b=139.....solved..!!!

the numbers are 12, 12r ,12 r r , 45/2 and the condition is 12. 12 r r = (12r + 45/2)/2 solving this we get r = 5/4 substituting this we get the numbers as 12 ,15 ,75/4 , 45/2 therefore the required sum is 135/4.

Okay, let us put this statement as 12, x, y, 45/2. Let us make the whole sequence whole numbers. So, it becomes 24, 2x, 2y, 45. Since 24 and 45 are divisible by 3, so let us divide the whole sequence by 3. So it becomes 8, 2x/3, 2y/3,15. Let 2x/3 be k, and 2y/3 be l. So the sequence is 8, k, l, 15. Through trial and error, we find that k = 10, and l = 12.5, as 8, 10, 12 have a geometric progression of scale factor 1.25, and that 10, 12.5, 15 is an arithmetic progression. So we put the sequence back originally, and it becomes 8 3/2, 10 3/2, 12.5 3/2, 15 3/2 = 12, 15, 18.75, 22.5. As 15+18.75 = 33.75, and that 33.75 can be written as a/b, where a and b are coprime positive integers, the only answer would be 135/4 = 33.75, as b cannot be 1, 2 and 3 because other integers will not result in a being a positive integer. So 135 + 4 = 139.

let the number sequence be 12, x, y, 45/2
the common ratio and common difference be r and d
then x=12r =45/2 -d
y =r^2=45/2 -d
solving the simultaneous equation
r=5/4 , d=15/4
then we get x=15, y=75/4
x+y=135/4
therefore a+b = 139

12, b, c, 45/2 b^2 = 12c 2c = b + 45/2 >> c = (b/2) + (45/4) substitution to b^2 = 12c b^2 = 12[(b/2) + (45/4)] we get: b = 15 c = 75/4

so 15 + 75/4 = 135/4 135+4 = 139

The first 3 numbers of the sequence are in geometric progression ( with common ratio r) \Rightarrow They can be writen as: a1 =12, a2 = 12r, a3 = 12r^2. The last 3 numbers of the sequence are in arithmetic progression (with common difference d) \Rightarrow They can be written as: a2 = 12r, a3 = 12r + d, a4 = 12r + 2d = \frac {45}{2}. \Rightarrow r= \frac {5}{4} or r=\frac {-3}{4} The 4 numbers are positive \Rightarrow r= \frac {5}{4} . Therefore d= \frac {15}{4}. Therefore a2 = 15, a3= \frac {75}{4}. \Rightarrow a2 + a3 = \frac {135}{4}. Therefore a + b = 135 +4 = 139

Let the sequence be 12, x, y, 45/2. (y/x = x/12) and (45/2 -b = b-a), using the facts that the first and last three are arithmetic and geometric progressions. Continuing the equations, (x^2 = 12y) and (2y= 45/2 +x) --> (12y= 135 +6x). Combining the two equations yields (x^2=135+6x)-->(x^2 -6x -135 =0). Factoring the quadratic into (x+9)(x-15)=0 has two values for x. If x is positive (15), then y is 15(15/12)=75/4. The progressions hold true, so x+y= 15+ 75/4 = 135/4 --> 139

LET X AND Y BE THE TWO MIDDLE NUMBERS. SERIES IS 12, X, Y, 22.5 (45/2) 12, X, Y ARE IN GEOMETRIC PROGRESSION. THEREFORE X/12=Y/X X^2=12Y. X, Y, 22.5 ARE IN ARITHMETIC PROGRESSION. THEREFORE Y-X=22.5-Y. 2Y=22.5+X WE HAVE NOW 2 EQUATIONS: X^2=12Y AND 2Y=22.5+X AS X^2=12Y, Y=X^2/12. SUBSTITUTING VALUE OF Y IN 2Y=22.5+X, WE GET X^2/6=22.5+X. MULTIPLYING BOTH SIDES BY 6 WE GET A QUADRATIC EQUATION. X^2-6X-135=0. (X-15)(X+9) ARE THE ROOTS. X,Y ARE POSITIVE NUMBERS. THEREFORE X-15=0, X=15 . THUS WE GET Y=75/4. NOW X+Y CAN BE WRITTEN AS A/B WHERE A AND B ARE CO PRIME. X+Y=15+(75/4)=135/4. THEREFORE A=135 AND B=4 . THEY ARE COPRIME. THEREFORE A+B=135+4=139

Considering the progressions, note that we can write the first $3$ integers as $12,12q$ and $12q^2$ , whereas the last $3$ integers would be $\frac{45}{2}-2d,\frac{45}{2}-d$ and $\frac{45}{2}$ .

From this, we can conclude that $\frac{45}{2}-2d=12q$ and $\frac{45}{2}-d=12q^2$ .

Now, considering that in any tripple in an arithmetic progression the middle integer is the semi-sum of the first and last integers ( $\frac{a+(a+2d)}{2}=\frac{2a+2d}{2}=a+d$ ).

Now since the last $3$ integers form an arithmetic progression (which can now be written as $12q,12q^2,\frac{45}{2}$ ) we can apply this rule, thus having

$12q+\frac{45}{2}=24q^2 \Rightarrow 24q+45=48q^2 \Rightarrow 48q^2-24q-45=0$ .

By solving the quadratic equation we get $2$ roots for $q$ , those being $q_1=\frac{5}{4}$ and $q_2=-\frac{3}{4}$ . However, taking into account that the integers we are dealing with have to be positive , $q_2$ is not a valid solution, leaving us with $q_1$ .

Since $q=\frac{5}{4}$ , we can easily find the $2$ integers we are looking for

$12q=12\cdot\frac{5}{4}=15$ and $12q^2=12\cdot\frac{25}{16}=\frac{75}{4}$ .

Their sum is $15+\frac{75}{4}=\frac{135}{4}$ .

Since we're looking for $\frac{a}{b}$ and $135$ and $4$ are both coprime positive integers, we can easily find the required sum

$a+b=135+4=139$ , which is also the solution to the problem.

First of all I consider this problem ugly and useless.

We can write the terms this way:

$12$ ; $12\times a$ ; $12\times a\times a$ ; $45/2$ ; or $12$ ; $12\times a$ ; $12\times a + b$ ; $45/2$ ; We therefor have 2 equations with 2 unknown variables : $12\times a\times a =12\times a + b$ and $12\times a+ 2\times b = 45/2$ so, by putting $b=\frac{45-24\times a}{4}$ in the second equation, we get a simple quadratic equation.

Let the four numbers be $x,y,z,w$ , then:

$x=12\;,\;y=12q\;;\;z=12q^2=y+d\;,\;w=\frac{45}2=y+2d$

and from here we get

$\frac{45}2=y+2d=12q+2d\implies 24q+4d=45\;\;\;(*)$

$z=12q^2=y+d=12q+d\implies d=12q^2-12q$

and substituting now on (*):

$24q+4(12q^2-12q)=45\implies 16q^2-8q-15=0\implies q=\frac54$

so finally

$y+z=12\frac54+12\frac{25}{16}=\frac{135}{4}\implies a+b=135+4=139$

let, the common ratio of the G.P. is = r

the common different of the A.P. is = d

In G.P. The 1st 3 numbers are 12, 12r & 12r^2

In A.P. the last 3 numbers are $45/2 - 2d$ , $45/2 - d$ & $45/2$

There the 2nd number = $45/2 - 2d = 12r$

&

The **3rd number are = $45/2 - d = 12r^2$

Solving these equations we get the value of r = $5/4$ {you may get the d value}

putting the value of r the 2nd and 3rd number are $15$ and $75/4$ respectively;

Adding the two number $15 + 75/4$ we get $a/b$ = $135/4$

Then a + b = $139$

Write down the sequence as

$12, ar, ar^2, \frac {45}{2}$

Since the first three numbers are on geometric progression, then: $\frac {ar}{12} = \frac {ar^2}{ar}$ So, we get a = 12

Since the last three numbers are on arithmetic progression, then : $ar^2 - ar = \frac {45}{2} - ar^2$ Substitute $a = 12$ into this equation, $16r^2 -8r -15=0 (4r+3)(4r-5)=0$ Because, the four numbers are positive take $r= \frac {5}{4}$ So, the middle numbers are: $ar + ar^2= 15+ \frac {75}{4}= \frac {135}{4}$

The final answer is $135 + 4 = 139$

LET us assume the sequence to be as such....12,A,B,(45/2).. Now the first three terms are in a G.P...hence...A/12 = B/A .or,A^2=12B....(1). The last three terms are in A.P..hence B-A=(45/2)-B...or,B=(45/4)+(A/2)...(2)

Substituting the value of B , in (1)..we get.. A^2=12{(45/4)+(A/2)} or,A^2=135+6A now, solving the quadratic.. we get A=15 or -9 BUT A cannot be negative...A=15 putting value of A in (2)... B=225/12. A+B=225/12 + 15 = 405/12 = 135/4..(which is a/b..in the question...) therefore,a+b=139....SOLVED

For the first three terms, let the GP be 12, 12r, 12 $r^{2}$ where r is the common ratio.

For the last three terms, let the AP be $\frac{45}{2}$ -2d, $\frac{45}{2}$ -d, $\frac{45}{2}$ where d is the common difference.

The corresponding terms are equal. Therefore, we can write:

12r= $\frac{45}{2}$ -2d...................................(1) , and

12 $r^{2}$ = $\frac{45}{2}$ -d............................(2)

Multiplying equation (2) with 2 and then subtracting from equation (1), we get

48 $r^{2}$ -24r=45, which on solving gives r= $\frac{5}{4}$ (ignoring the negative value as it doesn't satisfy the AP).

Hence, the sum of the two middle terms=12 $r^{2}$ +12r= $\frac{135}{4}$

Thus, a+b=135+4=139

Let the 4 nos. be 12 , x , y , 22.5

By Geometric Mean in first three nos. x^2 = 12*y.

y = (x^2)/12 ............(i)

And by applying Arithmetic Mean to last three nos. (x + 22.5)/2 = y ............(ii)

By equating (i) and (ii), we get quadratic equation,

x^2 - 6x - 135 = 0

(x - 15)(x + 9) = 0

gives x = 15, -9 (-9 is rejected because we are treating positive nos.)

Now by putting value of x in eqn. (ii)

we get value of y = (15 + 22.5)/2 = 18.75

Now we have to add x and y i.e. (15 + 18.75) = 33.75 which in fraction can be written as 135/4. Clearly, a = 135, b = 4 and a + b =139 (Ans.)

let the sequence be 12, 12 r , 12 r^{2} , and the fourth number is thus 12 r^{2} + [ 12 r^{2} -- 12 r ] . Now equating it with the given value of 45/2. thus we get two values for the variable r .. now ignoring the negative value we can find the sum of middle two terms as 139.

Let the 2 middle numbers be x and y .

Now, as first three are in G.P. so, x^{2}=12*y ....(1)

As the last three are in A.P. so, 2y=x+22.5 ....(2)

Solving equation (1) and (2) we get, x=15 and y=75/4

Hence, x+y=135/4 so, 135+4=139 (a+b) .

Let the seq. ( X , Y , Z , M ) X = 12 , Y = 12 r ,Z = 12 r^2 M = 22.5, Y = 22.5 + 2 d , Z = 22.5 + d 12 r = 22.5 + 2 d ( 1 ) 12 r ^2 = 22.5 + d ( 2 ) ( 1 ) - ( 2 ) d = 12 r - 12 r^2 ( 3 ) ( 3 ) in ( 2 ) 12 r^2 = 22.5 + 12 r - 12 r ^2 24 r^2 - 12 r - 22.5 = 0 r = 5/4 or r = -3/4 refused ( since all are positive ) Sum of middle terms Y + Z = 12 r + 12 r ^2 S = 12( 5 / 4 ) + 12 ( 5 / 4 )^2 = 135 / 4 = a / b a + b = 135 + 4 = 139

The first 3 numbers are a/r,a and ar (say)Then the fourth no. is ar+(ar-a).=2ar-a So,a/r=12 and 2ar-a=45/2 So,a=12r Therefore,2 12r r-12r=45/2 Solving this we get r=5/4 or -3/4 But if r= -3/4 then 2ar-a is negative So r=5/4 This yields the solution a/b=135/4 (this a is the a in the problem not the second term.) So a+b=135+4=139