Combined Arithmetic Geometric Progression

We let the terms in the Arithmetic progression be $a, a+r, a+2r$ and the terms in the geometric progression to be $b, bp, bp^2$ . Then, since $a+b = 120, a+r+bp=116, a+2r+bp^2=130$ , therefore, we can directly substitute $b=120-a$

$bp = 116-(a+r)$

$bp^2 = 130-(a+2r)$ .

After that, we solve the linear equation, as follows.

$120-a+116-a-r+130-a-2r=342$

$366-3a-3r=342$

$3a+3r=24$

$a+r=8$

This implies to $bp=108$ . Then, we left with $b+bp^2=234$ , implying that

$\frac{108}{p}+108p=234$

$\frac{1}{p}+p=\frac{13}{6}$

$\frac{p^2+1}{p}=\frac{13}{6}$

$6p^2+6=13p$

$6p^2-13p+6=0$

$(p-\frac{3}{2})(p-\frac{2}{3})=0$

Therefore, p could be either $\frac{3}{2}$ or $\frac{2}{3}$ . But since we need the largest term, so we multiply 108 with $\frac{3}{2}$ , to get our desired answer, which is $\boxed{162}$ .

Given:

$a+x = 120$

$b+y = 116$

$c+z = 130$

Adding the three,

$a+b+c+(x+y+z) = 366$

But $x+y + z = 342$ . So,

$a+b+c = 366 -342 = 24$

In an Arithmetic Progression, $2b = a+c$ .

$2b + b =24 \Longrightarrow b = 8$

If $b = 8$ , $y = 116 - 8 = 108$

Let $x = \dfrac{y}{r}$ and $z = yr$ , where $r$ is the common ratio of the Geometric Progression.

$\dfrac{y}{r} + y + yr = 342$

$\dfrac{108}{r} + 108 + 108r = 342$

$\Longrightarrow 54r^{2} - 117r + 54 = 0$

Solving for $r$ , we get $r = \dfrac{3}{2}$ or $r=\dfrac{2}{3}$

$x = \dfrac{y}{r} = \dfrac{108}{\frac{3}{2}} = 72$

$z = yr = 108 \times \dfrac{3}{2} = \boxed{162}$

Let the terms in A.P. be $(a-b),a,(a+b)$ . And the G.P. terms be $(A),(Ar)$ and $( Ar^2)$ . therefore $a-b+A=120$ (1), $a+Ar=116$ (2), $a+b+ Ar^2 =130$ (3). Adding (1),(2) and (3), we get $3a +A(1+r+ r^2)=366$ (4). Again, $A+Ar+ Ar^2=342$ . Substituting this in (4), we get $3a=366-342=24$ . So, $a=8$ (1),(2) reduces to $A=112+b$ (5) and $Ar=108$ (6). (6) divided by (5) gives $r = 108/(112+b)$ , and from (3) we have $8+b+Ar^2=130$ , so $b + (108*108)/(112+b)=122$ , or that $b^2 +112b+(108*108)=(112*122)+122b$ , which is equivalent to $b^2-10b-2000=0$ . On solving, we get $b=50$ or -40. So, the terms in AP are either -42,8,58 and the terms in GP are 162,108,72. Alternatively the terms in AP are 48,8,-32 and the terms in GP are 72,108,162. The answer is the greatest term in GP, i.e. 162.

[Latex Edits - Calvin]

use A,B,C to express terms of the AP and X,Y,Z to express terms of the GP.

first , it is to count the sum of the AP is $120+116+130-342=24$ , so B must $24/3=8$ and $Y=116-B=108$ .

so let $A=8-d,C=8+d,X=\frac{108}{q},Z=108q$ ,as $d$ is the common difference and $q$ is the commom ratio. also we have $A+X=120 ,C+Z=130$ .

solve the equation and we get two solution $q=\frac{2}{3},d=50 or , q=\frac{3}{2},d=-40$ .both of them show the largest term in the Geometric Progression is $108*\frac{3}{2}=162$

Firstly, a very important clue to note is that the sum of the corresponding terms of the two series first decreased, then increased ( $120,116, 130$ ). Since the difference $d$ $(>0)$ of the $AP$ - arithmetic progression - is always constant, the decrease then increase in the sum of the terms (above) must be due to the terms in the $GP$ . From here we can deduce that the GP and AP are progressing in "different direction" (i.e. one series is decreasing while the other increasing). In this question, the AP is decreasing while the GP is increasing (because the gradual increase in sum of terms occur due to the increase in the difference of terms in GP).

Secondly, let the $3$ terms (in decreasing order) in the AP be $a + d, a, a - d$ . Adding them up, we get $3a = 24$ and $a = 8$ . Thus, we know that the second term of the GP $= 116 - 8 = 108$ . Similarly, the first and third term of the GP are $120-(8+d)= 112 - d \text { and }130 -(8-d) = 122 +d$ respectively. Recall that in a GP, the ratio is common for all terms. Hence, $\frac {108}{112-d} = \frac{122+d}{108}$

Cross multiplying, we get: $108^2 = (122+d)(112-d) = 13664 -10d -d^2$

Then, $d^2 +10d -2000 = (d-40)(d+50) =0$ .

Thus, $d = 40 \text{ or } -50$ but since the difference is assumed to be $(>0)$ above, $d=40$ . The third term of the GP, which is the largest term is simply $130 - (8 - 40) = \boxed{162}$

Assume that, the number that is on arithmetic progression :

$(a - d), a, (a+d) \ldots \ldots \ldots (1)$

And assume that, the number that is on geometric progression :

$b, br, br^2 \ldots \ldots \ldots (2)$ . So,

Since, the sum of all the terms in the Geometric Progression is $342$ . And the sum of equation (1) and (2) is $(120 + 116 + 130 = 366)$ We obtain :

$((a - d) + a + (a+d)) + (a + ar + ar^2) = 366 \implies 3a + 342 = 366$ $3a = 24 \implies a = 8$ . So,

the number that is on arithmetic progression :

$(8 - d), 8, (8+d)$ . And,

Since, the sum second term is $116$ , we obtain :

$a + (br) = 116 \implies 8 + br = 116 \implies br = 108$ .

Since, the sum of all the terms in the Geometric Progression is $342$ . We obtain :

$b + br + br^2 = 342 \implies br( \frac{1}{r} + 1 + r ) = 342 \implies 108( \frac{1}{r} + 1 + r ) = 342$ $\frac{1}{r} + r = \frac{13}{6} \implies 6r^2 - 13r + 6 = 0 \implies (2r - 3)(3r - 2) = 0$

So, $r = \frac{2}{3} or \frac{3}{2}$ . And, WLOG(without loss of generality), assume that $r = \frac{3}{2}$ . So,

the number that is on geometric progression :

$72, 108, 162$ . So, the largest term in the geometric progression is $\boxed{162}$

Let the numbers in the Arithmetic Progression be a-b , a , a+b and the numbers in the Geometric Progression be x , xr , xr^2. Since a - b + x = 120 a + xr = 116, and a + b + xr^2 = 130, a-b + a + a+b + x + xr + xr^2 = 120 + 116 + 130 = 366. We also know that x + xr + xr^2 = 342, therefore, subtracting the geometric terms from a-b + a + a+b + x + xr + xr^2 =366, you get a-b + a + a+b = 24. Combine like terms and 3a = 24, and a = 8. From a + xr = 116, we can subtract a out and xr = 108. We can now find r by setting up a ratio: xr / (xr^2 + xr + x) = r / (r^2+r+1) = 108/342 = 6 / 19. Cross multiply you get 19r = 6r^2 + 6r + 6. Subtract 19r from both sides: 0 = 6r^2 - 13r + 6. Now, with the quadratics formula, r = 2/3 or 3/2. Plug r in xr = 108 and x = 162 or 72. Plug x and r in for xr^2 and xr^2 = 162 or 72. Therefore, the largest term is 162 no matter whether r is 2/3 or 3/2.

Let us take AP as x-d, x , and x+d and GP as a/r, a, ar Now, a(\­(\frac {1+r+r^2}{r}\­)) = 342 -------------(vii) and \­(\frac {a}{r}\­) + ar = 342 - a -------------(iv) and x-d+\­(\frac {a}{r}\­) = 120 -----------(i) x+a = 116 ------------(ii) x+d+ar -------------(iii) Adding eqn. (i) and (iii) 2x + \­(\frac {a}{r}\­) + ar = 250 ------------(v)

from (iv) and (v) 2x + 342 - a = 250 2x -a = -92 ----------------(vi) from (ii) and (vi) 3x= 24 => x=8 from (ii) a=116-8 = 108 Now putting the value of a in eqn (vii)

\­(\frac {1+r+r^2}{r}\­)= \­(\frac {342}{108}\­) \­(\frac {1+r+r^2}{r}\­)= \­(\frac {19}{6}\­) 6\­(r^2\­) -13r + 6 =0 solving the quardratic eq. r = \­(\frac{13 \pm \sqrt{13^2-466}}{12}\­) r =\­(\frac {13 \pm 5}{12}\­) => r= \­(\frac {2}{3}\­) or \­(\frac {3}{2}\­) for max ar = 108 * \­(\frac {3}{2}\­)

hence max ar = 162

Let the 3 Geometric Progression terms be $\frac{x}{a}$ , $x$ and $ax$ .

Let the 3 Arithmetic Progression terms be $m-n$ , $m$ and $m+n$ .

We have $\frac{x}{a}+x+ax=342$ and $\frac{x}{a}+x+ax+m-n+m+m+n=\frac{x}{a}+x+ax+3m=120+116+130=366$ , so $3m=366-342=24$ and $m=8$ .

Since $m+x=116$ , thus $x=116-8=108$ . Now we have $\frac{108}{a}+108+108a=108(\frac{1}{a}+1+a)=342$ and $\frac{1}{a}+1+a=\frac{19}{6}$ , thus $\frac{1}{a}+a=\frac{13}{6}$ .

Multiplying both sides by a will get $a^2+1=\frac{13}{6}a$ or $a^2-\frac{13}{6}a+1=0$ . Factorising gives $(a-\frac{3}{2})(a-\frac{2}{3})=0$ , so $a=\frac{3}{2} or \frac{2}{3}$ .

Now one of the Geometric Progression terms is $\frac{2}{3}\cdot 108=72$ , another is 108 and the last is $\frac{3}{2}\cdot 108=162$ .

So the largest of the Geometric Progression terms is 162.

Let the numbers in AP be a, a+d, a+2d and the numbers in GP be A, AR, AR^2.

By using the given information,

a+A=120......[1]

a+d+AR=116........[2]

a+2d+AR^2=130........[3]

A+AR+AR^2=342.......[4]

Adding 1,2,3 and substituting from 4,

a+d=8

Hence, the equation reduces to

AR=108

A(1+R^2)=234

Solving for A and R,

A=72;R=3/2

OR

A=162;R=2/3

Calculating AR^2 for both and taking larger value,

AR^2=162

Let $a-d, a, a+d$ be the terms of an arithmetic progression and $\frac{b}{r}, b, br$ be the terms of a geometric progression. Then we have $a-d+\frac{b}{r}=120, a+b=116, a+d+br=130$ . Thus, the sum of the terms of an arithmetic progression is equal to $120+116+130-342=24=3a$ , which gives $a=8$ . Thus, $b=116-a=108$ and $108(\frac{1}{r}+1+r)=342$ , which gives $\frac{1}{r}+r=\frac{13}{6}$ or $6r^2-13r+6=(3r-2)(2r-3)=0$ . Thus, with $r=\frac{3}{2}, \frac{2}{3}$ , the largest term in a geometric progression is $108\cdot\frac{3}{2}=162$ .

Hey guys! I feel that this problem is slightly over-rated. All it needs is four equations, manipulations and their solutions!

So let the three terms of the GP be $\frac{b}{x}$ , b and bx.

And the three terms of the AP be a-d, a and a+d.

Now with the given data, we can formulate four equations:

1-- a-d + $\frac{b}{x}$ = 120

2-- a + b = 116

3-- a+d + bx = 130

4-- $\frac{b}{x}$ + b + bx = 342

Adding equations 1 and 3, we get-- 2a + $\frac{b}{x}$ + bx = 250

Substituting for $\frac{b}{x}$ + bx from equation 4, we get equation--

5-- 2a - b = -92

Solving equations 2 and 5, we get a = 8 and b = 108.

Now plugging in the values for b in equation 4 and solving for x, we get x = $\frac{2}{3}$ or $\frac{3}{2}$ .

Thus, either way, we get the max. value of the GP to be 162! Hence the answer! Cheers!

solve 108(1/r + 1 + r) = 342, GP is 72, 108, 162. AP is - 42, 8, 58. We get 162 - 42 = 120, 108 + 8 = 116 and 58 + 72 = 130

Let $(A, B, C)$ and $(X, Y, Z)$ be the terms in arithmetic sequence and geometric sequence respectively.

Adding the sum of the corresponding terms of the AP and GP will give us:

$(A + X) + (B + Y) + (C + Z) = (A + B + C) + (X + Y + Z) = 120 + 116 + 130 = 366$

which is the sum of the terms of AP and GP.

Since the sum of the terms of GP is $342$ , subtracting it will give us the sum of the terms of AP. Thus $A + B + C = 366 - 342 = 24$ which implies, the middle term $B$ to be $8$ .

Now let, $r$ be the common ratio of the GP. Given the fact that $Y$ is the geometric mean, we can represent all the other terms of the geometric progression with respect to $r$ and $B$ thus, $\frac{Y}{r} = X$ and $Yr = Z$ . This also holds for the terms in Arithmetic Progression.

From that,

(1) $\frac{Y}{r} + 8 - d = 120$

(2) $Y + 8 = 116$

(3) $Yr + 8 + d = 130$

From (2), we get $Y = 108$ , by substituting it to (1) and (3) we will have the system of equations:

(4) $\frac{108}{r} - d = 112$

(5) $108r + d = 122$

which by elimination method, get rid of $d$

(6) $\frac{108}{r} + 108r = 234$

Solving for $r$ we'll get $r = \frac{3}{2}$

Thus by substitution, to $Z$ we'll get, $Z = (108)(\frac{3}{2}) = 162$ which is what we have been asked to find.

A,B, C are in ap and x,y,z are in gp...let x=a=first term of gp then y=ar and z=ar^2......2B=A+C...... A+x=120, B+y=116, C+z=130 Now adding all above three equations A+B+C+x+y+z=120+116+130=>366 A+B+C+342=366 2B+B=366-342, 3B=24 B=8 Now B+y=108,=>y=108 Now ar=108....therefore a=108/r Adding all gp terms, x+y+z=342 x+z=342-108 x+z=234 a+ar^2=234 Now putting a=108/r we get a quadratic in r whose solutions are 6/9 and 9/6.....put this values one by one in x,y,z and search for the largest value.....we will get two series in gp for two values of r....72,108,162 and 162,108,72...now largest value=162

Let, the three terms of A.P. series = $(p-d), p, (p+d)$ [where $d =$ common difference of A.P.]

And, the three terms of G.P. series = $\frac{a}{r}, a, ar$ [where $r =$ common ratio of G.P.]

Then, $(p - d) + \frac{a}{r} = 120 ....(i)$

$p + a = 116 .....(ii)$

$(p +d) + ar = 130 ....(iii)$

So, the sum of all terms in A.P. and G.P. is = $((p-d) + p + (p -d)) + ( \frac{a}{r} + a + ar) = 120 + 116 + 130 = 366$

And the sum of all terms of G.P. is $\frac{a}{r} + a + ar = 342$

So, the sum of all terms in A.P. = $((p-d) + p + (p -d) + \frac{a}{r} + a + ar)) - (\frac{a}{r} + a + ar) = 366 - 342$

$\implies (p - d) + p + (p + d) = 24$

$\implies 3p = 24$

$\implies p = 8$

In equation (ii) putting the value of $p$ $8 + a = 116$

$a = 108$

So, in equation (i) $(p - d) + \frac{a}{r} = 120$

$\implies \frac{108}{r} - d = 112 .... (iv)$

And in equation (iii) $(p +d) + ar = 130$

$\implies 108r + d = 122 .... (v)$

Solving equations (iv) and (v) we get the value of $r = \frac{3}{2} or \frac{2}{3}$ and $d = -40$

So, we get the terms of G.P. series = $72, 108, 162$

And get the largest term in G.P. $\implies \boxed{162}$

Arithmetic sequence: a, a+x, a+x^2 Geometric sequence: b, b y, b y^2

Equation 1: b + b y + b y^2 = 342

Equation 2: a+b = 120

Equation 3: a+x+by = 116

Equation 4: a+2x+by^2 = 130

Equation 4 - Equation 3: x + by(y-1) = 14

Equation 3 - Equation 2: x+ b(y-1) = -4

Equation 5: Combining the last two equations gives: b(y-1)^2 = 18

Combine equations 1 and 5.

b (1+y+y^2) / b (y-1)^2 = 342/18 = 19

18y^2 - 39y + 18 = 0;

y = 3/2 or 2/3 (both will end up working; they'll just reverse the order of the sequences).

If y=3/2, b = 72 and a = 48. The geometric sequence is 72, 108, 162. 162 is the answer.