Question

If a, b, c and d are four positive increasing integers such that a, b, c are in A.P. and b, c, d are in G.P. If d - a = 30. Find the four numbers.

• A. 18, 27, 36, 48
• B. 10, 20, 30, 40
• C. 5, 10, 15, 20
• D. 2, 4, 6, 8

Answer 1

Answer: (A)

18, 27, 36, 48

Answer 2

Let the four positive increasing integers be $a, b, c, d$.

Since $a, b, c$ are in Arithmetic Progression (A.P.), we can write $b = a+x$ and $c = a+2x$ for some common difference $x > 0$. Since $b, c, d$ are in Geometric Progression (G.P.), we can write $c = br$ and $d = cr = br^2$ for some common ratio $r > 1$.

For $b, c, d$ to be integers, the common ratio $r$ must be a rational number. Let $r = m/n$, where $m, n$ are coprime positive integers and $m > n$. Then $c = b(m/n)$ and $d = c(m/n) = b(m/n)^2$. For $b, c, d$ to be integers, let $b = jn^2$ for some integer $j$. Then $c = jn^2 (m/n) = jnm$. And $d = jnm (m/n) = jm^2$. So, $b=jn^2, c=jnm, d=jm^2$.

Since $a, b, c$ are in A.P., $a = 2b - c$. Substituting the expressions for $b$ and $c$: $a = 2(jn^2) - jnm = jn(2n - m)$.

The four numbers are $a = jn(2n-m)$, $b = jn^2$, $c = jnm$, $d = jm^2$.

For these numbers to be positive increasing integers:

1. $j, n, m$ must be positive integers.
2. $a < b \implies jn(2n-m) < jn^2 \implies 2n-m < n \implies n < m$.
3. $b < c \implies jn^2 < jnm \implies n < m$.
4. $c < d \implies jnm < jm^2 \implies n < m$.
5. $a > 0 \implies jn(2n-m) > 0$. Since $j>0, n>0$, we need $2n-m > 0 \implies m < 2n$.

So, the conditions are: $j \ge 1, n \ge 1, m > n, m < 2n$, and $gcd(m, n) = 1$.

We are given $d - a = 30$. $jm^2 - jn(2n-m) = 30$ $j(m^2 - 2n^2 + nm) = 30$ $j(m^2 + nm - 2n^2) = 30$ Factor the quadratic term: $j(m-n)(m+2n) = 30$.

Let $X = m-n$ and $Y = m+2n$. Then $jXY = 30$. From the constraints:

• $m > n \implies X = m-n \ge 1$.
• $m < 2n \implies m-n < n \implies X < n$.
• $Y = m+2n$. Since $m>n\ge 1$, $Y > 1+2(1) = 3$.
• $Y - X = (m+2n) - (m-n) = 3n$. So, $Y-X$ must be a positive multiple of 3.
• $gcd(m, n) = 1 \implies gcd(n+X, n) = gcd(X, n) = 1$.

We need to find integer factors $j, X, Y$ of 30 such that:

1. $j \ge 1, X \ge 1, Y \ge 1$.
2. $XY = 30/j$.
3. $Y > X$ (since $3n > 0$).
4. $Y-X$ is a positive multiple of 3.
5. $X < n$.
6. $gcd(X, n) = 1$.

Let's test possible values for $j$:

• If $j=3$: $XY = 30/3 = 10$. Possible $(X,Y)$ pairs with $Y>X$: (1,10), (2,5).

  • Case (1,10): $Y-X = 10-1 = 9$. This is a multiple of 3. $Y-X = 3n \implies 9 = 3n \implies n=3$. Check $X < n$: $1 < 3$. Yes. Check $gcd(X, n)=1$: $gcd(1, 3)=1$. Yes. Now find $m$: $m = n+X = 3+1 = 4$. Check $m < 2n$: $4 < 2(3)=6$. Yes. This gives a valid set of parameters: $j=3, n=3, m=4$. The four numbers are: $a = jn(2n-m) = 3 \times 3 \times (2 \times 3 - 4) = 9 \times (6-4) = 9 \times 2 = 18$. $b = jn^2 = 3 \times 3^2 = 3 \times 9 = 27$. $c = jnm = 3 \times 3 \times 4 = 36$. $d = jm^2 = 3 \times 4^2 = 3 \times 16 = 48$. The numbers are 18, 27, 36, 48. They are positive, increasing, $a,b,c$ are in AP (common diff 9), $b,c,d$ are in GP (common ratio 4/3), and $d-a = 48-18=30$. This is the correct solution.

  • Case (2,5): $Y-X = 5-2 = 3$. This is a multiple of 3. $Y-X = 3n \implies 3 = 3n \implies n=1$. Check $X < n$: $2 < 1$. No. This condition fails.

Other values of $j$ do not yield valid solutions. The four numbers are 18, 27, 36, and 48.