Question

Directions for questions 67 to 70: Seven integers A, B, C, D, E, F and G are to be arranged in increasing order such that:
(i) The first four numbers, in this increasing order, are in arithmetic progression (A.P.).
(ii) The last four numbers, in this increasing order, are in geometric progression (G.P.).
(iii) There is exactly one number between E and G, in the increasing order.
(iv) There is no number between A and B, in the increasing order.
(v) D is the smallest number and E is the greatest number.
(vi) \[ \frac{A}{D}=\frac{G}{C}=\frac{F}{A}>1 \]
(vii) E = 960

67. \(\dfrac{E}{A} = ?\)

• A. 2
• B. 3
• C. 4
• D. None of the above

Hint:

Work out the seven values in simple ratio units first (D = 1), then scale everything up so E equals 960.

Answer 1

The Correct Option is D

Step 1: Set up positions in increasing order.
Call the seven values in increasing order \(p_1<p_2<\dots<p_7\). From condition (v), \(D=p_1\) (the smallest) and \(E=p_7\) (the greatest).

Step 2: Place G using condition (iii).
"There is exactly one number between E and G" means, in the increasing order, G sits two places before E, with one term sitting in between. So \(G=p_5\), and \(p_6\) is the one term between G and E.

Step 3: Use the A.P./G.P. conditions and condition (iv).
The first four numbers \(p_1,p_2,p_3,p_4\) are in A.P., and the last four \(p_4,p_5,p_6,p_7\) are in G.P., with \(p_4\) common to both. That leaves \(p_2,p_3,p_4,p_6\) to be shared among A, B, C and F. Condition (iv), no number between A and B, means A and B sit at two consecutive positions. Checking this against condition (vi), \(\dfrac{A}{D}=\dfrac{G}{C}=\dfrac{F}{A}\), the only placement that is consistent with all seven conditions together is:
\[ D=p_1,\ \ C=p_2,\ \ B=p_3,\ \ A=p_4,\ \ G=p_5,\ \ F=p_6,\ \ E=p_7 \]
(A sits at the point where the A.P. and G.P. meet, B is right next to it as condition (iv) needs, C is the second A.P. term, and F is the last G.P. term before E.)

Step 4: Solve for the common difference and the ratio.
Let the A.P. have first term D and common difference d, so \(p_1=D,\ p_2=D+d,\ p_3=D+2d,\ p_4=D+3d\). Let the G.P. have ratio r, so \(p_5=p_4r,\ p_6=p_4r^2,\ p_7=p_4r^3\).
Condition (vi) sets three ratios equal to one value, call it k:
\[ \frac{A}{D}=\frac{D+3d}{D}=k,\qquad \frac{G}{C}=\frac{p_4r}{D+d}=k,\qquad \frac{F}{A}=\frac{p_4r^2}{p_4}=r^2=k \]
So \(k=r^2\). Combining the first two ratio equations and simplifying (the common factor \(D+3d\) cancels out) leads to \(d=D\), which in turn gives \(r=2\) and \(k=4\).

Step 5: Compute all seven values using E = 960.
With \(d=D\): \(p_1=D,\ p_2=2D,\ p_3=3D,\ p_4=4D,\ p_5=8D,\ p_6=16D,\ p_7=32D\). Since \(E=p_7=32D=960\), \(D=30\).
\[ D=30,\ \ C=60,\ \ B=90,\ \ A=120,\ \ G=240,\ \ F=480,\ \ E=960 \]
Quick check of every condition: the A.P. 30, 60, 90, 120 has a common difference of 30; the G.P. 120, 240, 480, 960 has a ratio of 2; exactly one number, 480, lies between G = 240 and E = 960; A = 120 and B = 90 have no number between them; D = 30 is the smallest and E = 960 is the greatest; and \(A/D=120/30=4\), \(G/C=240/60=4\), \(F/A=480/120=4\), all equal and greater than 1. Every condition checks out.

Step 6: Answer the question asked.
\[ \frac{E}{A}=\frac{960}{120}=8 \]
This value, 8, is not equal to 2, 3, 4 or 5, so none of the first four numeric choices match it.

Final Answer:
Since 8 is not among 2, 3, 4 or 5, the correct choice is "None of the above."
\[ \boxed{\dfrac{E}{A}=8,\ \text{so None of the above}} \]