Question

Directions: 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.
(ii) The last four numbers (in this increasing order) are in geometric progression.
(iii) There is exactly one number between E and G.
(iv) There is no number between A and B.
(v) D is the smallest number and E is the greatest.
(vi) \( \dfrac{A}{D} = \dfrac{G}{C} = \dfrac{F}{A} > 1 \)
(vii) E = 960

The position and value of A is:

• A. 5th highest and 100
• B. 4th highest and 100
• C. 4th highest and 110
• D. None of the above

Hint:

Build the seven numbers as one chain: an AP of four terms feeding into a GP of four terms that shares its last AP term. Use the equal-ratio clue to pin the common difference and ratio, then rank A among all seven values.

Answer 1

The Correct Option is D

Step 1: Name the seven values in sorted order.
Call the seven integers, once sorted from smallest to largest, \(s_1 < s_2 < \dots < s_7\). Clue (v) says D is the smallest and E is the greatest, so \(s_1 = D\) and \(s_7 = E = 960\).

Step 2: Place G using clue (iii).
"One number between E and G" means E and G sit two ranks apart in the sorted list. Since E is at rank 7, G must be at rank 5, so \(G = s_5\), with one value (\(s_6\)) sitting between them.

Step 3: Write the AP and the GP.
By clue (i), \(s_1,s_2,s_3,s_4\) form an AP with first term \(D\) and common difference \(d\): \(s_2=D+d,\ s_3=D+2d,\ s_4=D+3d\).
By clue (ii), \(s_4,s_5,s_6,s_7\) form a GP with common ratio \(r\): \(s_5=s_4r,\ s_6=s_4r^2,\ s_7=s_4r^3=960\).

Step 4: Place A, B, C, F using clue (iv).
D, G and E already occupy \(s_1, s_5, s_7\), so A, B, C, F fill the remaining ranks \(s_2,s_3,s_4,s_6\). Clue (iv) says A and B sit next to each other. Trying the possible adjacent pairs and checking which one is consistent with clue (vi) shows that A must sit at \(s_4\) and B at \(s_3\), which leaves C at \(s_2\) and F at \(s_6\).

Step 5: Solve clue (vi) for the common ratio \(k\).
Let \(k = A/D = G/C = F/A\). Since \(A=s_4=D+3d\), we get \(k = 1 + 3d/D\), so \(d = \dfrac{D(k-1)}{3}\).
Because \(F=s_6=s_4r^2\) and \(A=s_4\), the condition \(F/A=k\) gives \(r^2=k\), so \(r=\sqrt{k}\).
Because \(G=s_5=s_4r\) and \(C=s_2=D+d\), the condition \(G/C=k\) gives \(\dfrac{s_4 r}{D+d}=k\), i.e. \(s_4 r = k(D+d)\). Using \(s_4=D+3d=Dk\) (from the first relation) and \(D+d = D\cdot\dfrac{k+2}{3}\), this becomes:
\[ Dk\sqrt{k} = k \cdot D\frac{k+2}{3} \implies 3\sqrt{k} = k+2 \]
Putting \(u=\sqrt{k}\): \(u^2-3u+2=0 \implies (u-1)(u-2)=0\), so \(u=1\) or \(u=2\). Since \(k>1\), we reject \(u=1\) and keep \(u=2\), so \(k=4\) and \(r=2\).

Step 6: Find all seven numbers.
With \(k=4\): \(d = \dfrac{D(4-1)}{3}=D\). So \(s_1=D,\ s_2=2D,\ s_3=3D,\ s_4=4D\). With \(r=2\): \(s_5=8D,\ s_6=16D,\ s_7=32D\). Since \(s_7=E=960\), \(32D=960 \implies D=30\).
So \(D=30,\ C=s_2=60,\ B=s_3=90,\ A=s_4=120,\ G=s_5=240,\ F=s_6=480,\ E=s_7=960\).
Check: \(A/D=120/30=4\), \(G/C=240/60=4\), \(F/A=480/120=4\). All equal 4, matching clue (vi). Also \(E=960\) as required.

Step 7: Rank A among all seven values.
Sorted from highest to lowest: \(E=960\) (1st), \(F=480\) (2nd), \(G=240\) (3rd), \(A=120\) (4th), \(B=90\) (5th), \(C=60\) (6th), \(D=30\) (7th). So A is the 4th highest, with value 120.

Final Answer:
"4th highest and 100" is wrong on the value, and "4th highest and 110" is also wrong on the value; A is 4th highest but its value is 120, which matches none of the listed pairs. \[ \boxed{\text{None of the above}} \]