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Difficulty: Very hardArithmetic and Geometric Progressions (AP and GP)

The 1st1^{\text{st}}, 2nd2^{\text{nd}}, and 5th5^{\text{th}} terms of an arithmetic progression (A.P.) with a non-zero common difference form three consecutive terms of a geometric progression (G.P.). If the sum of the first 44 terms of the A.P. is 4040, what is the 5th5^{\text{th}} term of the G.P.?

  1. 4052\frac{405}{2}Answer
  2. B
    12152\frac{1215}{2}
  3. C
    162162
  4. D
    452\frac{45}{2}

Answer

4052\frac{405}{2}
By setting up the geometric mean property (a+d)2=a(a+4d)(a+d)^2 = a(a+4d), we find d=2ad = 2a, which establishes that the G.P. has a common ratio r=3r = 3. Substituting d=2ad = 2a into the A.P. sum formula S4=2[2a+3d]=16a=40S_4 = 2[2a + 3d] = 16a = 40 gives a=52a = \frac{5}{2}. Finally, evaluating the 5th5^{\text{th}} term of the G.P. using ar4=52×34a r^4 = \frac{5}{2} \times 3^4 yields 4052\frac{405}{2}.

Step-by-Step Solution

1
Express the given A.P. terms in terms of first term aa and common difference dd, and set up the G.P. condition.
The terms are T1=aT_1 = a, T2=a+dT_2 = a + d, and T5=a+4dT_5 = a + 4d. Since they form a G.P., (a+d)2=a(a+4d)(a + d)^2 = a(a + 4d).
Three terms x,y,zx, y, z form a G.P. if y2=xzy^2 = xz.
2
Solve for the relationship between dd and aa.
a2+2ad+d2=a2+4ad  ⟹  d2=2ad  ⟹  d=2aa^2 + 2ad + d^2 = a^2 + 4ad \implies d^2 = 2ad \implies d = 2a (since d≠0d \neq 0).
Expanding and simplifying the equation yields the ratio of dd to aa.
3
Determine the common ratio rr of the G.P.
r=T2T1=a+da=a+2aa=3r = \frac{T_2}{T_1} = \frac{a + d}{a} = \frac{a + 2a}{a} = 3.
The common ratio is the quotient of consecutive terms of the G.P.
4
Use the sum of the first 44 terms of the A.P. to find aa.
S4=42[2a+(4−1)d]=2[2a+3(2a)]=16a=40  ⟹  a=4016=52S_4 = \frac{4}{2}[2a + (4-1)d] = 2[2a + 3(2a)] = 16a = 40 \implies a = \frac{40}{16} = \frac{5}{2}.
Applying Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d] with S4=40S_4 = 40 allows solving for aa.
5
Calculate the 5th5^{\text{th}} term of the G.P.
G5=g1⋅r5−1=a⋅r4=52⋅34=52⋅81=4052G_5 = g_1 \cdot r^{5-1} = a \cdot r^4 = \frac{5}{2} \cdot 3^4 = \frac{5}{2} \cdot 81 = \frac{405}{2}.
The nthn^{\text{th}} term of a G.P. is gn=g1rn−1g_n = g_1 r^{n-1}.

Key Concept

Connecting Arithmetic and Geometric Progressions using term definitions and sum formulas.
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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