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Difficulty: MediumArithmetic and Geometric Progressions (AP and GP)

The second term of a geometric progression (G.P.) with positive terms is 66 and the fifth term is 4848. What is the sum of the first 66 terms of the progression?

  1. A
    9393
  2. B
    9696
  3. 189189Answer
  4. D
    192192

Answer

The sum of the first 66 terms is 189189.
Using the nn-th term formula Tn=arn−1T_n = a r^{n-1}, we form two equations: ar=6a r = 6 and ar4=48a r^4 = 48. Dividing the fifth term equation by the second term equation gives r3=8r^3 = 8, so the common ratio r=2r = 2. Substituting r=2r = 2 back gives the first term a=3a = 3. Finally, applying the sum formula S6=3(26−1)2−1S_6 = \frac{3(2^6 - 1)}{2 - 1} gives 3×63=1893 \times 63 = 189.

Step-by-Step Solution

1
Set up equations for the given terms using the nn-th term formula Tn=arn−1T_n = a r^{n-1}.
T2=ar=6T_2 = a r = 6 and T5=ar4=48T_5 = a r^4 = 48.
The nn-th term of a G.P. is defined by Tn=arn−1T_n = a r^{n-1}.
2
Divide the equation for T5T_5 by the equation for T2T_2 to find the common ratio rr.
\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2.
Dividing eliminates the first term aa and allows solving for rr directly.
3
Substitute r=2r = 2 into T2=6T_2 = 6 to find the first term aa.
a(2) = 6 \implies a = 3.
Knowing rr allows calculating aa from any known term.
4
Calculate the sum of the first 66 terms using Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}.
S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = 189.
Applying the G.P. sum formula for n=6n = 6, a=3a = 3, and r=2r = 2 gives the total sum.

Key Concept

Geometric Progression term formula (Tn=arn−1T_n = a r^{n-1}) and sum of nn terms formula (Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}).
Estimated Time:1m 30s
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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