Question

Difficulty: MediumArithmetic and Geometric Progressions (AP and GP)

The 3rd3^{\text{rd}} term of a geometric progression is 1818 and the 6th6^{\text{th}} term is 486486. What is the sum of the first 55 terms of the progression?

  1. 242242Answer
  2. B
    162162
  3. C
    8080
  4. D
    728728

Answer

The sum of the first 55 terms is 242242.
Using Tn=arn1T_n = a r^{n-1}, we set up ar2=18a r^2 = 18 and ar5=486a r^5 = 486. Dividing these gives r3=27r^3 = 27, so r=3r = 3, which leads to a=2a = 2. Applying S5=a(r51)r1S_5 = \frac{a(r^5 - 1)}{r - 1} yields 2(2431)2=242\frac{2(243 - 1)}{2} = 242.

Step-by-Step Solution

1
Express the given terms using the geometric progression nthn^{\text{th}} term formula Tn=arn1T_n = a r^{n-1}.
T3=ar2=18T_3 = a r^2 = 18 and T6=ar5=486T_6 = a r^5 = 486.
This establishes a system of equations in terms of the first term aa and common ratio rr.
2
Divide the expression for T6T_6 by T3T_3 to determine rr.
\frac{a r^5}{a r^2} = \frac{486}{18} \implies r^3 = 27 \implies r = 3.
Dividing the equations eliminates aa and allows direct solution for the common ratio rr.
3
Substitute r=3r = 3 back into T3=ar2=18T_3 = a r^2 = 18 to solve for aa.
a (3)^2 = 18 \implies 9a = 18 \implies a = 2.
Determining the first term aa is required to evaluate the sum.
4
Calculate the sum of the first 55 terms using Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}.
S_5 = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242.
Applying the GP sum formula yields the required value.

Key Concept

Geometric Progression nth term and sum formulas
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