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Question

If the first term of a geometric progression is 2 and the common ratio is 3, then what will be the fifth term of the geometric progression?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
162

Geometric Progression Solution

Problem Setup

We are asked to find the fifth term of a geometric progression (GP). We are given the first term ($a$) and the common ratio ($r$).

  • First term ($a$): 2
  • Common ratio ($r$): 3
  • Term to find ($n$): 5

Geometric Progression Formula

The formula to calculate the $n$-th term ($a_n$) of a geometric progression is:

$a_n = a \cdot r^{(n-1)}$

Calculation Steps

Substitute the given values into the formula:

  1. Identify the values: $a = 2$, $r = 3$, $n = 5$.
  2. Apply the formula: $a_5 = a \cdot r^{(5-1)}$.
  3. Simplify the exponent: $a_5 = 2 \cdot r^4$.
  4. Substitute the common ratio: $a_5 = 2 \cdot 3^4$.
  5. Calculate the power: $3^4 = 3 \times 3 \times 3 \times 3 = 81$.
  6. Perform the final multiplication: $a_5 = 2 \times 81$.
  7. Result: $a_5 = 162$.

Final Answer

The fifth term of the geometric progression is 162.

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Important Questions from Geometric Progression

  1. In a school of 750 boys, 34% are Muslims, 42% Hindus, 12% Sikhs and the rest belong to other communities. How many boys are from other community?

  2. Ms. Tina has 2 parents, 4 grand parents, 8 great grand parents and so on. Find the number of her ancestors during 10 (ten) generations preceeding her own.

  3. A square is drawn by joining the midpoints of the sides of a given square. In the same way, this process continues indefinitely. If side of the first square is 4 cm, than the sum of the area of all the squares are equal to

  4. Third term of sequence an = (-1)n(n − 7) is :

  5. 811/3 × 811/9 × 811/27 × 811/81 × ______ up to infinite term will be equal to

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