Question 30 of 51intermediate🔧 ApplyShort Answer2 marks

Can you find a recursive rule for the formula tn=3×10n−1t_n = 3 \times 10^{n-1} that generates the geometric progression 3, 30, 300, 3000, …?

Correct Answer

To find the recursive rule for the given geometric progression 3, 30, 300, 3000, …, we first identify the first term and the common ratio. The first term of the sequence is t1=3t_1 = 3.

Next, we find the common ratio by dividing any term by its previous term. For example, 30/3=1030/3 = 10, 300/30=10300/30 = 10, and so on. The common ratio is 10. This means each term is obtained by multiplying the previous term by 10. Describing a sequence by relating terms to previous terms is known as a recursive rule.

Therefore, the recursive formula for this sequence is t1=3t_1 = 3 and tn=10tn−1t_n = 10t_{n-1} for n≥2n \geq 2.

Exercise: Exercise | Q: 2 | (Page 15)
For More Understanding

Explanation

The textbook explains that a recursive rule relates a term to its previous term. For a geometric progression with the nth term formula tn=a×rn−1t_n = a \times r^{n-1}, the recursive formula is t1=at_1 = a and tn=r×tn−1t_n = r \times t_{n-1} for n≥2n \geq 2. Here, a=3a = 3 and r=10r = 10, so the recursive rule is t1=3t_1 = 3 and tn=10tn−1t_n = 10t_{n-1} for n≥2n \geq 2.

Solution Steps

  1. Step 1: Identify the first term of the sequence, t1=3t_1 = 3.

  2. Step 2: Find the common ratio by dividing consecutive terms, 30/3=1030/3 = 10.

  3. Step 3: Write the recursive rule using the first term and common multiplier: t1=3t_1 = 3 and tn=10tn−1t_n = 10t_{n-1} for n≥2n \geq 2.

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