Question 34 of 51intermediate🔧 ApplyNumerical4 marks

Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nthn^{th} term.

Correct Answer

(Explicit formula) a_n = 2 ×\times 3n−13^{n-1}

(Recursive formula) t_1 = 2, t_n = 3t_{n-1} for n ≥\geq 2

(Which term) 8th term

Exercise: Exercise Set 8.3 | Q: 4 | (Page 20)
For More Understanding

Explanation

To find which term of the GP equals 4374, we first identify the first term 'a' and the common ratio 'r'. From the context, the nth term of a GP is given by a × r^(n-1), and the recursive formula is given as t_1 = a and t_n = r × t_{n-1} for n ≥\geq 2. By setting the explicit formula equal to 4374, we can solve for n.

Solution Steps

  1. Identify a and r: For the GP 2, 6, 18, …, the first term a = 2. The common ratio r = 6/2 = 3.

  2. Set up the equation: Let the nth term be 4374. Using the explicit formula a × r^(n-1), we have 2×32 \times 3^(n-1) = 4374.

  3. Solve for n: Divide both sides by 2: 3^(n-1) = 4374 / 2 = 2187. Since 2187 = 373^{7}, we get 3^(n-1) = 373^{7}. Equating the powers, n - 1 = 7, so n = 8. Thus, the 8th term is 4374.

  4. Write explicit formula: Substituting a = 2 and r = 3 into the general explicit formula a × r^(n-1), we get a_n = 2×32 \times 3^(n-1).

  5. Write recursive formula: Following the format t_1 = a and t_n = r × t_{n-1}, we get t_1 = 2 and t_n = 3t_{n-1} for n ≥\geq 2.

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