Question 33 of 51advanced🔧 ApplyNumerical4 marks

A sequence is given by the recursive rule t1=2t_1 = 2, tn+1=3tn−2t_{n+1} = 3t_n - 2 for n≥1n \geq 1. Which term of the sequence is 730?

Correct Answer

7th term

Exercise: Exercise Set 8.3 | Q: 3 | (Page 20)
For More Understanding

Explanation

Following the method shown in the context for recursive sequences, we successively substitute the values of n to compute each term until we reach the value 730.

Solution Steps

  1. Step 1: Given t1=2t_1 = 2.

  2. Step 2: Calculate t2=3t1−2=3(2)−2=4t_2 = 3t_1 - 2 = 3(2) - 2 = 4.

  3. Step 3: Calculate t3=3t2−2=3(4)−2=10t_3 = 3t_2 - 2 = 3(4) - 2 = 10.

  4. Step 4: Calculate t4=3t3−2=3(10)−2=28t_4 = 3t_3 - 2 = 3(10) - 2 = 28.

  5. Step 5: Calculate t5=3t4−2=3(28)−2=82t_5 = 3t_4 - 2 = 3(28) - 2 = 82.

  6. Step 6: Calculate t6=3t5−2=3(82)−2=244t_6 = 3t_5 - 2 = 3(82) - 2 = 244.

  7. Step 7: Calculate t7=3t6−2=3(244)−2=730t_7 = 3t_6 - 2 = 3(244) - 2 = 730. Thus, 730 is the 7th term of the sequence.