Writing x = ar³, y = ar⁹ and z = ar¹⁵, we compute y² = (ar⁹)² = a²r¹⁸. Also x x z = (ar³) x (ar¹⁵) = a²r¹⁸. Since y² = x x z, x, y, z form a GP.
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Starting with 30 bacteria and doubling factor 2, the count after n hours is 30 x 2ⁿ. Thus, after 2 hours there are 120, after 4 hours 480 and after n hours 30 x 2ⁿ.
Since t7 – t5 = 12, we have 2d = 12, giving d = 6. With third term a + 2d = 16, substituting d gives a = 4. The required AP is 4, 10, 16, 22, ….
Starting with t1 equal to 2, the recursive rule gives the sequence 2, 4, 10, 28, 82, 244, 730. Therefore, 730 occurs at the seventh position. Hence, 730 is the seventh term of the sequence.
The first term is 5 and the common ratio is 5. Using the GP formula, tn equals a multiplied by r raised to n minus 1. Therefore, tn equals 5 raised to n. The 10th term is 9765625.