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, ….
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Given a + 10d = 38 and a + 15d = 73, subtracting gives 5d = 35, so d = 7 and a = -32. Using tn = a + (n – 1)d, the 31st term is -32 + 30 ...
The bounce heights form a GP with first term 48 metres and common ratio 0.6. Therefore, the fifth bounce height is 48 multiplied by 0.6 raised to 4, which equals 6.2208 metres. Hence, the required height is 6.2208 metres.
The first term is 2 and the common ratio is 3. The explicit formula is tn equals 2 multiplied by 3 raised to n minus 1. Since 4374 equals 2 multiplied by 3 raised to 7, it is the eighth ...
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.