# Next term of the series

Find the next term of the series $$79,95,25,39,99,$$____

$$(1)$$ $$117$$

$$(2)$$ $$81$$

$$(3)$$ $$243$$

Source: IBPS PO mains exam in India

• Don't forget to reference where this came from
Jul 22, 2019 at 12:19
• Actually this came in IBPS PO mains exam in India. Jul 22, 2019 at 12:22
• @mathmaniac. Which book did you use to get a source of all the passt year questions of SBI PO and IBPS PO? Apr 12, 2021 at 4:06
• @Rajarshi Koyal sorry for the very late reply. I was doing all these questions just for fun from gradeup in the semester break just for fun. Actually I am in academics and I am willing to do research at some day. Now I am pursuing Masters from ISI.:) Jul 23, 2021 at 16:17

I think the answer could be

$$81$$

Reasoning

For a 2-digit number $$N$$, let $$S(N)$$ be the sum of its digits and let $$T(N)$$ be the result of subtracting the first digit from the second digit.
The formula for the $$n$$th term of the sequence $$a_n$$ is given recursively by
$$a_n = a_{n-1} + \left[S\left(a_{n-1}\right) \times \left(T\left(a_{n-1}\right) - 1\right)\right]$$

Examples

$$79 \rightarrow 79 + \left[(7+9) \times (9-7-1)\right] = 79 + 16 = 95$$
$$95 \rightarrow 95 + \left[(9+5) \times (5-9-1)\right] = 95 - 70 = 25$$
$$25 \rightarrow 25 + \left[(2+5) \times (5-2-1)\right] = 25 + 14 = 39$$
$$39 \rightarrow 39 + \left[(3+9) \times (9-3-1)\right] = 39 + 60 = 99$$
$$99 \rightarrow 99 + \left[(9+9) \times (9-9-1)\right] = 99 - 18 = 81$$