3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Don't forget to reference where this came from $\endgroup$
    – Adam
    Jul 22, 2019 at 12:19
  • $\begingroup$ Actually this came in IBPS PO mains exam in India. $\endgroup$ Jul 22, 2019 at 12:22
  • $\begingroup$ @mathmaniac. Which book did you use to get a source of all the passt year questions of SBI PO and IBPS PO? $\endgroup$ Apr 12, 2021 at 4:06
  • $\begingroup$ @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.:) $\endgroup$ Jul 23, 2021 at 16:17

1 Answer 1

3
$\begingroup$

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

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.