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
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$\begingroup$ Don't forget to reference where this came from $\endgroup$– AdamCommented Jul 22, 2019 at 12:19
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$\begingroup$ Actually this came in IBPS PO mains exam in India. $\endgroup$– math maniac.Commented Jul 22, 2019 at 12:22
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$\begingroup$ @mathmaniac. Which book did you use to get a source of all the passt year questions of SBI PO and IBPS PO? $\endgroup$– Rajorshi KoyalCommented Apr 12, 2021 at 4:06
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$\begingroup$ @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.:) $\endgroup$– math maniac.Commented Jul 23, 2021 at 16:17
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1 Answer
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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
$79 \rightarrow 79 + \left[(7+9) \times (9-7-1)\right] = 79 + 16 = 95$
$95 \rightarrow 95 + \left[(9+5) \times (5-9-1)\right] = 95 - 70 = 25$
$25 \rightarrow 25 + \left[(2+5) \times (5-2-1)\right] = 25 + 14 = 39$
$39 \rightarrow 39 + \left[(3+9) \times (9-3-1)\right] = 39 + 60 = 99$
$99 \rightarrow 99 + \left[(9+9) \times (9-9-1)\right] = 99 - 18 = 81$
answered Jul 22, 2019 at 14:33