ISRO EC 2023 Q.92: solution and answer

ISRO 2023 · ECQ.92 · 2 marks · MCQ

Find out the wrong number in the series.
7, 8, 18, 57, 228, 1165, 6996

(A)
7
(B)
8
(C)
6996
(D)
228

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The wrong number is $228$ , option (D); the term that fits the rule is $232$ .

Official answer: DSource: ISRO's revised and final answer key of 16 February 2024

Check how each term follows the one before it: $7 \times 1 + 1 = 8$ , $8 \times 2 + 2 = 18$ , $18 \times 3 + 3 = 57$ .
The rule is $a_{n + 1} = n a_{n} + n = n (a_{n} + 1)$ for $n = 1, 2, 3, \dots$
Apply it to the fourth term: $a_{5} = 4 \times 57 + 4 = 4 \times 58 = 232$ . The series prints $228$ , so $228$ breaks the rule.
Continue from the corrected value $232$ : $a_{6} = 5 \times 232 + 5 = 5 \times 233 = 1165$ , which matches the printed term.
Then $a_{7} = 6 \times 1165 + 6 = 6 \times 1166 = 6996$ , which also matches the printed term.
So $7$ , $8$ , $18$ , $57$ , $1165$ and $6996$ all fit the rule, and only $228$ is out of place (it should be $232$ ).

KEY IDEA

When each term is a multiple of the previous one plus a growing amount, find the rule from the early terms, then test whether the later terms continue from it. The wrong term is the one that fails while the terms after it still agree with the rule applied to the correct value.

WATCH OUT FOR

Seeing that the terms roughly multiply by $2, 3, 4, 5, 6$ and stopping there, without computing the exact value ( $232$ , not $228$ ). Another slip is to blame $1165$ or $6996$ , because they are large, when both follow exactly from $232$ .

From the ISRO EC 2023 paper. Also Q.92 of the ISRO Scientist/Engineer Computer Science 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 5 ISRO Numerical questions, with notes

ISRO EC 2023 Q.92: solution and answer

Answer and solution

More ISRO Numerical questions