3, 2, 8, 9, 13, 22, 18, 32, 23, 42

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Solution

The correct option is B 9
The given sequence is a combination of two series:
I. 3, 8, 13, 18, 23 and II. 2, 9, 22, 32, 42
The pattern in I is + 5, and the pattern in II is + 10. So, in II, 9 is wrong and must be replaced by (2 + 10) i.e. 12.

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Similar questions

Q. In following questions, one term in number series is incorrect.
Find out the incorrect number 3, 2, 8, 9, 13, 22, 18, 32, 23, 42

Q. Simplify:
(i)

(3^{2} + 2^{2}) \times {(\frac{1}{2})}^{3}

(ii)

(3^{2} - 2^{2}) \times {(\frac{2}{3})}^{- 3}

(iii)

[{(\frac{1}{3})}^{- 3} - {(\frac{1}{2})}^{- 3}] \div {(\frac{1}{4})}^{- 3}

(iv)

(2^{2} + 3^{2} - 4^{2}) \div {(\frac{3}{2})}^{2}

In this case how many time intervals can Sid calculate?

$\begin{matrix} B u s & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 T i m e (h r : m i n) & 8 : 15 & 8 : 18 & 8 : 23 & 8 : 29 & 8 : 32 & 8 : 40 & 8 : 45 & 8 : 52 & 8 : 54 & 9 : 00 & 9 : 07 & 9 : 13 & 9 : 20 & 9 : 25 & 9 : 32 \end{matrix}$

1^{3} + 1^{2} + 1 + 2^{3} + 2^{2} + 2 + 3^{3} + 3^{2} + 3 + . . . 3 n

terms

=

Q. The sum

\frac{3 \times 1^{3}}{1^{2}} + \frac{5 \times (1^{3} + 2^{3})}{1^{2} + 2^{2}} + \frac{7 \times (1^{3} + 2^{3} + 3^{3})}{1^{2} + 2^{2} + 3^{2}} + \dots

upto

10^{th}

term, is :

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3, 2, 8, 9, 13, 22, 18, 32, 23, 42

The correct option is B 9The given sequence is a combination of two series:I. 3, 8, 13, 18, 23 and II. 2, 9, 22, 32, 42The pattern in I is + 5, and the pattern in II is + 10. So, in II, 9 is wrong and must be replaced by (2 + 10) i.e. 12.

The correct option is B 9
The given sequence is a combination of two series:
I. 3, 8, 13, 18, 23 and II. 2, 9, 22, 32, 42
The pattern in I is + 5, and the pattern in II is + 10. So, in II, 9 is wrong and must be replaced by (2 + 10) i.e. 12.