Find out the wrong term 3- 2- 8- 9- 13- 22- 18- 32- 23- 42- Maths - Q-A

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Find out the wrong term $3, 2, 8, 9, 13, 22, 18, 32, 23, 42$

$8$

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$9$

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$13$

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$22$

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Similar questions

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

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

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

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

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

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

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

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

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

=

\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

10^{th}

\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 terms

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