In each of the following questions, a number series is given with one term missing. Choose the correct alternative that will continue the same pattern.
13, 35, 57, 79, 911, ?
Answer: C
Explanation:
The terms of the given series are numbers formed by joining together consecutive
odd numbers in order i.e. 1 and 3, 3 and 5, 5 and 7, 7 and 9, 9 and 11, .....
So, missing term = number formed by joining 11 and 13 = 1113.
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1, 4, 10, 22, 46, ?
Answer: C
Correct answer: 94
Each term is twice the one before, plus two: \( 2(1)+2=4 \), \( 2(4)+2=10 \), \( 2(10)+2=22 \), \( 2(22)+2=46 \), so the next is \[ 2(46)+2 = \mathbf{94} \]
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120, 99, 80, 63, 48, ?
Answer: A
Correct answer: 35
Each term is one less than a square, counting downwards: \[ 11^2-1=120,\; 10^2-1=99,\; 9^2-1=80,\; 8^2-1=63,\; 7^2-1=48 \] so the next is \( 6^2 - 1 = \mathbf{35} \).
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589654237, 89654237, 8965423, 965423, ?
Answer: D
Correct answer: 96542
A digit is struck off each time, alternately from the front and from the back: 589654237 loses its first digit, the result loses its last, and so on. Taking the last digit off 965423 gives 96542.
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3, 10, 101,?
Answer: C
Correct answer: 10202
Each term is the square of the one before it, plus one: \[ 3^2+1 = 10,\quad 10^2+1 = 101,\quad 101^2+1 = \mathbf{10202} \]
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In the series 2, 6, 18, 54, …… what will be the 8th term ?
Answer: B
Correct answer: 4374
The series multiplies by 3 each time, so the \( n \)th term is \( 2 \times 3^{\,n-1} \). The eighth term is \[ 2 \times 3^{7} = 2 \times 2187 = \mathbf{4374} \]
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125,80,45,20,?
Answer: A
Correct answer: 5
The amounts taken away shrink by ten each time: \( 125-80=45 \), \( 80-45=35 \), \( 45-20=25 \). The next subtraction is 15, giving \[ 20 - 15 = \mathbf{5} \]
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66, 36, 18, ?
Answer: C
Correct answer: 8
Each term is the product of the digits of the one before it: \( 6 \times 6 = 36 \), \( 3 \times 6 = 18 \), and then \[ 1 \times 8 = \mathbf{8} \]
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In the series 3, 9, 15, ...... what will be the 21st term?
Answer: C
Correct answer: 123
The series rises by 6 each time, so the \( n \)th term is \( 3 + (n-1)\times 6 \). For the 21st term, \[ 3 + 20 \times 6 = \mathbf{123} \]
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28, 33, 31, 36, ?, 39
Answer: B
Correct answer: 34
The series alternately adds 5 and subtracts 2: \( 28 + 5 = 33 \), \( 33 - 2 = 31 \), \( 31 + 5 = 36 \), so the next step subtracts 2: \[ 36 - 2 = \mathbf{34} \] and 34 + 5 = 39, as the series shows.
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