Progressions (A.P., G.P., H.P.) Questions and Answers

Q: The sum of the following infinite series is\[ \begin{aligned} &2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} \\ &+ \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \cdots \end{aligned} \]

The sum is 7/2.Apart from the leading 2, the terms are the powers of 1/2 and the powers of 1/3 interleaved. Summing each geometric series separately:\[ \begin{aligned} \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots &= \frac{1/2}{1 - 1/2} = 1 \\ \tfrac{1}{3} + \tfrac{1}{9} + \tfrac{1}{27} + \cdots &= \frac{1/3}{1 - 1/3} = \tfrac{1}{2} \\ 2 + 1 + \tfrac{1}{2} &= \tfrac{7}{2} \end{aligned} \]Rearranging is allowed because every term is positive. 11/3, 13/4 and 9/2 come from mis-summing one of the two series.

Q: If the sum of the first 20 consecutive positive odd numbers is divided by \(20^{2}\), the result is

The result is 1.The sum of the first \(n\) odd numbers is \(n^{2}\): \(1 + 3 + 5 + \cdots + (2n - 1) = n^{2}\). For \(n = 20\) the sum is 400, and \(400/20^{2} = 1\).Equivalently, the 20 odd numbers from 1 to 39 have average \((1 + 39)/2 = 20\), so their sum is \(20 \times 20\). The options 20, 2 and 1/2 come from dividing by 20 instead of \(20^{2}\) or from miscounting the terms.

Q: The 100th common term between the series 3 + 5 + 7 + 9 + ... and 3 + 6 + 9 + 12 + ... is

The 100th common term is 597.The first series is the odd numbers from 3, and the second is the multiples of 3. Their common terms are the odd multiples of 3: 3, 9, 15, 21, ..., an arithmetic progression with first term 3 and common difference 6.\[ \begin{aligned} T_n &= a + (n-1)d \\ T_{100} &= 3 + 99 \times 6 \\ &= 597 \end{aligned} \]Using \(n\) instead of \(n-1\) gives \(3 + 600 = 603\). 595 and 607 are not odd multiples of 3, and 605 is not a multiple of 3, so none of them can be a common term.

Q: Thirty-six vehicles are parked in a parking area in a single row. After the first car, there is one scooter. After the second car, there are two scooters and so on. Work out the number of scooters in the second half of the row.

There are 15 scooters in the second half.Each car is followed by a growing number of scooters: car 1 then 1 scooter, car 2 then 2 scooters, and so on. After the 7th car and its 7 scooters, the row holds 7 + (1 + 2 + ... + 7) = 7 + 28 = 35 vehicles, so the 36th vehicle is the 8th car.Numbering the positions: cars stand at 1, 3, 6, 10, 15, 21, 28 and 36, and every other position holds a scooter. The second half is positions 19 to 36. Its scooters are at 19 and 20 (2, after car 5), 22 to 27 (6, after car 6) and 29 to 35 (7, after car 7): 2 + 6 + 7 = 15.As a check, the row has 28 scooters in all, so the first half holds 13, and 13 + 15 = 28.

Q: In a flower bed, there are 23 rose plants in the first row, 21 in the second row, 19 in the third row and 5 rose plants in the last row. The number of rows in the flower bed is :

The flower bed has 10 rows.The rows form an arithmetic progression with first term 23 and common difference -2, ending at 5. The number of terms is:\[ \begin{aligned} n &= \frac{\text{first} - \text{last}}{\text{difference}} + 1 \\ &= \frac{23 - 5}{2} + 1 = 9 + 1 = 10 \end{aligned} \]Check: 23, 21, 19, 17, 15, 13, 11, 9, 7, 5 is ten rows. Forgetting the +1 gives 9; 15, 17 and 18 do not fit the sequence.

Q: Which of the following number is in the list of numbers 5, 11, 17, 23, ... ?

The number in the list is 179.The list 5, 11, 17, 23, ... increases by 6, so every term is one less than a multiple of 6: the nth term is 6n - 1. A number belongs to the list if adding 1 gives a multiple of 6.\[ \begin{aligned} 179 + 1 &= 180 = 6 \times 30 \\ 189 + 1 &= 190,\ 169 + 1 = 170,\ 159 + 1 = 160 \end{aligned} \]Only 180 is divisible by 6, so 179 is the 30th term. 190, 170 and 160 leave remainders when divided by 6, so 189, 169 and 159 are not in the list.

Q: The number of all three-digit numbers that are divisible by 7 is :

There are 128 three-digit numbers divisible by 7.The smallest three-digit multiple of 7 is 105 (7 × 15) and the largest is 994 (7 × 142). Counting the multiples from the 15th to the 142nd:\[ \begin{aligned} n &= \frac{994 - 105}{7} + 1 \\ &= 127 + 1 = 128 \end{aligned} \]Equivalently, 142 - 15 + 1 = 128. Forgetting to add 1 gives 127, the most common slip; 129 and 130 overcount.

Q: What will be the sum of the first 20 odd natural numbers ?

The sum is 400.The sum of the first n odd natural numbers is always \(n^{2}\): 1 = 1, 1 + 3 = 4, 1 + 3 + 5 = 9, and so on. For n = 20:\[ 1 + 3 + 5 + \dots + 39 = 20^{2} = 400 \]The same result comes from the arithmetic series formula: 20 terms with average (1 + 39)/2 = 20 give 20 × 20 = 400. 210 is the sum of the first 20 natural numbers, and 420 is the sum of the first 20 even numbers, 2 + 4 + ... + 40.

Q: Sum to \(n\) terms of an Arithmetic progression is \(2n^{2}+n\) then eighth term is

The eighth term is 31.The \(n\)th term is the difference of consecutive sums:\[ \begin{aligned} T_{8} &= S_{8} - S_{7} \\ &= (2 \cdot 64 + 8) - (2 \cdot 49 + 7) \\ &= 136 - 105 = 31 \end{aligned} \]Check: \(S_{1} = 3\) and \(S_{2} = 10\), so the progression is 3, 7, 11, ... with common difference 4, and \(T_{8} = 3 + 7 \times 4 = 31\). 136 is \(S_{8}\), the sum of eight terms, not the eighth term.

Q: \(a, b, c\) are said to be in Harmonic Progression if their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in Arithmetic progression. What would be the value of \(x\) for which 3, \(x\), 6 are in Harmonic Progression.

The value of \(x\) is 4.If 3, \(x\), 6 are in harmonic progression, their reciprocals are in arithmetic progression, so \(\frac{1}{x}\) is the average of \(\frac{1}{3}\) and \(\frac{1}{6}\):\[ \begin{aligned} \frac{2}{x} &= \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \\ x &= 4 \end{aligned} \]Equivalently, the harmonic mean is \(\frac{2ab}{a+b} = \frac{36}{9} = 4\). 4½ is the arithmetic mean of 3 and 6, the trap; 5 and 5½ fit neither mean.

Important Instructions

The sum of the following infinite series is

\[ \begin{aligned} &2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} \\ &+ \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \cdots \end{aligned} \]

A.11/3
B.7/2
C.13/4
D.9/2

Exam Year(s)

GATE Chemistry - 2024,

Answer: B

The sum is 7/2.

Apart from the leading 2, the terms are the powers of 1/2 and the powers of 1/3 interleaved. Summing each geometric series separately:

\[ \begin{aligned} \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots &= \frac{1/2}{1 - 1/2} = 1 \\ \tfrac{1}{3} + \tfrac{1}{9} + \tfrac{1}{27} + \cdots &= \frac{1/3}{1 - 1/3} = \tfrac{1}{2} \\ 2 + 1 + \tfrac{1}{2} &= \tfrac{7}{2} \end{aligned} \]

Rearranging is allowed because every term is positive. 11/3, 13/4 and 9/2 come from mis-summing one of the two series.

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If the sum of the first 20 consecutive positive odd numbers is divided by \(20^{2}\), the result is

A.1
B.20
C.2
D.1/2

Exam Year(s)

GATE Mathematics - 2024,

Answer: A

The result is 1.

The sum of the first \(n\) odd numbers is \(n^{2}\): \(1 + 3 + 5 + \cdots + (2n - 1) = n^{2}\). For \(n = 20\) the sum is 400, and \(400/20^{2} = 1\).

Equivalently, the 20 odd numbers from 1 to 39 have average \((1 + 39)/2 = 20\), so their sum is \(20 \times 20\). The options 20, 2 and 1/2 come from dividing by 20 instead of \(20^{2}\) or from miscounting the terms.

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The 100^th common term between the series 3 + 5 + 7 + 9 + ... and 3 + 6 + 9 + 12 + ... is

A.595
B.597
C.605
D.607

Exam Year(s)

Kerala PSC KAS - 2025,

Answer: B

The 100th common term is 597.

The first series is the odd numbers from 3, and the second is the multiples of 3. Their common terms are the odd multiples of 3: 3, 9, 15, 21, ..., an arithmetic progression with first term 3 and common difference 6.

\[ \begin{aligned} T_n &= a + (n-1)d \\ T_{100} &= 3 + 99 \times 6 \\ &= 597 \end{aligned} \]

Using \(n\) instead of \(n-1\) gives \(3 + 600 = 603\). 595 and 607 are not odd multiples of 3, and 605 is not a multiple of 3, so none of them can be a common term.

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Thirty-six vehicles are parked in a parking area in a single row. After the first car, there is one scooter. After the second car, there are two scooters and so on. Work out the number of scooters in the second half of the row.

A.10
B.12
C.15
D.17

Exam Year(s)

KPSC KAS - 2020,

Answer: C

There are 15 scooters in the second half.

Each car is followed by a growing number of scooters: car 1 then 1 scooter, car 2 then 2 scooters, and so on. After the 7th car and its 7 scooters, the row holds 7 + (1 + 2 + ... + 7) = 7 + 28 = 35 vehicles, so the 36th vehicle is the 8th car.

Numbering the positions: cars stand at 1, 3, 6, 10, 15, 21, 28 and 36, and every other position holds a scooter. The second half is positions 19 to 36. Its scooters are at 19 and 20 (2, after car 5), 22 to 27 (6, after car 6) and 29 to 35 (7, after car 7): 2 + 6 + 7 = 15.

As a check, the row has 28 scooters in all, so the first half holds 13, and 13 + 15 = 28.

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In a flower bed, there are 23 rose plants in the first row, 21 in the second row, 19 in the third row and 5 rose plants in the last row. The number of rows in the flower bed is :

A.10
B.15
C.17
D.18

Exam Year(s)

MPPSC State Service - 2024, MPPSC Forest Service - 2024,

Answer: A

The flower bed has 10 rows.

The rows form an arithmetic progression with first term 23 and common difference -2, ending at 5. The number of terms is:

\[ \begin{aligned} n &= \frac{\text{first} - \text{last}}{\text{difference}} + 1 \\ &= \frac{23 - 5}{2} + 1 = 9 + 1 = 10 \end{aligned} \]

Check: 23, 21, 19, 17, 15, 13, 11, 9, 7, 5 is ten rows. Forgetting the +1 gives 9; 15, 17 and 18 do not fit the sequence.

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Which of the following number is in the list of numbers 5, 11, 17, 23, ... ?

A.179
B.189
C.169
D.159

Exam Year(s)

MPPSC State Service - 2024, MPPSC Forest Service - 2024,

Answer: A

The number in the list is 179.

The list 5, 11, 17, 23, ... increases by 6, so every term is one less than a multiple of 6: the nth term is 6n - 1. A number belongs to the list if adding 1 gives a multiple of 6.

\[ \begin{aligned} 179 + 1 &= 180 = 6 \times 30 \\ 189 + 1 &= 190,\ 169 + 1 = 170,\ 159 + 1 = 160 \end{aligned} \]

Only 180 is divisible by 6, so 179 is the 30th term. 190, 170 and 160 leave remainders when divided by 6, so 189, 169 and 159 are not in the list.

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The number of all three-digit numbers that are divisible by 7 is :

A.127
B.128
C.129
D.130

Exam Year(s)

MPPSC State Service - 2024, MPPSC Forest Service - 2024,

Answer: B

There are 128 three-digit numbers divisible by 7.

The smallest three-digit multiple of 7 is 105 (7 × 15) and the largest is 994 (7 × 142). Counting the multiples from the 15th to the 142nd:

\[ \begin{aligned} n &= \frac{994 - 105}{7} + 1 \\ &= 127 + 1 = 128 \end{aligned} \]

Equivalently, 142 - 15 + 1 = 128. Forgetting to add 1 gives 127, the most common slip; 129 and 130 overcount.

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What will be the sum of the first 20 odd natural numbers ?

A.210
B.300
C.400
D.420

Exam Year(s)

MPPSC State Service - 2023, MPPSC Forest Service - 2023,

Answer: C

The sum is 400.

The sum of the first n odd natural numbers is always \(n^{2}\): 1 = 1, 1 + 3 = 4, 1 + 3 + 5 = 9, and so on. For n = 20:

\[ 1 + 3 + 5 + \dots + 39 = 20^{2} = 400 \]

The same result comes from the arithmetic series formula: 20 terms with average (1 + 39)/2 = 20 give 20 × 20 = 400. 210 is the sum of the first 20 natural numbers, and 420 is the sum of the first 20 even numbers, 2 + 4 + ... + 40.

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Sum to \(n\) terms of an Arithmetic progression is \(2n^{2}+n\) then eighth term is

A.136
B.36
C.131
D.31

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: D

The eighth term is 31.

The \(n\)th term is the difference of consecutive sums:

\[ \begin{aligned} T_{8} &= S_{8} - S_{7} \\ &= (2 \cdot 64 + 8) - (2 \cdot 49 + 7) \\ &= 136 - 105 = 31 \end{aligned} \]

Check: \(S_{1} = 3\) and \(S_{2} = 10\), so the progression is 3, 7, 11, ... with common difference 4, and \(T_{8} = 3 + 7 \times 4 = 31\). 136 is \(S_{8}\), the sum of eight terms, not the eighth term.

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10.

\(a, b, c\) are said to be in Harmonic Progression if their reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in Arithmetic progression. What would be the value of \(x\) for which 3, \(x\), 6 are in Harmonic Progression.

A.4½
B.4
C.5
D.5½

Exam Year(s)

TNPSC Group 1 - 2017, TNPSC - 2017,

Answer: B

The value of \(x\) is 4.

If 3, \(x\), 6 are in harmonic progression, their reciprocals are in arithmetic progression, so \(\frac{1}{x}\) is the average of \(\frac{1}{3}\) and \(\frac{1}{6}\):

\[ \begin{aligned} \frac{2}{x} &= \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \\ x &= 4 \end{aligned} \]

Equivalently, the harmonic mean is \(\frac{2ab}{a+b} = \frac{36}{9} = 4\). 4½ is the arithmetic mean of 3 and 6, the trap; 5 and 5½ fit neither mean.

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Tags: A.P. G.P. H.P. Progressions Questions and Answers || Arithmetic Progression MCQ Questions || Geometric Progression GK Questions || Harmonic Progression MCQ || Quantitative Aptitude Progressions GK

Progressions (A.P., G.P., H.P.) Questions and Answers

Important Instructions

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Other Subjects

Tags: A.P. G.P. H.P. Progressions Questions and Answers || Arithmetic Progression MCQ Questions || Geometric Progression GK Questions || Harmonic Progression MCQ || Quantitative Aptitude Progressions GK

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