Arithmetic & Geometric Progression — Multiple Choice Questions

The 10th term of the A.P. 5, 8, 11, 14, .... is

1. 32

2. 35

3. 38

4. 185

Answer

The above series is an A.P. with first term = a = 5 and common difference = d = 8 - 5 = 3.

We know that

    a_n = a + (n - 1)d
∴ a₁₀ = 5 + (10 - 1) × 3 = 5 + 9 × 3 = 5 + 27 = 32.

Hence, Option 1 is the correct option.

The 30th term of the A.P. 10, 7, 4, ... is

1. 87

2. 77

3. -77

4. -87

Answer

The above series is an A.P. with first term = a = 10 and common difference = d = 7 - 10 = -3.

We know that

    a_n = a + (n - 1)d
∴ a₃₀ = 10 + (30 - 1) × (-3) = 10 + 29 × (-3) = 10 - 87 = -77.

Hence, Option 3 is the correct option.

The 11th term of the A.P. $-3, -\dfrac{1}{2}, 2, ...$ is

1. 28

2. 22

3. -38

4. $-48\dfrac{1}{2}$

Answer

The above series is an A.P. with first term = a = -3 and common difference = $-\dfrac{1}{2} - (-3) = -\dfrac{1}{2} + 3 = -\dfrac{-1 + 6}{2} = \dfrac{5}{2}$.

We know that

a_n = a + (n - 1)d

$\therefore a_{11} = -3 + (11 - 1) \times \dfrac{5}{2} \\[1em] = -3 + 10 \times \dfrac{5}{2} \\[1em] = -3 + 25 \\[1em] = 22.$

Hence, Option 2 is the correct option.

The 15th term from the last of the A.P. 7, 10, 13, ... , 130 is

1. 49

2. 85

3. 88

4. 110

Answer

Here, common difference = d = 10 - 7 = 3 and last term = l = 130.

We know that the nth term from the end is given by the formula:

    nth term from end = l - (n - 1)d
∴ 11th term from the end = 130 - (15 - 1) × 3 = 130 - 14 × 3 = 130 - 42 = 88.

Hence, Option 3 is the correct option.

If the common difference of the A.P. is 5, then a₁₈ - a₁₃ is

1. 5

2. 20

3. 25

4. 30

Answer

Given d = 5,

We know that

    a_n = a + (n - 1)d
∴ a₁₈ = a + (18 - 1) × 5 = a + 17 × 5 = a + 85.
    a₁₃ = a + (13 - 1) × 5 = a + 12 × 5 = a + 60.

∴ a₁₈ - a₁₃ = a + 85 - (a + 60) = a - a + 85 - 60 = 25.

Hence, Option 3 is the correct option.

In an A.P., if a₁₈ - a₁₄ = 32 then the common difference is

1. 8

2. -8

3. -4

4. 4

Answer

Given, a₁₈ - a₁₄ = 32.

We know that

    a_n = a + (n - 1)d
∴ a₁₈ = a + (18 - 1) × d = a + 17d.
    a₁₄ = a + (14 - 1) × d = a + 13d.

∴ a₁₈ - a₁₄ = a + 17d - (a + 13d) = a - a + 17d - 13d = 4d.

⇒ 4d = 32
⇒ d = 8.

Hence, Option 1 is the correct option.

In an A.P., if d = -4, n = 7, a_n = 4, then a is

1. 6

2. 7

3. 20

4. 28

Answer

We know that

    a_n = a + (n - 1)d
∴ 4 = a + (7 - 1) × (-4)
⇒ 4 = a + 6 × (-4)
⇒ 4 = a - 24
⇒ a = 24 + 4
⇒ a = 28.

Hence, Option 4 is the correct option.

In an A.P., if a = 3.5, d = 0, n = 101, then a_n will be

1. 0

2. 3.5

3. 103.5

4. 104.5

Answer

We know that

    a_n = a + (n - 1)d
∴ a_n = 3.5 + (101 - 1) × 0
⇒ a_n = 3.5 + 100 × 0
⇒ a_n = 3.5

Hence, Option 2 is the correct option.

Which term of the A.P. 21, 42, 63, 84, ... is 210?

1. 9th

2. 10th

3. 11th

4. 12th

Answer

Here, a = 21, d = 42 - 21 = 21.

Let nth term be 210, so a_n = 210.

We know that

    a_n = a + (n - 1)d
∴ 210 = 21 + (n - 1) × 21
⇒ 210 = 21 + 21n - 21
⇒ 21n = 210
⇒ n = 10

Hence, Option 2 is the correct option.

If the last term of A.P. 5, 3, 1, -1, .... is -41, then the A.P. consists of

1. 46 terms

2. 25 terms

3. 24 terms

4. 23 terms

Answer

Here, a = 5, d = 3 - 5 = -2.

Let nth term be -41, so a_n = -41.

We know that

    a_n = a + (n - 1)d
∴ -41 = 5 + (n - 1) × (-2)
⇒ -41 = 5 - 2n + 2
⇒ 7 - 2n = -41
⇒ 2n = 7 + 41
⇒ 2n = 48
⇒ n = 24.

Hence, Option 3 is the correct option.

If k - 1, k + 1 and 2k + 3 are in A.P. , then the value of k is

1. -2

2. 0

3. 2

4. 4

Answer

We know that in an A.P.,

Common difference = d = any term - preceding term

∴ 2k + 3 - (k + 1) = k + 1 - (k - 1)
⇒ 2k - k + 3 - 1 = k - k + 1 - (-1)
⇒ k + 2 = 2
⇒ k = 0.

Hence, Option 2 is the correct option.

The 21st term of an A.P. whose first two terms are -3 and 4 is

1. 17

2. 137

3. 143

4. -143

Answer

Given a = -3 and a₂ = 4

We know that

    a_n = a + (n - 1)d
∴ a₂ = -3 + (2 - 1) × d
⇒ 4 = -3 + d
⇒ d = 4 + 3
⇒ d = 7.

    a₂₁ = -3 + (21 - 1) × 7
⇒ a₂₁ = -3 + 20 × 7
⇒ a₂₁ = -3 + 140
⇒ a₂₁ = 137.

Hence, Option 2 is the correct option.

If the first term of an A.P. is -5 and the common difference is 2, then the sum of its first 6 terms is

1. 0

2. 5

3. 6

4. 15

Answer

Given, a = -5 and d = 2.

We know that,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_6 = \dfrac{6}{2}[2 \times -5 + (6 - 1) \times 2] \\[1em] = 3[-10 + 5 \times 2] \\[1em] = 3[-10 + 10] \\[1em] = 3 \times 0 \\[1em] = 0.$

Hence, Option 1 is the correct option.

The sum of 25 terms of the A.P., $-\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{2}{3}, ...$ is

1. 0

2. $-\dfrac{2}{3}$

3. $-\dfrac{50}{3}$

4. -50

Answer

The above series is an A.P. with first term = $-\dfrac{2}{3}$ and common difference = d = $-\dfrac{2}{3} - \Big(-\dfrac{2}{3}\Big) = 0$

We know that,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_{25} = \dfrac{25}{2}\Big[2 \times -\dfrac{2}{3} + (25 - 1) \times 0\Big] \\[1em] = \dfrac{25}{2}\Big[-\dfrac{4}{3} + 24 \times 0\Big] \\[1em] = \dfrac{25}{2} \times -\dfrac{4}{3} \\[1em] = -\dfrac{50}{3}.$

Hence, Option 3 is the correct option.

In an A.P. if a = 1, a_n = 20 and S_n = 399, then n is

1. 19

2. 21

3. 38

4. 42

Answer

Given a = 1, a_n = 20 and S_n = 399.

We know that

    a_n = a + (n - 1)d
∴ a_n = 1 + (n - 1) × d
⇒ 20 = 1 + (n - 1)d
⇒ (n - 1)d = 20 - 1
⇒ (n - 1)d = 19
⇒ d = $\dfrac{19}{n - 1}$.

The formula for sum of A.P. is given by,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_n = \dfrac{n}{2}[2 \times 1 + (n - 1) \times d] \\[1em] \Rightarrow 399 = \dfrac{n}{2}[2 + (n - 1) \times \dfrac{19}{n - 1}] \\[1em] \Rightarrow 399 = \dfrac{n}{2}[2 + 19] \\[1em] \Rightarrow \dfrac{n}{2} \times 21 = 399 \\[1em] \Rightarrow n = \dfrac{399 \times 2}{21} \\[1em] \Rightarrow n = \dfrac{798}{21} \\[1em] \Rightarrow n = 38.$

Hence, Option 3 is the correct option.

In an A.P., if a = -5, l = 21 and S = 200, then n is equal to

1. 50

2. 40

3. 32

4. 25

Answer

We know that

    l = a + (n - 1)d
∴ 21 = -5 + (n - 1) × d
⇒ (n - 1)d = 21 + 5
⇒ (n - 1)d = 26
⇒ d = $\dfrac{26}{n - 1}$.

The formula for sum of A.P. is given by,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_n = \dfrac{n}{2}[2 \times (-5) + (n - 1) \times \dfrac{26}{n - 1}] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[-10 + (n - 1) \times \dfrac{26}{n - 1}] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[-10 + 26] \\[1em] \Rightarrow \dfrac{n}{2} \times 16 = 200 \\[1em] \Rightarrow 8n = 200 \\[1em] \Rightarrow n = \dfrac{200}{8} \\[1em] \Rightarrow n = 25.$

Hence, Option 4 is the correct option.

The sum of first five multiples of 3 is

1. 45

2. 55

3. 65

4. 75

Answer

First five multiples of 3 are : 3, 6, 9, 12, 15.

The above series is an A.P. with first term = a = 3 and common difference = d = 3.

The formula for sum of A.P. is given by,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S_5 = \dfrac{5}{2}[2 \times 3 + (5 - 1) \times 3] \\[1em] = \dfrac{5}{2}[6 + 4 \times 3] \\[1em] = \dfrac{5}{2}[6 + 12] \\[1em] = \dfrac{5}{2} \times 18 \\[1em] = 45.$

Hence, Option 1 is the correct option.

The number of two digit numbers which are divisible by 3 is

1. 33

2. 31

3. 30

4. 29

Answer

Two digit numbers which are divisible by 3 are 12, 15, 18, 21, ..... , 99.

The above series is an A.P. with first term = a = 12 and common difference = d = 15 - 12 = 3.

Let 99 be the nth term, we know that

    a_n = a + (n - 1)d
∴ 99 = 12 + (n - 1) × 3
⇒ 99 = 12 + 3n - 3
⇒ 3n + 9 = 99
⇒ 3n = 90
⇒ n = 30.

Hence, Option 3 is the correct option.

The number of multiples of 4 that lie between 10 and 250 is

1. 62

2. 60

3. 59

4. 55

Answer

The multiples of 4 that lie between 10 and 250 are 12, 16, 20, ....., 248.

The above series is an A.P. with first term = a = 12 and common difference = d = 16 - 12 = 4.

Let 248 be the nth term, we know that

    a_n = a + (n - 1)d
∴ 248 = 12 + (n - 1) × 4
⇒ 248 = 12 + 4n - 4
⇒ 4n = 248 - 8
⇒ 4n = 240
⇒ n = 60.

Hence, Option 2 is the correct option.

The sum of first 10 even whole numbers is

1. 110

2. 90

3. 55

4. 45

Answer

The first 10 even whole numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16 , 18.

Sum of an A.P. is given by,

$S_n = \dfrac{n}{2}(a + l) \\[1em] \therefore S_{10} = \dfrac{10}{2}(0 + 18) \\[1em] = 5 \times 18 \\[1em] = 90.$

Hence, Option 2 is the correct option.

The 11th term of the G.P. $\dfrac{1}{8}, -\dfrac{1}{4}, 2, -1, ....$ is

1. 64

2. -64

3. 128

4. -128

Answer

Given, a = $\dfrac{1}{8}$ and common ratio = r = $\dfrac{-\dfrac{1}{4}}{\dfrac{1}{8}} = -2.$

We know that

    a_n = ar^{n - 1}

$\therefore a_{11} = \dfrac{1}{8}(-2)^{11 - 1} \\[1em] = \dfrac{1}{8}(-2)^{10} \\[1em] = \dfrac{1}{8} \times 2^{10} \\[1em] = \dfrac{2^{10}}{2^3} \\[1em] = 2^7 \\[1em] = 128.$

Hence, Option 3 is the correct option.

The 5th term from the end of the G.P. 2, 6, 18, ...., 13122 is

1. 162

2. 486

3. 54

4. 1458

Answer

The above series is a G.P. with a = 2 and d = $\dfrac{6}{2} = 3.$

nth from end = $l\Big(\dfrac{1}{r}\Big)^{n - 1}$

∴ 5th term from end = $13122 \times \Big(\dfrac{1}{3}\Big)^{5 - 1}$

$= 13122 \times \dfrac{1}{3^4} \\[1em] = \dfrac{13122}{81} \\[1em] = 162.$

Hence, Option 1 is the correct option.

If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is

1. -1

2. -4

3. 1

4. 4

Answer

In a G.P.,

$\dfrac{\text{ any term }}{\text{ preceding term }} = \text{ common ratio }. \\[1em] \therefore \dfrac{2(k + 1)}{k} = \dfrac{3(k + 1)}{2(k + 1)} \text{ (Eq 1) } \\[1em] \Rightarrow (2k + 2)(2k + 2) = k(3k + 3) \\[1em] \Rightarrow 4k^2 + 4k + 4k + 4 = 3k^2 + 3k \\[1em] \Rightarrow 4k^2 - 3k^2 + 8k - 3k + 4 = 0 \\[1em] \Rightarrow k^2 + 5k + 4 = 0 \\[1em] \Rightarrow k^2 + k + 4k + 4 = 0 \\[1em] \Rightarrow k(k + 1) + 4(k + 1) = 0 \\[1em] \Rightarrow (k + 1)(k + 4) =$

But $k + 1$ ≠ 0 as that will not satisfy Eq 1

$\Rightarrow k + 4 = 0 \\[1em] \Rightarrow k = -4.$

Hence, Option 2 is the correct option.