Arithmetic and Geometric Progressions (AP and GP) Practice Questions — JAMB UTME — Page 2

Question 21Question →

An arithmetic progression (A.P.) has a first term of $5$ and a common difference of $-3$ . What is the $7^{\text{th}}$ term of the progression?

-13

-16

-19

23

Show answer & explanation

Answer: $-13$

Answer

-13

Using the

n^{\text{th}}

term formula

T_n = a + (n - 1)d

with

a = 5

,

d = -3

, and

n = 7

, we get

T_7 = 5 + (6)(-3) = -13

.

Step-by-Step Solution

1

Identify the given parameters of the A.P.

First term

a = 5

, common difference

d = -3

, and position

n = 7

.

These parameters are directly specified in the problem.

2

Apply the

n^{\text{th}}

term formula

T_n = a + (n - 1)d

.

T_7 = 5 + (7 - 1)(-3) = 5 + 6(-3)

.

The common difference is added

(n-1)

times to the first term.

3

Perform the multiplication and addition.

T_7 = 5 - 18 = -13

.

Simplifying the arithmetic gives the exact value of the

7^{\text{th}}

term.

Key Concept

Calculating the

n^{\text{th}}

term of an Arithmetic Progression

Question 22Question →

An arithmetic progression (A.P.) has a fifth term of $17$ and a common difference of $3$ . What is the first term of the progression?

5

2

-1

29

Show answer & explanation

Answer: $5$

Answer

The first term of the progression is

5

.

Using the

n^{\text{th}}

term formula for an arithmetic progression,

T_n = a + (n - 1)d

, substituting

T_5 = 17

,

n = 5

, and

d = 3

gives

17 = a + 4(3) \implies 17 = a + 12 \implies a = 5

.

Step-by-Step Solution

1

Identify the given parameters and formula for the

n^{\text{th}}

term of an A.P.

Formula:

T_n = a + (n - 1)d

, where

T_5 = 17

,

n = 5

, and

d = 3

.

The standard formula connects the

n^{\text{th}}

term, first term, term position, and common difference.

2

Substitute the given values into the formula.

17 = a + (5 - 1) \times 3 \implies 17 = a + 4 \times 3 \implies 17 = a + 12

.

Evaluating

(n - 1)d

gives the total difference added to the first term.

3

Solve for the first term

a

.

a = 17 - 12 = 5

.

Subtracting

12

from both sides isolates

a

.

Key Concept

Arithmetic Progression

n^{\text{th}}

term calculation

Estimated Time:45s

Question 23Question →

The $3^{\text{rd}}$ term of a geometric progression (G.P.) is $18$ and its common ratio is $3$ . What is the first term of the progression?

Show answer & explanation

Answer: 2

Answer

The first term of the geometric progression is 2.

In a geometric progression, the

n^{\text{th}}

term is given by

T_n = a r^{n-1}

. For the

3^{\text{rd}}

term (

n = 3

) with common ratio

r = 3

and term value

18

, the equation is

18 = a \cdot 3^{2} = 9a

. Dividing by

9

yields the first term

a = 2

.

Step-by-Step Solution

1

Identify the formula for the

n^{\text{th}}

term of a geometric progression.

T_n = a r^{n-1}

This formula connects the

n^{\text{th}}

term

T_n

to the first term

a

, common ratio

r

, and term index

n

.

2

Substitute

T_3 = 18

,

r = 3

, and

n = 3

into the formula.

18 = a \cdot 3^{3-1} \implies 18 = 9a

Evaluating

3^{3-1} = 3^2 = 9

simplifies the equation.

3

Solve for the first term

a

.

a = \frac{18}{9} = 2

Dividing both sides of the equation by 9 isolates the first term.

Key Concept

Geometric Progression nth term calculation

Question 24Question →

An arithmetic progression (A.P.) has a first term of $7$ and a common difference of $5$ . What is the $12^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 62

Answer

The

12^{\text{th}}

term of the progression is

62

.

Using the formula for the

n^{\text{th}}

term of an arithmetic progression,

T_n = a + (n - 1)d

, with

a = 7

,

d = 5

, and

n = 12

, we calculate

T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62

.

Step-by-Step Solution

1

Identify known parameters from the question

a = 7

,

d = 5

,

n = 12

These parameters are given in the problem statement.

2

Use the general formula for the

n^{\text{th}}

term of an arithmetic progression

T_n = a + (n - 1)d

This formula connects the first term, common difference, and term number to the value of the term.

3

Substitute values and evaluate

T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62

Performing basic arithmetic gives the final result.

Key Concept

Finding the

n^{\text{th}}

term of an Arithmetic Progression

Estimated Time:45s

Question 25Question →

An arithmetic progression (A.P.) has a first term of $8$ and a common difference of $6$ . What is the $9^{\text{th}}$ term of the progression?

56

62

50

68

Show answer & explanation

Answer: $56$

Answer

The

9^{\text{th}}

term of the arithmetic progression is

56

.

The

n^{\text{th}}

term of an arithmetic progression is given by

T_n = a + (n - 1)d

. Substituting

a = 8

,

d = 6

, and

n = 9

yields

T_9 = 8 + (9 - 1) \times 6 = 8 + 48 = 56

.

Step-by-Step Solution

1

Identify the given terms from the problem

First term

a = 8

, common difference

d = 6

, and term position

n = 9

.

These are the parameters required for the

n^{\text{th}}

term formula of an A.P.

2

Apply the

n^{\text{th}}

term formula

T_n = a + (n - 1)d

T_9 = 8 + (9 - 1) \times 6

The

n^{\text{th}}

term requires multiplying the common difference by

(n - 1)

.

3

Simplify the expression

T_9 = 8 + 8 \times 6 = 8 + 48 = 56

Perform multiplication before addition according to standard order of operations.

Key Concept

General term of an Arithmetic Progression

Question 26Question →

The sum of the first $n$ terms of an arithmetic progression (A.P.) is given by $S_n = 2n^2 + 3n$ . What is the $5^{\text{th}}$ term of the progression?

21

65

44

25

Show answer & explanation

Answer: $21$

Answer

The

5^{\text{th}}

term of the arithmetic progression is

21

.

The

n^{\text{th}}

term of a sequence can be determined from its sum formula using

T_n = S_n - S_{n-1}

. Evaluating

S_5 = 2(5)^2 + 3(5) = 65

and

S_4 = 2(4)^2 + 3(4) = 44

, the difference

T_5 = 65 - 44 = 21

gives the correct value of the

5^{\text{th}}

term.

Step-by-Step Solution

1

Calculate the sum of the first 5 terms (

S_5

)

S_5 = 2(5)^2 + 3(5) = 2(25) + 15 = 65

To find the sum up to the

5^{\text{th}}

term using the given formula

S_n = 2n^2 + 3n

.

2

Calculate the sum of the first 4 terms (

S_4

)

S_4 = 2(4)^2 + 3(4) = 2(16) + 12 = 44

To find the cumulative total of all terms prior to the

5^{\text{th}}

term.

3

Subtract

S_4

from

S_5

to isolate the

5^{\text{th}}

term (

T_5

)

T_5 = S_5 - S_4 = 65 - 44 = 21

The

n^{\text{th}}

term of any sequence is given by the relation

T_n = S_n - S_{n-1}

.

Key Concept

Relationship between the sum of

n

terms (

S_n

) and the

n^{\text{th}}

term (

T_n

) in an Arithmetic Progression

Estimated Time:1m 30s

Question 27Question →

An arithmetic progression (A.P.) and a geometric progression (G.P.) both have a first term of $2$ . The common difference of the A.P. is $4$ . If the $5^{\text{th}}$ term of the A.P. is equal to the $3^{\text{rd}}$ term of the G.P., and the common ratio of the G.P. is positive, what is the $4^{\text{th}}$ term of the G.P.?

54

162

18

14

Show answer & explanation

Answer: 54

Answer

54

First, evaluate the 5th term of the A.P. using

T_5 = 2 + (5-1)4 = 18

. Next, set the 3rd term of the G.P. equal to 18:

2 r^2 = 18

, giving

r^2 = 9

and

r = 3

. Finally, find the 4th term of the G.P. using

T_4 = 2 \times 3^3 = 54

.

Step-by-Step Solution

1

Calculate the 5th term of the A.P.

T_5^{(A.P.)} = a + (5-1)d = 2 + 4(4) = 18

The formula for the

n^{\text{th}}

term of an A.P. is

T_n = a + (n-1)d

.

2

Find the common ratio

r

of the G.P. by equating the 3rd term of the G.P. to 18

T_3^{(G.P.)} = a r^{3-1} = 2 r^2 = 18 \implies r^2 = 9 \implies r = 3

(since

r > 0

)

The formula for the

n^{\text{th}}

term of a G.P. is

T_n = a r^{n-1}

.

3

Compute the 4th term of the G.P.

T_4^{(G.P.)} = a r^{4-1} = 2 \times 3^3 = 2 \times 27 = 54

Substitute

a = 2

,

r = 3

, and

n = 4

into

T_n = a r^{n-1}

.

Key Concept

Connecting terms of Arithmetic and Geometric Progressions using their

n^{\text{th}}

term formulas

Question 28Question →

In an Arithmetic Progression (A.P.), the sum of the first $10$ terms is $120$ and the sum of the next $10$ terms is $320$ . What is the common difference of the progression?

1

2

4

20

Show answer & explanation

Answer: $2$

Answer

The common difference of the progression is

2

.

The sum of an A.P. is

S_n = \frac{n}{2}[2a + (n-1)d]

. For the first 10 terms,

S_{10} = 5(2a + 9d) = 120

, simplifying to

2a + 9d = 24

. The sum of the first 20 terms is

120 + 320 = 440

, so

S_{20} = 10(2a + 19d) = 440

, simplifying to

2a + 19d = 44

. Subtracting the two equations gives

10d = 20

, which yields the common difference

d = 2

.

Step-by-Step Solution

1

Express the sum of the first 10 terms using the formula

S_n = \frac{n}{2}[2a + (n-1)d]

.

S_{10} = \frac{10}{2}[2a + 9d] = 5(2a + 9d) = 120 \implies 2a + 9d = 24

.

Relating the given sum of the first 10 terms to the first term

a

and common difference

d

generates the first linear equation.

2

Determine the sum of the first 20 terms (

S_{20}

) and set up the second linear equation.

S_{20} = S_{10} + \text{sum of next 10 terms} = 120 + 320 = 440

. Thus,

S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d) = 440 \implies 2a + 19d = 44

.

The sum of the next 10 terms added to the sum of the first 10 terms gives the total sum of the first 20 terms.

3

Solve the system of simultaneous linear equations for

d

.

(2a + 19d) - (2a + 9d) = 44 - 24 \implies 10d = 20 \implies d = 2

.

Subtracting equation (1) from equation (2) eliminates

a

to solve directly for the common difference

d

.

Key Concept

Sum of an Arithmetic Progression and Simultaneous Equations

Arithmetic and Geometric Progressions (AP and GP)

Answer

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