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

Question 1Question →

Evaluate the numerical value of each of the following sequence and series expressions, and arrange the items in ascending order (from smallest to largest value):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is: the sum to infinity of

18 + 6 + 2 + \dots

(

27

), followed by the 4th term of the geometric progression (

48

), then the 15th term of the arithmetic progression (

59

), and finally the sum of the first 5 terms of the arithmetic progression (

70

).

Evaluating each expression gives values of

27

,

48

,

59

, and

70

. Arranging these from smallest to largest places the sum to infinity (

27

) first, the GP term (

48

) second, the 15th AP term (

59

) third, and the AP sum (

70

) last.

Step-by-Step Solution

1

Calculate the sum to infinity of the given geometric progression

For

18 + 6 + 2 + \dots

, first term

a = 18

and common ratio

r = \frac{6}{18} = \frac{1}{3}

. Using

S_\infty = \frac{a}{1-r}

, we get

S_\infty = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 27

.

The sum to infinity formula for a convergent geometric series is

S_\infty = \frac{a}{1-r}

where

|r| < 1

.

2

Calculate the 4th term of the geometric progression

Given

T_2 = ar = 12

and

T_5 = ar^4 = 96

. Dividing

T_5

by

T_2

yields

r^3 = \frac{96}{12} = 8 \implies r = 2

. Thus

a = \frac{12}{2} = 6

. The 4th term is

T_4 = ar^3 = 6 \times 2^3 = 48

.

Using the

n

-th term formula of a GP,

T_n = ar^{n-1}

, to solve simultaneous equations for

a

and

r

.

3

Calculate the 15th term of the arithmetic progression

Given

T_3 = a + 2d = 11

and

T_8 = a + 7d = 31

. Subtracting the equations gives

5d = 20 \implies d = 4

. Substituting back yields

a = 11 - 2(4) = 3

. The 15th term is

T_{15} = a + 14d = 3 + 14(4) = 59

.

Using the

n

-th term formula of an AP,

T_n = a + (n-1)d

, to find the first term

a

and common difference

d

.

4

Calculate the sum of the first 5 terms of the arithmetic progression

With

a = 4

,

d = 5

, and

n = 5

,

S_5 = \frac{5}{2}[2(4) + (5-1)5] = \frac{5}{2}[8 + 20] = \frac{5}{2}(28) = 70

.

Applying the AP sum formula

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

.

5

Order the evaluated quantities from smallest to largest

Comparing numerical results

27 < 48 < 59 < 70

gives the sequence: item 1, item 2, item 4, item 3.

Sorting the calculated values in ascending numerical order.

Key Concept

Evaluation and comparison of terms and sums in Arithmetic and Geometric Progressions

Estimated Time:2m 30s

Question 2Question →

What is the $10^{\text{th}}$ term of the arithmetic progression $3, 7, 11, 15, \dots$ ?

Show answer & explanation

Answer: 39

Answer

The

10^{\text{th}}

term of the arithmetic progression is

39

.

By applying the

n^{\text{th}}

term formula for an arithmetic progression

T_n = a + (n - 1)d

with first term

a = 3

, common difference

d = 4

, and term index

n = 10

, the calculation yields

T_{10} = 3 + (10 - 1) \times 4 = 3 + 36 = 39

.

Step-by-Step Solution

1

Identify the key parameters of the arithmetic progression from the given sequence.

First term

a = 3

, common difference

d = 7 - 3 = 4

, and number of terms

n = 10

.

These parameters are required to use the

n^{\text{th}}

term formula of an A.P.

2

Substitute the values into the formula

T_n = a + (n - 1)d

.

T_{10} = 3 + (10 - 1) \times 4

The formula relates the

n^{\text{th}}

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

3

Evaluate the mathematical expression.

T_{10} = 3 + 9 \times 4 = 3 + 36 = 39

Perform multiplication before addition according to standard order of operations.

Key Concept

nth term of an Arithmetic Progression

Question 3Question →

The first term of a geometric progression (GP) is $2$ and its common ratio is $3$ . What is the $4^{\text{th}}$ term of the progression?

54

162

11

24

Show answer & explanation

Answer: $54$

Answer

The

4^{\text{th}}

term of the geometric progression is

54

.

For a geometric progression with first term

a

and common ratio

r

, the

n^{\text{th}}

term is given by

T_n = a r^{n-1}

. Substituting

a = 2

,

r = 3

, and

n = 4

yields

T_4 = 2 \times 3^3 = 2 \times 27 = 54

.

Step-by-Step Solution

1

Identify the given parameters of the geometric progression.

First term

a = 2

, common ratio

r = 3

, and term position

n = 4

.

These values are directly provided in the question statement.

2

Apply the general formula for the

n^{\text{th}}

term of a geometric progression,

T_n = a r^{n-1}

.

T_4 = 2 \times 3^{4-1} = 2 \times 3^3

.

The exponent of the common ratio is always one less than the term index

n

.

3

Evaluate the exponent and multiply by the first term.

3^3 = 27

, so

T_4 = 2 \times 27 = 54

.

Performing standard arithmetic yields the exact term value.

Key Concept

Formula for the nth term of a Geometric Progression:

T_n = a r^{n-1}

Estimated Time:45s

Question 4Question →

The $3^{\text{rd}}$ , $6^{\text{th}}$ , and $11^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $15$ , what is the $4^{\text{th}}$ term of the geometric progression?

Show answer & explanation

Answer: 125

Answer

The 4th term of the geometric progression is 125.

By writing the 3rd, 6th, and 11th terms of the AP as

15+2d

,

15+5d

, and

15+10d

, we utilize the geometric mean property

(15+5d)^2 = (15+2d)(15+10d)

to find

d=6

. This yields the GP terms

27, 45, 75

, giving a common ratio of

5/3

. Multiplying the third term

75

by

5/3

gives the 4th GP term as

125

.

Step-by-Step Solution

1

Write down the AP term expressions

T_3 = 15 + 2d

,

T_6 = 15 + 5d

,

T_{11} = 15 + 10d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

with initial term

a = 15

.

2

Apply the geometric progression condition

(15 + 5d)^2 = (15 + 2d)(15 + 10d)

If three terms

A, B, C

are in GP, then

B^2 = A \cdot C

.

3

Expand and solve the quadratic equation for the common difference

d

d = 6

Expanding gives

225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6

because

d \neq 0

.

4

Find the terms and common ratio of the GP

G_1 = 27

,

G_2 = 45

,

G_3 = 75

, and common ratio

r = \frac{5}{3}

Substituting

d = 6

gives the GP terms, and dividing consecutive terms gives

r = \frac{45}{27} = \frac{5}{3}

.

5

Calculate the 4th term of the GP

G_4 = 125

Multiplying the 3rd term by the common ratio yields

75 \times \frac{5}{3} = 125

.

Key Concept

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties

Estimated Time:2m 30s

Question 5Question →

The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = 2n^2 + 3n$ . The $3^{\text{rd}}$ term of this AP is equal to the $2^{\text{nd}}$ term of a geometric progression (GP). If the common ratio of the GP is $2$ , what is the sum of the first $4$ terms of the GP?

97.5

127.5

48.75

45.5

Show answer & explanation

Answer: $97.5$

Answer

97.5

Evaluating

S_3 - S_2

gives the 3rd AP term as

27 - 14 = 13

. Setting the 2nd GP term

a(2) = 13

yields

a = 6.5

. The sum of the first 4 terms of the GP is

6.5 \times (2^4 - 1) = 6.5 \times 15 = 97.5

.

Step-by-Step Solution

1

Find the 3rd term (

T_3

) of the AP using the given sum formula

S_n = 2n^2 + 3n

T_3 = S_3 - S_2 = [2(3)^2 + 3(3)] - [2(2)^2 + 3(2)] = [18 + 9] - [8 + 6] = 27 - 14 = 13

The

n

-th term of a sequence is equal to

S_n - S_{n-1}

.

2

Determine the first term (

a

) of the GP

Since

G_2 = 13

and common ratio

r = 2

,

a \cdot r^{2-1} = 13 \implies 2a = 13 \implies a = 6.5

The

n

-th term of a GP is given by

G_n = a r^{n-1}

.

3

Calculate the sum of the first 4 terms of the GP

S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{6.5(2^4 - 1)}{2 - 1} = 6.5 \times 15 = 97.5

The sum of the first

n

terms of a GP with

r > 1

is

S_n = \frac{a(r^n - 1)}{r - 1}

.

Key Concept

Combining Arithmetic Progression sum formula with Geometric Progression term and sum formulas

Estimated Time:2m 0s

Question 6Question →

The sum of the first three terms of an increasing geometric progression of positive real numbers is $21$ , and the sum of their squares is $189$ . What is the common ratio of this progression?

Show answer & explanation

Answer: 2

Answer

The common ratio of the geometric progression is 2.

Let the terms be

a

,

ar

, and

ar^2

. The given conditions yield

a(1+r+r^2) = 21

and

a^2(1+r^2+r^4) = 189

. Squaring the first equation gives

a^2(1+r+r^2)^2 = 441

. Dividing the sum of squares equation by this squared equation gives

\frac{1-r+r^2}{1+r+r^2} = \frac{189}{441} = \frac{3}{7}

. Simplifying

7(1-r+r^2) = 3(1+r+r^2)

results in

2r^2 - 5r + 2 = 0

, which factors into

(2r-1)(r-2) = 0

. Because the progression is increasing,

r > 1

, making

r = 2

the correct common ratio.

Step-by-Step Solution

1

Formulate algebraic expressions for the sum of terms and sum of squares.

a(1 + r + r^2) = 21

and

a^2(1 + r^2 + r^4) = 189

The first three terms of any geometric progression can be expressed as

a

,

ar

, and

ar^2

.

2

Eliminate the first term

a

by squaring the first equation and dividing.

\frac{a^2(1 + r^2 + r^4)}{a^2(1 + r + r^2)^2} = \frac{189}{441} \implies \frac{(1 + r + r^2)(1 - r + r^2)}{(1 + r + r^2)^2} = \frac{3}{7}

Using the algebraic factorization

1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)

allows cancellation of

a^2

and

(1 + r + r^2)

.

3

Solve the resulting equation for the common ratio

r

.

7(1 - r + r^2) = 3(1 + r + r^2) \implies 4r^2 - 10r + 4 = 0 \implies 2r^2 - 5r + 2 = 0

Cross-multiplying reduces the ratio to a standard quadratic equation.

4

Factor the quadratic equation and select the correct root.

(2r - 1)(r - 2) = 0 \implies r = 2 \text{ or } r = 0.5

Since the geometric progression is specified as increasing, the common ratio must be greater than 1 (

r = 2

).

Key Concept

Geometric progression term representations, sum formulas, and algebraic identity factorization.

Alternative Method

Find

a

and

r

by testing factors of

21

:

21 = 3 \times 7

, so the terms could be

3, 6, 12

(

a=3, r=2

). Check squares:

3^2 + 6^2 + 12^2 = 9 + 36 + 144 = 189

, which confirms

r = 2

.

Estimated Time:1m 30s

Question 7Question →

The $3^{\text{rd}}$ term of an arithmetic progression (AP) is $10$ and the $7^{\text{th}}$ term is $22$ . What is the sum of the first $12$ terms of the progression?

228

246

264

444

Show answer & explanation

Answer: $246$

Answer

The sum of the first

12

terms of the arithmetic progression is

246

.

Using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

, the equations

a + 2d = 10

and

a + 6d = 22

yield

d = 3

and

a = 4

. Substituting these values into

S_{12} = \frac{12}{2}[2(4) + 11(3)]

gives

6 \times 41 = 246

.

Step-by-Step Solution

1

Set up simultaneous equations using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

.

a + 2d = 10

and

a + 6d = 22

.

The

3^{\text{rd}}

term corresponds to

n=3

and the

7^{\text{th}}

term corresponds to

n=7

.

2

Subtract the first equation from the second to find the common difference

d

.

4d = 12 \implies d = 3

.

Subtracting eliminates the first term

a

.

3

Substitute

d = 3

back into the first equation to find the first term

a

.

a + 2(3) = 10 \implies a = 4

.

Determining the first term is necessary to calculate the sum.

4

Calculate the sum of the first

12

terms using

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

.

S_{12} = \frac{12}{2}[2(4) + (12-1)(3)] = 6[8 + 33] = 6(41) = 246

.

Applying the AP sum formula with

n = 12

,

a = 4

, and

d = 3

.

Key Concept

Arithmetic Progression: Finding common difference, first term, and sum of terms

Estimated Time:1m 30s

Question 8Question →

An arithmetic progression (AP) has a first term of $2$ and a common difference of $3$ . A geometric progression (GP) has a first term of $1$ and a common ratio of $2$ . Arrange the following quantities in ascending order of their numerical values:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is the 4th term of the GP (8), followed by the 4th term of the AP (11), the sum of the first 3 terms of the AP (15), and the 5th term of the GP (16).

Evaluating each term individually gives: the 4th term of the GP equals 8, the 4th term of the AP equals 11, the sum of the first 3 terms of the AP equals 15, and the 5th term of the GP equals 16. Arranging these calculated values from smallest to largest gives the order 8, 11, 15, 16.

Step-by-Step Solution

1

Calculate the 4th term of the AP

T_4 = 2 + (4 - 1) \times 3 = 2 + 9 = 11

Using the AP

n^{\text{th}}

term formula

T_n = a + (n-1)d

.

2

Calculate the 5th term of the GP

T_5 = 1 \times 2^{5-1} = 2^4 = 16

Using the GP

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

3

Calculate the sum of the first 3 terms of the AP

S_3 = \frac{3}{2}[2(2) + (3-1)3] = \frac{3}{2}[4 + 6] = 15

Using the AP sum formula

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

.

4

Calculate the 4th term of the GP

T_4 = 1 \times 2^{4-1} = 2^3 = 8

Using the GP

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

5

Compare and arrange the values in ascending order

8 < 11 < 15 < 16

Arranging from smallest to largest numerical value.

Key Concept

Nth term and sum formulas of Arithmetic and Geometric Progressions

Question 9Question →

The $2^{\text{nd}}$ term of a geometric progression (GP) is $6$ and the $5^{\text{th}}$ term is $48$ . What is the sum of the first $6$ terms of the progression?

189

93

162

192

Show answer & explanation

Answer: $189$

Answer

The sum of the first 6 terms of the geometric progression is 189.

By using the GP term formula

T_n = a r^{n-1}

, we establish

a r = 6

and

a r^4 = 48

. Dividing the fifth term by the second term yields

r^3 = 8

, giving

r = 2

. Substituting

r = 2

into

a r = 6

gives

a = 3

. Finally, using the sum formula

S_n = \frac{a(r^n - 1)}{r - 1}

, we evaluate

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(63) = 189

.

Step-by-Step Solution

1

Set up equations for the given terms using the nth term formula

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

.

The nth term of a GP is defined by

T_n = a r^{n-1}

.

2

Solve for the common ratio

r

by dividing

T_5

by

T_2

.

\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2

.

Dividing the terms eliminates the first term

a

.

3

Find the first term

a

.

a(2) = 6 \implies a = 3

.

Substitute

r = 2

back into the equation for

T_2

.

4

Calculate the sum of the first 6 terms using

S_n = \frac{a(r^n - 1)}{r - 1}

.

S_6 = \frac{3(2^6 - 1)}{2 - 1} = 3(64 - 1) = 3 \times 63 = 189

.

Apply the sum formula for a GP with

r > 1

.

Key Concept

Geometric Progression nth term and sum formulas

Question 10Question →

A geometric progression has a first term of $5$ and a common ratio of $2$ . What is the $6^{\text{th}}$ term of this progression?

160

320

60

15

Show answer & explanation

Answer: $160$

Answer

The

6^{\text{th}}

term of the geometric progression is

160

.

The

n^{\text{th}}

term of a geometric progression is given by

T_n = a r^{n-1}

. Substituting

a = 5

,

r = 2

, and

n = 6

yields

T_6 = 5 \times 2^5 = 5 \times 32 = 160

, which makes

160

the correct value.

Step-by-Step Solution

1

Identify the given values from the problem statement.

First term

a = 5

, common ratio

r = 2

, and term index

n = 6

.

These parameters are required for the

n^{\text{th}}

term formula of a geometric progression.

2

State the formula for the

n^{\text{th}}

term of a Geometric Progression (GP).

T_n = a r^{n-1}

The

n^{\text{th}}

term of a GP is obtained by multiplying the initial term by the common ratio raised to the power of

n-1

.

3

Substitute the values into the formula and evaluate.

T_6 = 5 \times 2^{6-1} = 5 \times 2^5 = 5 \times 32 = 160

Evaluating

2^5 = 32

and then multiplying by

5

yields the correct

6^{\text{th}}

term.

Key Concept

Geometric Progression

n^{\text{th}}

Term

Question 11Question →

Calculate the numerical values of the following sequence and progression quantities, then arrange them in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order of the numerical values is: the common ratio of the GP (value = 3), the common difference of the AP (value = 4), the 5th term of the AP (value = 8), and the sum to infinity of the GP (value = 10).

Evaluating each sequence property yields numerical values: the GP common ratio equals 3, the AP common difference equals 4, the AP 5th term equals 8, and the GP sum to infinity equals 10. Ordering these values from least to greatest produces the sequence 3, 4, 8, 10.

Step-by-Step Solution

1

Calculate the value for the first quantity (AP common difference)

d = 4

From

a + 6d = 27

and

a + 2d = 11

, subtract to find

4d = 16 \implies d = 4

.

2

Calculate the value for the second quantity (GP common ratio)

r = 3

From

ar^4 = 162

and

ar = 6

, divide to obtain

r^3 = 27 \implies r = 3

.

3

Calculate the value for the third quantity (GP sum to infinity)

S_{\infty} = 10

Apply the sum to infinity formula

S_{\infty} = \frac{a}{1-r} = \frac{5}{1 - 0.5} = 10

.

4

Calculate the value for the fourth quantity (AP 5th term)

T_5 = 8

Apply the nth term formula

T_5 = a + 4d = 2 + 4(1.5) = 8

.

5

Sort the calculated values in ascending order

3 < 4 < 8 < 10

Comparing the values gives

3

(GP common ratio),

4

(AP common difference),

8

(AP 5th term), and

10

(GP sum to infinity).

Key Concept

Nth term, common difference, common ratio, and sum to infinity calculations for AP and GP sequences.

Question 12Question →

Three consecutive terms of an arithmetic progression are $x + 2$ , $3x - 1$ , and $4x + 1$ . What is the $10^{\text{th}}$ term of the progression?

63

77

70

35

Show answer & explanation

Answer: 70

Answer

The

10^{\text{th}}

term of the progression is 70.

Equating the differences between consecutive terms gives

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

, which simplifies to

2x - 3 = x + 2

, giving

x = 5

. The first term is

a = 5 + 2 = 7

and the common difference is

d = 14 - 7 = 7

. Substituting these into

T_n = a + (n - 1)d

for

n = 10

yields

T_{10} = 7 + 9(7) = 70

.

Step-by-Step Solution

1

Set up an equation using the common difference property of an arithmetic progression.

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

In any arithmetic progression, the difference between consecutive terms is constant (

T_2 - T_1 = T_3 - T_2

).

2

Simplify and solve for

x

.

2x - 3 = x + 2 \implies x = 5

Subtract

x

from both sides and add 3 to both sides.

3

Find the first term

a

and the common difference

d

.

First term

a = 5 + 2 = 7

; second term

T_2 = 3(5) - 1 = 14

; common difference

d = 14 - 7 = 7

.

Substitute

x = 5

into the expressions for the terms.

4

Calculate the

10^{\text{th}}

term using

T_n = a + (n - 1)d

.

T_{10} = 7 + (10 - 1)(7) = 7 + 9(7) = 7 + 63 = 70

Apply

n = 10

,

a = 7

, and

d = 7

to the

n^{\text{th}}

term formula.

Key Concept

Arithmetic Progression - Consecutive terms and nth term evaluation

Question 13Question →

An arithmetic progression has a first term of $5$ and a common difference of $4$ . What is the $12^{\text{th}}$ term of this progression?

49

53

57

45

Show answer & explanation

Answer: $49$

Answer

49

The

n^{\text{th}}

term of an arithmetic progression is determined using the formula

T_n = a + (n - 1)d

. Substituting

a = 5

,

d = 4

, and

n = 12

yields

T_{12} = 5 + 11 \times 4 = 49

, making

49

the correct value.

Step-by-Step Solution

1

Identify the given values from the problem statement

First term

a = 5

, common difference

d = 4

, and position

n = 12

These are the standard variables required for calculating terms in an arithmetic progression.

2

Apply the

n^{\text{th}}

term formula for an arithmetic progression

T_{12} = 5 + (12 - 1) \times 4

The standard formula for the

n^{\text{th}}

term of an AP is

T_n = a + (n - 1)d

.

3

Simplify the expression to find the final value

T_{12} = 5 + 11 \times 4 = 5 + 44 = 49

Perform multiplication before addition according to standard order of operations.

Key Concept

Arithmetic Progression

n^{\text{th}}

Term Formula

Estimated Time:45s

Question 14Question →

The $5^{\text{th}}$ term of an arithmetic progression (AP) is $17$ and its common difference is $3$ . What is the first term of the progression?

5

2

29

-1

Show answer & explanation

Answer: $5$

Answer

The first term of the progression is

5

.

The

n^{\text{th}}

term formula of an arithmetic progression is

T_n = a + (n - 1)d

. Substituting

T_5 = 17

,

n = 5

, and

d = 3

gives

17 = a + 4(3) = a + 12

. Isolating

a

yields

a = 17 - 12 = 5

.

Step-by-Step Solution

1

Identify the given values and formula for the nth term of an AP.

The formula is

T_n = a + (n - 1)d

, with

T_5 = 17

,

n = 5

, and

d = 3

.

This formula connects the nth term, the first term, the number of terms, and the common difference.

2

Substitute the known values into the equation.

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

.

Subtracting 1 from the term index 5 gives 4, and multiplying by 3 gives 12.

3

Solve for the first term

a

.

a = 17 - 12 = 5

.

Subtracting 12 from both sides isolates

a

.

Key Concept

n-th term of an Arithmetic Progression

Estimated Time:1m 0s

Question 15Question →

The sum of the first $n$ terms of a sequence is given by $S_n = 2n^2 + 3n$ . Find the $5^{\text{th}}$ term of a geometric progression whose first term is the $3^{\text{rd}}$ term of this sequence, and whose common ratio is equal to the common difference of this sequence.

3328

13312

6912

1280

Show answer & explanation

Answer: $3328$

Answer

The 5th term of the geometric progression is 3328.

Evaluating

S_1

,

S_2

, and

S_3

gives sequence terms

T_1 = 5

,

T_2 = 9

, and

T_3 = 13

. The common difference is

d = 9 - 5 = 4

. Using

T_3 = 13

as the first term of the GP and

r = 4

as the common ratio, the 5th term of the GP is

13 \times 4^{5-1} = 13 \times 256 = 3328

.

Step-by-Step Solution

1

Calculate the first few terms of the sequence using the sum formula

S_n = 2n^2 + 3n

.

S_1 = 2(1)^2 + 3(1) = 5

,

S_2 = 2(2)^2 + 3(2) = 14

,

S_3 = 2(3)^2 + 3(3) = 27

.

The sum formula gives cumulative sums, from which individual terms can be derived.

2

Find the 3rd term (

T_3

) and the common difference (

d

) of the arithmetic progression.

T_1 = 5

,

T_2 = S_2 - S_1 = 9

,

T_3 = S_3 - S_2 = 13

. Common difference

d = T_2 - T_1 = 4

.

The difference between consecutive cumulative sums gives the individual sequence terms, and their constant difference gives the common difference.

3

Define the parameters of the geometric progression (GP).

First term of GP

a = T_3 = 13

, common ratio

r = d = 4

.

The problem specifies that the first term of the GP is the 3rd term of the sequence and the common ratio equals the common difference.

4

Compute the 5th term of the geometric progression using

G_n = a \cdot r^{n-1}

.

G_5 = 13 \cdot 4^{5-1} = 13 \cdot 4^4 = 13 \cdot 256 = 3328

.

Applying the standard nth term formula for a GP with

n = 5

.

Key Concept

Arithmetic Progression sum formula to term conversion and Geometric Progression nth term evaluation

Estimated Time:2m 0s

Question 16Question →

The $3^{\text{rd}}$ term of a geometric progression is $18$ and the $6^{\text{th}}$ term is $486$ . What is the sum of the first $5$ terms of the progression?

242

162

80

728

Show answer & explanation

Answer: $242$

Answer

The sum of the first

5

terms is

242

.

Using

T_n = a r^{n-1}

, we set up

a r^2 = 18

and

a r^5 = 486

. Dividing these gives

r^3 = 27

, so

r = 3

, which leads to

a = 2

. Applying

S_5 = \frac{a(r^5 - 1)}{r - 1}

yields

\frac{2(243 - 1)}{2} = 242

.

Step-by-Step Solution

1

Express the given terms using the geometric progression

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

T_3 = a r^2 = 18

and

T_6 = a r^5 = 486

.

This establishes a system of equations in terms of the first term

a

and common ratio

r

.

2

Divide the expression for

T_6

by

T_3

to determine

r

.

\frac{a r^5}{a r^2} = \frac{486}{18} \implies r^3 = 27 \implies r = 3.

Dividing the equations eliminates

a

and allows direct solution for the common ratio

r

.

3

Substitute

r = 3

back into

T_3 = a r^2 = 18

to solve for

a

.

a (3)^2 = 18 \implies 9a = 18 \implies a = 2.

Determining the first term

a

is required to evaluate the sum.

4

Calculate the sum of the first

5

terms using

S_n = \frac{a(r^n - 1)}{r - 1}

.

S_5 = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242.

Applying the GP sum formula yields the required value.

Key Concept

Geometric Progression nth term and sum formulas

Question 17Question →

The first, third, and seventh terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $4$ , what is the sum of the first $4$ terms of the geometric progression?

60

28

124

92

Show answer & explanation

Answer: 60

Answer

The sum of the first

4

terms of the geometric progression is

60

.

The first three terms of the GP are

T_1 = 4

,

T_3 = 4 + 2d

, and

T_7 = 4 + 6d

. Equating

(4 + 2d)^2 = 4(4 + 6d)

yields

4d^2 - 8d = 0

, giving

d = 2

. The first four terms of the GP are

4, 8, 16, 32

, which sum to

4 + 8 + 16 + 32 = 60

.

Step-by-Step Solution

1

Express the terms of the arithmetic progression in terms of first term

a

and common difference

d

.

First term

T_1 = 4

, third term

T_3 = 4 + 2d

, seventh term

T_7 = 4 + 6d

.

The

n^{\text{th}}

term of an AP is defined as

T_n = a + (n-1)d

.

2

Set up the geometric progression condition

(T_3)^2 = T_1 \cdot T_7

to solve for

d

.

(4 + 2d)^2 = 4(4 + 6d) \implies 16 + 16d + 4d^2 = 16 + 24d \implies 4d^2 - 8d = 0 \implies d = 2

(since the AP is non-constant,

d \neq 0

).

Three terms

x, y, z

form a GP if and only if

y^2 = xz

.

3

Determine the terms and common ratio

r

of the GP.

First term

G_1 = 4

, second term

G_2 = 4 + 2(2) = 8

. Thus,

r = \frac{8}{4} = 2

.

The common ratio

r

is the quotient of consecutive terms of the GP.

4

Calculate the sum of the first

4

terms of the GP using

S_n = \frac{a(r^n - 1)}{r - 1}

.

S_4 = \frac{4(2^4 - 1)}{2 - 1} = \frac{4(16 - 1)}{1} = 60

.

Formula for the sum of the first

n

terms of a geometric progression.

Key Concept

Arithmetic and Geometric Progression Inter-relationships

Question 18Question →

The $2^{\text{nd}}$ , $5^{\text{th}}$ , and $14^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the sum of the first $6$ terms of the AP is $72$ , calculate the common difference of the AP.

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

Equating the square of the middle GP term

(a+4d)^2

to the product of the outer terms

(a+d)(a+13d)

yields

3d^2 = 6ad

, which simplifies to

d = 2a

. Substituting

2a = d

into the AP sum formula

S_6 = 3(2a+5d) = 72

gives

3(6d) = 72 \implies 18d = 72

, so

d = 4

.

Step-by-Step Solution

1

Express the

2^{\text{nd}}

,

5^{\text{th}}

, and

14^{\text{th}}

terms of the AP algebraically

T_2 = a + d

,

T_5 = a + 4d

,

T_{14} = a + 13d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

.

2

Apply the consecutive terms property of a GP

(a + 4d)^2 = (a + d)(a + 13d)

For three consecutive terms of a GP, the square of the middle term equals the product of the first and third terms.

3

Simplify the quadratic equation to find the relationship between

a

and

d

a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \implies 3d^2 = 6ad \implies d = 2a

Since the AP is non-constant,

d \neq 0

, allowing division by

3d

.

4

Formulate the sum of the first

6

terms of the AP

S_6 = 3(2a + 5d) = 72 \implies 2a + 5d = 24

The sum of the first

n

terms of an AP is

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

.

5

Substitute

2a = d

into the linear equation and solve for

d

d + 5d = 24 \implies 6d = 24 \implies d = 4

Replacing

2a

with

d

reduces the equation to a single variable.

Key Concept

Relating non-consecutive terms of an Arithmetic Progression to form a Geometric Progression

Question 19Question →

Arrange the following values related to arithmetic and geometric progressions in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from smallest to largest value is: the common ratio of the GP (4), the common difference of the AP (8), the 4th term of the AP (11), and the sum of the first 3 terms of the GP (13).

Evaluating each progression property gives numerical values of 4, 8, 11, and 13 respectively. Ordering these from least to greatest results in the order: common ratio of the GP (4), common difference of the AP (8), 4th term of the AP (11), and sum of the first 3 terms of the GP (13).

Step-by-Step Solution

1

Calculate the value for the first item (common ratio

r

)

Using

T_4 = T_2 \cdot r^2

, we have

80 = 5 r^2 \implies r^2 = 16 \implies r = 4

.

The terms of a GP follow

T_n = a r^{n-1}

.

2

Calculate the value for the second item (common difference

d

)

Using

T_5 = a + 4d

, we get

35 = 3 + 4d \implies 4d = 32 \implies d = 8

.

The nth term of an AP is given by

T_n = a + (n-1)d

.

3

Calculate the value for the third item (

4^{\text{th}}

term of AP)

T_4 = 2 + (4 - 1)(3) = 2 + 9 = 11

.

Direct application of the AP nth term formula.

4

Calculate the value for the fourth item (Sum of first

3

terms of GP)

S_3 = 1 + 3 + 9 = 13

(or using

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1(3^3 - 1)}{3 - 1} = 13

).

Sum of a finite geometric sequence.

5

Compare and arrange the calculated numerical values in ascending order

4 < 8 < 11 < 13

.

Ordering the numbers establishes the correct item sequence.

Key Concept

Evaluation of terms, differences, ratios, and sums in Arithmetic and Geometric Progressions.

Question 20Question →

The sum of the first $4$ terms of a geometric progression (GP) with a common ratio of $2$ is $45$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 96

Answer

The

6^{\text{th}}

term of the geometric progression is

96

.

Using the sum formula

S_4 = \frac{a(2^4 - 1)}{2 - 1} = 45

gives

15a = 45

, so the first term

a

is

3

. Substituting

a = 3

and

r = 2

into the term formula

T_6 = a r^5

gives

3 \times 32 = 96

.

Step-by-Step Solution

1

Express the sum of the first 4 terms using the GP sum formula to find the first term

a

.

Setting up

45 = \frac{a(2^4 - 1)}{2 - 1}

yields

15a = 45

, so

a = 3

.

The sum formula

S_n = \frac{a(r^n - 1)}{r - 1}

allows us to isolate the unknown initial term

a

when

S_4

and

r

are given.

2

Calculate the

6^{\text{th}}

term

T_6

using the

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

T_6 = 3 \times 2^{6-1} = 3 \times 32 = 96

.

The exponent for the common ratio in the

n^{\text{th}}

term formula is

n - 1

, giving

5

as the exponent.

Key Concept

Sum and

n^{\text{th}}

term of a Geometric Progression

Arithmetic and Geometric Progressions (AP and GP)

Answer

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