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

Question 1Question →

In a geometric progression of positive terms, the sum of the first two terms is $12$ and the sum of the third and fourth terms is $48$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 128

Answer

The 6th term of the geometric progression is 128.

Dividing

ar^2(1+r) = 48

by

a(1+r) = 12

yields

r^2 = 4

, so

r = 2

for positive terms. Substituting

r = 2

into

a(1+r) = 12

gives

a = 4

. Using

T_n = a r^{n-1}

for

n=6

, we get

T_6 = 4 \times 2^5 = 128

.

Step-by-Step Solution

1

Set up algebraic equations for the given sums using first term

a

and common ratio

r

.

a(1+r) = 12

and

ar^2(1+r) = 48

The terms of a geometric progression are given by

T_n = a r^{n-1}

.

2

Divide the equation for the third and fourth terms by the equation for the first and second terms.

r^2 = 4 \implies r = 2

Dividing eliminates

a

and

(1+r)

, giving

r^2 = 4

. Since terms are positive,

r > 0

.

3

Substitute

r = 2

into

a(1+r) = 12

to solve for

a

.

a = 4

3a = 12

leads directly to

a = 4

.

4

Evaluate the 6th term using the formula

T_6 = a r^{5}

.

T_6 = 4 \times 2^5 = 128

Applying the general term formula

T_n = a r^{n-1}

with

n=6

.

Key Concept

Geometric Progression term relations and finding the common ratio from consecutive term pairs

Question 2Question →

The second term of a geometric progression (G.P.) with positive terms is $6$ and the fifth term is $48$ . What is the sum of the first $6$ terms of the progression?

93

96

189

192

Show answer & explanation

Answer: $189$

Answer

The sum of the first

6

terms is

189

.

Using the

n

-th term formula

T_n = a r^{n-1}

, we form two equations:

a r = 6

and

a r^4 = 48

. Dividing the fifth term equation by the second term equation gives

r^3 = 8

, so the common ratio

r = 2

. Substituting

r = 2

back gives the first term

a = 3

. Finally, applying the sum formula

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

gives

3 \times 63 = 189

.

Step-by-Step Solution

1

Set up equations for the given terms using the

n

-th term formula

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

.

The

n

-th term of a G.P. is defined by

T_n = a r^{n-1}

.

2

Divide the equation for

T_5

by the equation for

T_2

to find the common ratio

r

.

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

Dividing eliminates the first term

a

and allows solving for

r

directly.

3

Substitute

r = 2

into

T_2 = 6

to find the first term

a

.

a(2) = 6 \implies a = 3.

Knowing

r

allows calculating

a

from any known term.

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} = \frac{3(64 - 1)}{1} = 3 \times 63 = 189.

Applying the G.P. sum formula for

n = 6

,

a = 3

, and

r = 2

gives the total sum.

Key Concept

Geometric Progression term formula (

T_n = a r^{n-1}

) and sum of

n

terms formula (

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

).

Estimated Time:1m 30s

Question 3Question →

The second, fourth, and eighth terms of an arithmetic progression (AP) with a non-zero common difference form three consecutive terms of a geometric progression (GP). If the sum of the first $5$ terms of the AP is $45$ , what is the first term of the AP?

3

5

9

15

Show answer & explanation

Answer: 3

Answer

The first term of the AP is 3.

By expressing the 2nd, 4th, and 8th terms as

a+d

,

a+3d

, and

a+7d

, the geometric mean condition

(a+3d)^2 = (a+d)(a+7d)

reduces to

d = a

. Substituting

d = a

into the sum formula

S_5 = 5(a+2d) = 45

gives

15a = 45

, yielding a first term of 3.

Step-by-Step Solution

1

Express the 2nd, 4th, and 8th terms of the AP in terms of first term

a

and common difference

d

T_2 = a + d

,

T_4 = a + 3d

, and

T_8 = a + 7d

The

n

-th term of an AP is given by

T_n = a + (n-1)d

.

2

Apply the condition for consecutive terms of a GP

(a + 3d)^2 = (a + d)(a + 7d) \implies a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2 \implies 2d^2 = 2ad \implies d = a

If three terms

x, y, z

form a GP, then

y^2 = xz

. Since

d \neq 0

, dividing by

2d

gives

d = a

.

3

Use the sum of the first 5 terms of the AP to set up an equation for

a

and

d

S_5 = \frac{5}{2}[2a + 4d] = 45 \implies 5(a + 2d) = 45 \implies a + 2d = 9

The sum of the first

n

terms of an AP is

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

.

4

Substitute

d = a

into the sum equation to solve for

a

a + 2(a) = 9 \implies 3a = 9 \implies a = 3

Substituting

d = a

simplifies the linear equation to solve directly for

a

.

Key Concept

Combining Arithmetic and Geometric Progression properties to set up and solve simultaneous equations.

Question 4Question →

An arithmetic progression (A.P.) has a first term of $5$ and a common difference of $3$ . What is the $8^{\text{th}}$ term of this progression?

26

29

32

23

Show answer & explanation

Answer: $26$

Answer

The

8^{\text{th}}

term of the arithmetic progression is

26

.

The

8^{\text{th}}

term is calculated using the standard formula

T_n = a + (n - 1)d

. Substituting

a = 5

,

d = 3

, and

n = 8

gives

T_8 = 5 + 7(3) = 26

.

Step-by-Step Solution

1

Identify the given parameters from the problem

First term

a = 5

, common difference

d = 3

, and term position

n = 8

.

These are the values required for substitution into the

n^{\text{th}}

term formula of an A.P.

2

Write down the general formula for the

n^{\text{th}}

term of an arithmetic progression

T_n = a + (n - 1)d

This formula defines any term in an arithmetic progression based on its position.

3

Substitute the values into the formula and simplify

T_8 = 5 + (8 - 1) \times 3 = 5 + 7 \times 3 = 5 + 21 = 26

Performing the multiplication before addition yields the value of the

8^{\text{th}}

term.

Key Concept

Arithmetic Progression (A.P.)

n^{\text{th}}

term formula

Estimated Time:45s

Question 5Question →

The 3rd term of an arithmetic progression (AP) is $14$ and its 7th term is $34$ . If the $n$ -th term of this AP is equal to the 4th term of a geometric progression (GP) whose first term is $2$ and common ratio is $3$ , what is the value of $n$ ?

10

11

12

9

Show answer & explanation

Answer: 11

Answer

The value of

n

is

11

.

The AP has first term

a = 4

and common difference

d = 5

, giving

T_n = 4 + 5(n-1) = 5n - 1

. The 4th term of the GP is

2 \times 3^{3} = 54

. Setting

5n - 1 = 54

gives

5n = 55

, so

n = 11

.

Step-by-Step Solution

1

Find the first term

a

and common difference

d

of the arithmetic progression.

a = 4

and

d = 5

The

n

-th term of an AP is given by

T_n = a + (n-1)d

. Using given terms:

T_3 = a + 2d = 14

and

T_7 = a + 6d = 34

. Subtracting the first equation from the second yields

4d = 20 \Rightarrow d = 5

. Substituting

d = 5

into

a + 2(5) = 14

gives

a = 4

.

2

Calculate the 4th term of the geometric progression (

G_4

).

G_4 = 54

The

m

-th term of a GP is given by

G_m = a_{gp} \cdot r^{m-1}

. With first term

a_{gp} = 2

and ratio

r = 3

,

G_4 = 2 \cdot 3^{4-1} = 2 \cdot 3^3 = 2 \cdot 27 = 54

.

3

Equate

T_n

to

G_4

and solve for

n

.

n = 11

Set

T_n = G_4 \Rightarrow 4 + (n-1)5 = 54

. Simplifying gives

(n-1)5 = 50 \Rightarrow n-1 = 10 \Rightarrow n = 11

.

Key Concept

Solving simultaneous AP/GP equations using the

n

-th term formulas

T_n = a + (n-1)d

and

G_n = a r^{n-1}

.

Estimated Time:2m 0s

Question 6Question →

The fourth term of an arithmetic progression (A.P.) is $15$ and the ninth term is $35$ . What is the common difference of the progression?

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

Using the A.P. term formula

T_n = a + (n-1)d

, the fourth term gives

a + 3d = 15

and the ninth term gives

a + 8d = 35

. Subtracting the two equations yields

5d = 20

, which simplifies directly to

d = 4

.

Step-by-Step Solution

1

Express the given terms using the n-th term formula

T_n = a + (n-1)d

a + 3d = 15

and

a + 8d = 35

The

n^{\text{th}}

term formula relates any term to the first term (

a

) and common difference (

d

).

2

Subtract the equation for the fourth term from the ninth term

5d = 20

Subtracting eliminates the first term

a

and leaves a simple equation in terms of

d

.

3

Divide by

5

to solve for

d

d = 4

Dividing both sides of

5d = 20

by

5

gives the common difference.

Key Concept

Finding the common difference of an Arithmetic Progression given two non-consecutive terms

Question 7Question →

The sum of the first three terms of an arithmetic progression (AP) with a positive common difference $d$ is $21$ . If $2$ is added to the first term, $3$ is added to the second term, and $9$ is added to the third term, the resulting three numbers form consecutive terms of a geometric progression (GP). What is the sum of the first $10$ terms of this arithmetic progression?

Show answer & explanation

Answer: 210

Answer

The sum of the first 10 terms of the arithmetic progression is 210.

Representing the AP terms as

7-d, 7, 7+d

and adding the specified values produces GP terms

9-d, 10, 16+d

. Solving

10^2 = (9-d)(16+d)

yields

d=4

. Consequently, the first term of the AP is

3

. Using

S_{10} = \frac{10}{2}[2(3) + 9(4)]

gives the final answer 210.

Step-by-Step Solution

1

Express AP terms symmetrically and solve for the middle term.

The middle term is

a = 7

, making the terms

7-d

,

7

, and

7+d

.

Choosing terms

a-d, a, a+d

allows the sum equation

3a = 21

to directly isolate the middle term.

2

Set up the geometric progression relation to determine common difference

d

.

The GP terms are

9-d

,

10

, and

16+d

. Solving

10^2 = (9-d)(16+d)

gives

d^2 + 7d - 44 = 0

, yielding

d = 4

.

In any geometric progression, the square of the middle term equals the product of the first and third terms.

3

Determine the first term

a_1

and calculate

S_{10}

.

The first term is

a_1 = 7 - 4 = 3

, and the sum

S_{10} = \frac{10}{2}[2(3) + (10-1)(4)] = 210

.

Applying the AP sum formula

S_n = \frac{n}{2}[2a_1 + (n-1)d]

with

n=10

,

a_1=3

, and

d=4

.

Key Concept

Integrating AP and GP structural relationships to solve for sequence parameters and evaluate finite sums

Question 8Question →

The $2^{\text{nd}}$ , $5^{\text{th}}$ , and $14^{\text{th}}$ terms of a non-constant arithmetic progression (A.P.) form three consecutive terms of a geometric progression (G.P.). If the first term of the A.P. is $3$ , what is the sum of the first $6$ terms of the A.P.?

108

126

99

90

Show answer & explanation

Answer: 108

Answer

108

With first term

a=3

, the terms

T_2 = 3+d

,

T_5 = 3+4d

, and

T_{14} = 3+13d

form a geometric progression. Therefore,

(3+4d)^2 = (3+d)(3+13d)

. Expanding gives

9 + 24d + 16d^2 = 9 + 42d + 13d^2

, which simplifies to

3d^2 - 18d = 0

. Since the sequence is non-constant (

d \neq 0

),

d = 6

. Using the A.P. sum formula

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

,

S_6 = 3[2(3) + 5(6)] = 3[6 + 30] = 108

.

Step-by-Step Solution

1

Express the

2^{\text{nd}}

,

5^{\text{th}}

, and

14^{\text{th}}

terms of the A.P. in terms of the first term

a=3

and common difference

d

.

T_2 = 3 + d

,

T_5 = 3 + 4d

, and

T_{14} = 3 + 13d

.

The

n^{\text{th}}

term of an A.P. is given by

T_n = a + (n-1)d

.

2

Set up the condition for consecutive terms of a G.P. and solve for

d

.

(3 + 4d)^2 = (3 + d)(3 + 13d) \implies 9 + 24d + 16d^2 = 9 + 42d + 13d^2 \implies 3d^2 - 18d = 0 \implies d = 6

.

For consecutive terms in a G.P., the middle term squared equals the product of the adjacent terms (

T_5^2 = T_2 \times T_{14}

).

3

Calculate the sum of the first

6

terms of the A.P. using

a=3

and

d=6

.

S_6 = \frac{6}{2}[2(3) + (6-1)(6)] = 3[6 + 30] = 3 \times 36 = 108

.

The sum formula for an A.P. is

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

.

Key Concept

Simultaneous conditions connecting A.P. and G.P. terms combined with sequence sum formulas.

Estimated Time:2m 0s

Question 9Question →

The sum of the first $n$ terms of an arithmetic progression is given by $S_n = 2n^2 + 3n$ . What is the value of the $7^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 29

Answer

The

7^{\text{th}}

term of the arithmetic progression is

29

.

For any sequence, the

n

-th term is found using

T_n = S_n - S_{n-1}

. Substituting

n = 7

gives

S_7 = 2(7)^2 + 3(7) = 119

and

S_6 = 2(6)^2 + 3(6) = 90

. Thus,

T_7 = 119 - 90 = 29

.

Step-by-Step Solution

1

State the relationship between the

n

-th term

T_n

and the sum of first

n

terms

S_n

T_n = S_n - S_{n-1}

The sum of the first

n

terms minus the sum of the first

n-1

terms equals the

n

-th term.

2

Calculate the sum of the first 7 terms (

S_7

)

S_7 = 2(7)^2 + 3(7) = 119

Substitute

n = 7

into the sum formula

S_n = 2n^2 + 3n

.

3

Calculate the sum of the first 6 terms (

S_6

)

S_6 = 2(6)^2 + 3(6) = 90

Substitute

n = 6

into the sum formula

S_n = 2n^2 + 3n

.

4

Compute the

7^{\text{th}}

term (

T_7

)

T_7 = 119 - 90 = 29

Subtract

S_6

from

S_7

.

Key Concept

Relationship between the nth term and the sum of first n terms of an AP

Question 10Question →

The sum of the first $n$ terms of an arithmetic progression (A.P.) is $210$ . If the first term is $3$ and the last term is $39$ , what is the value of $n$ ?

10

11

5

9

Show answer & explanation

Answer: 10

Answer

The number of terms

n

is

10

.

Using the standard sum formula

S_n = \frac{n}{2}(a + l)

for an A.P. with known first term

a = 3

and last term

l = 39

, we set

210 = \frac{n}{2}(3 + 39) = 21n

. Solving for

n

gives

n = 10

.

Step-by-Step Solution

1

Identify the given parameters of the arithmetic progression.

First term

a = 3

, last term

l = 39

, and sum

S_n = 210

.

These values are required to apply the sum formula for an A.P.

2

Apply the sum formula

S_n = \frac{n}{2}(a + l)

.

210 = \frac{n}{2}(3 + 39) = \frac{n}{2}(42) = 21n

.

The sum of

n

terms in an A.P. with a known first and last term is given by

\frac{n}{2}(a + l)

.

3

Solve for

n

.

n = \frac{210}{21} = 10

.

Dividing the total sum by

21

gives the exact number of terms.

Key Concept

Sum of an Arithmetic Progression using first and last terms

Estimated Time:1m 30s

Question 11Question →

The sum of the first four terms of an arithmetic progression (A.P.) is $32$ , and the sum of the next four terms is $96$ . What is the common difference of the progression?

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

The sum of the first four terms yields

2a + 3d = 16

. The sum of the first eight terms is

32 + 96 = 128

, which gives

2a + 7d = 32

. Subtracting these two linear equations gives

4d = 16

, leading to

d = 4

.

Step-by-Step Solution

1

Formulate an equation for the sum of the first 4 terms.

2a + 3d = 16

The sum of the first

n

terms of an A.P. is

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

. Substituting

n = 4

and

S_4 = 32

gives

2(2a + 3d) = 32

, which simplifies to

2a + 3d = 16

.

2

Formulate an equation for the sum of the first 8 terms.

2a + 7d = 32

The total sum of the first 8 terms is the sum of the first 4 terms plus the sum of the next 4 terms (

S_8 = 32 + 96 = 128

). Substituting

n = 8

gives

4(2a + 7d) = 128

, which simplifies to

2a + 7d = 32

.

3

Solve the system of simultaneous linear equations for

d

.

d = 4

Subtracting

(2a + 3d = 16)

from

(2a + 7d = 32)

eliminates

2a

, resulting in

4d = 16

, which yields

d = 4

.

Key Concept

Sum of an Arithmetic Progression

Question 12Question →

The $3^{\text{rd}}$ and $6^{\text{th}}$ terms of an Arithmetic Progression (A.P.) are $13$ and $28$ respectively. What is the sum of the first $10$ terms of the progression?

255

280

205

48

Show answer & explanation

Answer: $255$

Answer

The sum of the first

10

terms of the progression is

255

.

The

3^{\text{rd}}

term is

a + 2d = 13

and the

6^{\text{th}}

term is

a + 5d = 28

. Subtracting these equations gives

3d = 15

, so

d = 5

, which leads to

a = 3

. Using the sum formula

S_{10} = \frac{10}{2}[2(3) + (10 - 1)5]

, we obtain

5(6 + 45) = 255

.

Step-by-Step Solution

1

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

T_n = a + (n - 1)d

.

a + 2d = 13

and

a + 5d = 28

.

The

3^{\text{rd}}

term corresponds to

n=3

and the

6^{\text{th}}

term corresponds to

n=6

.

2

Solve the simultaneous equations for

a

(first term) and

d

(common difference).

Subtracting the first equation from the second gives

3d = 15 \Rightarrow d = 5

. Substituting

d = 5

into the first equation yields

a + 2(5) = 13 \Rightarrow a = 3

.

To find any property of an A.P., the first term

a

and common difference

d

must be determined.

3

Calculate the sum of the first

10

terms using

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

.

S_{10} = \frac{10}{2}[2(3) + (10 - 1)5] = 5[6 + 45] = 5(51) = 255

.

Applying the formula for the sum of the first

n

terms with

n = 10

,

a = 3

, and

d = 5

.

Key Concept

Finding the sum of the first

n

terms of an Arithmetic Progression given two specific terms.

Question 13Question →

If $x - 1$ , $x + 2$ , and $3x$ are three consecutive terms of a geometric progression (G.P.) with positive terms, what is the common ratio of the progression?

2

4

3

\frac{1}{2}

Show answer & explanation

Answer: $2$

Answer

The common ratio of the progression is

2

.

For any three consecutive terms in a G.P., the square of the middle term equals the product of the first and third terms. Solving

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

gives

2x^2 - 7x - 4 = 0

, which yields

x = 4

for positive terms. Substituting

x = 4

gives the terms

3, 6, 12

, which have a common ratio of

6 \div 3 = 2

.

Step-by-Step Solution

1

Set up the condition for consecutive terms in a Geometric Progression.

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

For three consecutive terms

a, b, c

in G.P., the middle term squared equals the product of the outer terms (

b^2 = ac

).

2

Expand both sides and rearrange into a quadratic equation.

x^2 + 4x + 4 = 3x^2 - 3x \implies 2x^2 - 7x - 4 = 0

Expanding allows gathering all terms on one side to solve for

x

.

3

Factorize the quadratic equation to find

x

.

(2x + 1)(x - 4) = 0 \implies x = 4

(since terms are positive,

x > 1

).

The solution

x = -\frac{1}{2}

gives negative terms, so

x = 4

is chosen.

4

Find the consecutive terms and calculate the common ratio

r

.

Terms are

4 - 1 = 3

,

4 + 2 = 6

, and

3(4) = 12

. Common ratio

r = \frac{6}{3} = 2

.

Dividing the second term by the first term gives the common ratio

r

.

Key Concept

Geometric Progression Consecutive Terms Property (

b^2 = ac

)

Estimated Time:1m 30s

Question 14Question →

The second term of a geometric progression (G.P.) is $6$ and its fifth term is $48$ . What is the sum of the first $6$ terms of the progression?

189

93

96

378

Show answer & explanation

Answer: 189

Answer

189

Using the geometric progression terms

T_2 = a r = 6

and

T_5 = a r^4 = 48

, dividing

T_5

by

T_2

gives

r^3 = 8

, so

r = 2

. Substituting

r = 2

into

a r = 6

yields

a = 3

. The sum of the first 6 terms is calculated as

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

.

Step-by-Step Solution

1

Set up equations using the

n^{\text{th}}

term formula for a G.P.,

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

Relate given terms to the first term

a

and common ratio

r

.

2

Divide the equation for

T_5

by the equation for

T_2

to find the common ratio

r

.

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

Eliminate the variable

a

to solve for

r

.

3

Substitute

r = 2

back into

a r = 6

to find the first term

a

.

a(2) = 6 \implies a = 3

Determine the first term of the progression.

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(63) = 189

Apply the sum formula for a finite geometric progression.

Key Concept

Geometric Progression (G.P.)

n^{\text{th}}

term and sum of finite terms

Question 15Question →

The sum to infinity of a geometric progression (G.P.) with positive terms is $18$ , and the sum of its first two terms is $16$ . What is the first term of the progression?

Show answer & explanation

Answer: 12

Answer

The first term of the progression is 12.

Using the sum to infinity formula

S_{\infty} = \frac{a}{1 - r} = 18

, we express the first term as

a = 18(1 - r)

. Combining this with the sum of the first two terms

S_2 = a(1 + r) = 16

yields

18(1 - r)(1 + r) = 16 \implies 18(1 - r^2) = 16

. Solving for

r

gives

r^2 = \frac{1}{9}

, so

r = \frac{1}{3}

for a sequence with positive terms. Substituting

r = \frac{1}{3}

back into

a = 18(1 - r)

gives

a = 12

.

Step-by-Step Solution

1

Express the sum to infinity in terms of the first term

a

and common ratio

r

.

a = 18(1 - r)

The sum to infinity formula for a convergent G.P. is

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

.

2

Write the expression for the sum of the first two terms.

a(1 + r) = 16

The sum of the first two terms is

T_1 + T_2 = a + ar = a(1 + r)

.

3

Substitute

a = 18(1 - r)

into the sum of the first two terms equation.

18(1 - r^2) = 16

Applying the difference of two squares identity

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

.

4

Solve for the common ratio

r

.

r = \frac{1}{3}

Rearranging gives

1 - r^2 = \frac{8}{9} \implies r^2 = \frac{1}{9}

. Since all terms are positive,

r

must be positive.

5

Calculate the first term

a

.

a = 12

Substitute

r = \frac{1}{3}

into

a = 18(1 - r)

to get

a = 18 \times \frac{2}{3} = 12

.

Key Concept

Geometric Progression sum to infinity and partial sums

Estimated Time:1m 30s

Question 16Question →

The $4^{\text{th}}$ term of an arithmetic progression (A.P.) is $15$ and the $9^{\text{th}}$ term is $35$ . Calculate the sum of the first $10$ terms of the progression.

Show answer & explanation

Answer: 210

Answer

The sum of the first 10 terms of the progression is 210.

Using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

, the equations

a + 3d = 15

and

a + 8d = 35

yield common difference

d = 4

and first term

a = 3

. Substituting these into the sum formula

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

for

n = 10

yields

S_{10} = 5[2(3) + 9(4)] = 210

.

Step-by-Step Solution

1

Set up equations for the given terms

a + 3d = 15

and

a + 8d = 35

The

n^{\text{th}}

term of an A.P. is given by

T_n = a + (n-1)d

.

2

Solve for the common difference

d

d = 4

Subtracting

(a + 3d = 15)

from

(a + 8d = 35)

gives

5d = 20

, so

d = 4

.

3

Solve for the first term

a

a = 3

Substituting

d = 4

into

a + 3(4) = 15

yields

a = 15 - 12 = 3

.

4

Calculate the sum of the first 10 terms

S_{10} = 210

Applying

S_{10} = \frac{10}{2}[2(3) + 9(4)] = 5(6 + 36) = 210

.

Key Concept

Finding terms and sums of an Arithmetic Progression using simultaneous linear equations

Estimated Time:1m 30s

Question 17Question →

The sum of the first $n$ terms of an arithmetic progression (A.P.) is given by $S_n = 3n^2 + 5n$ . A geometric progression (G.P.) has a first term of $1$ and a common ratio of $2$ . If the $5^{\text{th}}$ term of the A.P. is equal to the $k^{\text{th}}$ term of the G.P., what is the value of $k$ ?

5

6

4

7

Show answer & explanation

Answer: 6

Answer

The value of

k

is 6.

Evaluating the

5^{\text{th}}

term of the A.P. gives

32

. Equating this to the

k^{\text{th}}

term formula of the G.P.,

1 \cdot 2^{k-1} = 32

, leads to

2^{k-1} = 2^5

, which gives

k = 6

.

Step-by-Step Solution

1

Find the general formula for the

n^{\text{th}}

term

T_n

of the A.P. using

S_n

T_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2

The

n^{\text{th}}

term of a series is the difference between the sum of the first

n

terms and the sum of the first

n-1

terms.

2

Calculate the

5^{\text{th}}

term of the A.P.

T_5 = 6(5) + 2 = 32

Substitute

n = 5

into the derived expression for

T_n

.

3

Set up the equation for the

k^{\text{th}}

term of the G.P.

G_k = a \cdot r^{k-1} = 1 \cdot 2^{k-1} = 2^{k-1}

The standard formula for the

k^{\text{th}}

term of a G.P. is

G_k = a \cdot r^{k-1}

.

4

Equate

T_5

and

G_k

to solve for

k

2^{k-1} = 32 \implies 2^{k-1} = 2^5 \implies k - 1 = 5 \implies k = 6

Since the bases are equal (

2

), equate the exponents to find

k

.

Key Concept

Relating Arithmetic Progression sum formulas to term values and solving indexed Geometric Progression equations.

Estimated Time:2m 0s

Question 18Question →

The first term of an arithmetic progression (A.P.) is $7$ and its common difference is $4$ . What is the value of the $12^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 51

Answer

The

12^{\text{th}}

term of the arithmetic progression is

51

.

Applying the arithmetic progression

n^{\text{th}}

term formula

T_n = a + (n - 1)d

with first term

a = 7

, common difference

d = 4

, and term index

n = 12

gives

T_{12} = 7 + (12 - 1) \times 4 = 7 + 44 = 51

.

Step-by-Step Solution

1

Identify given variables

a = 7

,

d = 4

, and

n = 12

These are the necessary components to calculate the required term position in an arithmetic sequence.

2

Apply the

n^{\text{th}}

term formula for an AP

T_{12} = 7 + (12 - 1) \times 4

The standard formula

T_n = a + (n - 1)d

determines the value of any term in an AP.

3

Compute the numerical result

T_{12} = 7 + 44 = 51

Multiplying the common difference by 11 and adding the first term yields the correct value.

Key Concept

Calculating the nth term of an Arithmetic Progression

Question 19Question →

An infinite geometric progression of positive terms has a sum to infinity of $16$ , and the sum of its first two terms is $12$ . An arithmetic progression has its first term equal to the first term of this geometric progression, and its $5^{\text{th}}$ term equal to the sum to infinity of the geometric progression. Calculate the $10^{\text{th}}$ term of the arithmetic progression.

Show answer & explanation

Answer: 26

Answer

The 10th term of the arithmetic progression is 26.

By solving the geometric progression system, we find the common ratio

r = \frac{1}{2}

and first term

a = 8

. Using

a = 8

as the first term of the arithmetic progression and setting its

5^{\text{th}}

term

A_5 = 16

, we determine the common difference

d = 2

. Calculating

A_{10} = 8 + 9(2)

yields

26

.

Step-by-Step Solution

1

Formulate equations for the geometric progression using the sum to infinity and sum of the first two terms

a = 16(1 - r)

and

a(1 + r) = 12

The standard formula for the sum to infinity of a GP is

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

and the sum of the first two terms is

S_2 = a + ar = a(1+r)

.

2

Solve for the common ratio

r

and first term

a

of the geometric progression

r = 0.5

and

a = 8

Substituting

a = 16(1-r)

yields

16(1-r^2) = 12 \implies r^2 = \frac{1}{4} \implies r = \frac{1}{2}

. Then

a = 16(1 - 0.5) = 8

.

3

Determine the common difference

d

of the arithmetic progression

d = 2

The first term of the AP is

A_1 = a = 8

and the 5th term is

A_5 = S_\infty = 16

. Using

A_5 = A_1 + 4d \implies 8 + 4d = 16 \implies d = 2

.

4

Calculate the 10th term of the arithmetic progression

A_{10} = 26

Using the AP

n^{\text{th}}

term formula

A_n = A_1 + (n-1)d

:

A_{10} = 8 + 9(2) = 26

.

Key Concept

Combining geometric progression parameters (sum to infinity and sum of terms) with arithmetic progression term formulas

Question 20Question →

The $1^{\text{st}}$ , $2^{\text{nd}}$ , and $5^{\text{th}}$ terms of an arithmetic progression (A.P.) with a non-zero common difference form three consecutive terms of a geometric progression (G.P.). If the sum of the first $4$ terms of the A.P. is $40$ , what is the $5^{\text{th}}$ term of the G.P.?

\frac{405}{2}

\frac{1215}{2}

162

\frac{45}{2}

Show answer & explanation

Answer: $\frac{405}{2}$

Answer

\frac{405}{2}

By setting up the geometric mean property

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

, we find

d = 2a

, which establishes that the G.P. has a common ratio

r = 3

. Substituting

d = 2a

into the A.P. sum formula

S_4 = 2[2a + 3d] = 16a = 40

gives

a = \frac{5}{2}

. Finally, evaluating the

5^{\text{th}}

term of the G.P. using

a r^4 = \frac{5}{2} \times 3^4

yields

\frac{405}{2}

.

Step-by-Step Solution

1

Express the given A.P. terms in terms of first term

a

and common difference

d

, and set up the G.P. condition.

The terms are

T_1 = a

,

T_2 = a + d

, and

T_5 = a + 4d

. Since they form a G.P.,

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

.

Three terms

x, y, z

form a G.P. if

y^2 = xz

.

2

Solve for the relationship between

d

and

a

.

a^2 + 2ad + d^2 = a^2 + 4ad \implies d^2 = 2ad \implies d = 2a

(since

d \neq 0

).

Expanding and simplifying the equation yields the ratio of

d

to

a

.

3

Determine the common ratio

r

of the G.P.

r = \frac{T_2}{T_1} = \frac{a + d}{a} = \frac{a + 2a}{a} = 3

.

The common ratio is the quotient of consecutive terms of the G.P.

4

Use the sum of the first

4

terms of the A.P. to find

a

.

S_4 = \frac{4}{2}[2a + (4-1)d] = 2[2a + 3(2a)] = 16a = 40 \implies a = \frac{40}{16} = \frac{5}{2}

.

Applying

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

with

S_4 = 40

allows solving for

a

.

5

Calculate the

5^{\text{th}}

term of the G.P.

G_5 = g_1 \cdot r^{5-1} = a \cdot r^4 = \frac{5}{2} \cdot 3^4 = \frac{5}{2} \cdot 81 = \frac{405}{2}

.

The

n^{\text{th}}

term of a G.P. is

g_n = g_1 r^{n-1}

.

Key Concept

Connecting Arithmetic and Geometric Progressions using term definitions and sum formulas.

Arithmetic and Geometric Progressions (AP and GP)

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