Progressions - Solutions

Progressions - 1 Mark Problems

Important Formulas:

Arithmetic Progression (AP):
• nth term: aₙ = a + (n-1)d
• Sum of first n terms: Sₙ = n/2 [2a + (n-1)d]
• Common difference: d = aₙ - aₙ₋₁

Geometric Progression (GP):
• nth term: aₙ = a · rⁿ⁻¹
• Common ratio: r = aₙ / aₙ₋₁
• Sum of first n terms: Sₙ = a(1-rⁿ)/(1-r) for r≠1

The hand-bore well dealer charges Rs 200/- for the first one meter only and raises drilling charges at the rate of rupees 30/- for every subsequent meter. Write a progression for the above data. (M'15)

Step 1: Identify the pattern

First meter: Rs 200

Second meter: Rs 200 + 30 = Rs 230

Third meter: Rs 230 + 30 = Rs 260

And so on...

∴ The progression is: 200, 230, 260, 290, ...

In a flower garden, there are 23 plants in the first row, 21 plants in the second row, 19 plants in the third row and so on. If there are 10 rows in that flower garden, then find the total number of plants in the last row with the help of the formula tₙ = a + (n – 1)d. (M'15)

Step 1: Identify values

First term (a) = 23

Common difference (d) = 21 - 23 = -2

Number of terms (n) = 10

Step 2: Apply the formula

tₙ = a + (n - 1)d

t₁₀ = 23 + (10 - 1)(-2)

t₁₀ = 23 + 9(-2)

t₁₀ = 23 - 18 = 5

∴ There are 5 plants in the last row.

Write the common difference of an Arithmetic Progression, whose nth term is given by tₙ = 3n + 7. (J'15)

Step 1: Find first term

t₁ = 3(1) + 7 = 3 + 7 = 10

Step 2: Find second term

t₂ = 3(2) + 7 = 6 + 7 = 13

Step 3: Calculate common difference

d = t₂ - t₁ = 13 - 10 = 3

∴ The common difference is 3.

Find the sum of first 200 natural numbers. (M'16)

Step 1: Use the formula for sum of first n natural numbers

Sₙ = n(n+1)/2

Step 2: Substitute n = 200

S₂₀₀ = 200(200+1)/2

S₂₀₀ = 200 × 201 / 2

S₂₀₀ = 100 × 201

S₂₀₀ = 20,100

∴ The sum of first 200 natural numbers is 20,100.

Is 'zero' a term of the Arithmetic Progression 31, 28, 25, ...? Justify your answer. (J'16)

Step 1: Identify values

First term (a) = 31

Common difference (d) = 28 - 31 = -3

Step 2: Check if 0 is a term

tₙ = a + (n-1)d

0 = 31 + (n-1)(-3)

0 = 31 - 3(n-1)

3(n-1) = 31

n-1 = 31/3 ≈ 10.33

n = 11.33

Step 3: Interpret the result

Since n is not a natural number (n = 11.33), 0 is not a term of this AP.

∴ No, zero is not a term of this AP since n is not a natural number.

In a G.P. tₙ = (-1)ⁿ·2017. Find the common ratio. (M'17)

Step 1: Find first term

t₁ = (-1)¹·2017 = -2017

Step 2: Find second term

t₂ = (-1)²·2017 = 2017

Step 3: Calculate common ratio

r = t₂ / t₁ = 2017 / (-2017) = -1

∴ The common ratio is -1.

The nth term of AP is 6n + 2. Find the common difference (n∈ N). (J'17)

Step 1: Find first term

t₁ = 6(1) + 2 = 6 + 2 = 8

Step 2: Find second term

t₂ = 6(2) + 2 = 12 + 2 = 14

Step 3: Calculate common difference

d = t₂ - t₁ = 14 - 8 = 6

∴ The common difference is 6.

The sequence √3, √6, √9, √12, ... form an Arithmetic Progression? Give reason. (M'18)

Step 1: Check for common difference

First term = √3

Second term = √6

Third term = √9 = 3

Fourth term = √12 = 2√3

Step 2: Calculate differences

d₁ = √6 - √3

d₂ = 3 - √6

d₃ = 2√3 - 3

Step 3: Compare differences

√6 - √3 ≈ 2.449 - 1.732 = 0.717

3 - √6 ≈ 3 - 2.449 = 0.551

Since 0.717 ≠ 0.551, the differences are not equal.

∴ No, it does not form an AP because the common difference is not constant.

For the A.P.: -3, -7, -11, ... ; can we find directly a₃₀ - a₂₀ without actually finding a₃₀ and a₂₀? (J'18)

Step 1: Use the formula for nth term

aₙ = a + (n-1)d

a₃₀ = a + 29d

a₂₀ = a + 19d

Step 2: Find the difference

a₃₀ - a₂₀ = (a + 29d) - (a + 19d) = 10d

Step 3: Find common difference

d = -7 - (-3) = -4

Step 4: Calculate a₃₀ - a₂₀

a₃₀ - a₂₀ = 10d = 10(-4) = -40

∴ Yes, we can find directly: a₃₀ - a₂₀ = 10d = -40.

Radha says "1, 1, 1, ... are in A.P. and also in G.P". Do you agree with Radha? Give reason. (M'19)

Step 1: Check for AP

Common difference d = 1 - 1 = 0

Since d is constant, it is an AP.

Step 2: Check for GP

Common ratio r = 1/1 = 1

Since r is constant, it is a GP.

∴ Yes, I agree with Radha. The sequence 1, 1, 1, ... is both an AP (with d=0) and a GP (with r=1).

Find the 11th term of the A.P.: 15, 12, 9, ... (J'19)

Step 1: Identify values

First term (a) = 15

Common difference (d) = 12 - 15 = -3

n = 11

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₁ = 15 + (11-1)(-3)

t₁₁ = 15 + 10(-3)

t₁₁ = 15 - 30 = -15

∴ The 11th term is -15.

Find the 10th term of the arithmetic progression 3, 5, 7, ... (May 2022)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 5 - 3 = 2

n = 10

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₀ = 3 + (10-1)(2)

t₁₀ = 3 + 9(2)

t₁₀ = 3 + 18 = 21

∴ The 10th term is 21.

Find the 10th term of an arithmetic progression 5, 9, 13, ... (Additional Problem)

Step 1: Identify values

First term (a) = 5

Common difference (d) = 9 - 5 = 4

n = 10

Step 2: Apply the formula

tₙ = a + (n-1)d

t₁₀ = 5 + (10-1)(4)

t₁₀ = 5 + 9(4)

t₁₀ = 5 + 36 = 41

∴ The 10th term is 41.

Progressions - 2 Mark Problems

Important Formulas:

If 7 times of 7th term of an Arithmetic Progression is equal to the 11 times of 11th term of it, then find the 18th term of that Arithmetic Progression. (J'15)

Step 1: Write the given condition

7 × a₇ = 11 × a₁₁

Step 2: Express terms using AP formula

a₇ = a + 6d

a₁₁ = a + 10d

7(a + 6d) = 11(a + 10d)

Step 3: Solve for a in terms of d

7a + 42d = 11a + 110d

7a - 11a = 110d - 42d

-4a = 68d

a = -17d

Step 4: Find the 18th term

a₁₈ = a + 17d = -17d + 17d = 0

∴ The 18th term of the AP is 0.

Measures of sides of a triangle are in Arithmetic Progression. Its perimeter is 30 cm, the difference between the longest and shortest side is 4 cm; then find the measures of the sides. (M'16)

Step 1: Let the sides be in AP

Let the sides be: a-d, a, a+d

(Where a is the middle term and d is the common difference)

Step 2: Apply perimeter condition

Perimeter = (a-d) + a + (a+d) = 30

3a = 30

a = 10

Step 3: Apply difference condition

Longest side - Shortest side = 4

(a+d) - (a-d) = 4

2d = 4

d = 2

Step 4: Find the sides

First side = a-d = 10-2 = 8 cm

Second side = a = 10 cm

Third side = a+d = 10+2 = 12 cm

∴ The sides of the triangle are 8 cm, 10 cm, and 12 cm.

Explain the terms in the formula Sₙ = n/2 [2a + (n − 1)d]. (J'16)

Step 1: Write the formula

Sₙ = n/2 [2a + (n − 1)d]

Step 2: Explain each term

Sₙ = Sum of first n terms of the AP

n = Number of terms

a = First term of the AP

d = Common difference of the AP

Step 3: Explain the expression inside brackets

2a + (n-1)d = Sum of first and last terms

(Since last term = a + (n-1)d)

∴ Sₙ represents the sum of first n terms, n is number of terms, a is first term, and d is common difference.

Find the sum of the first 10 terms of an A.P. 3, 15, 27, 39, .............. (M'17)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 15 - 3 = 12

Number of terms (n) = 10

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₁₀ = 10/2 [2(3) + (10-1)(12)]

S₁₀ = 5 [6 + 9(12)]

S₁₀ = 5 [6 + 108]

S₁₀ = 5 × 114 = 570

∴ The sum of first 10 terms is 570.

Find the value of 'k' so that k + 2, 4k – 6 and 3k – 2 are the three consecutive terms of an A.P. (J'17)

Step 1: Condition for three consecutive terms in AP

If a, b, c are in AP, then 2b = a + c

Step 2: Apply the condition

2(4k - 6) = (k + 2) + (3k - 2)

8k - 12 = k + 2 + 3k - 2

8k - 12 = 4k

Step 3: Solve for k

8k - 4k = 12

4k = 12

k = 3

∴ The value of k is 3.

Find the 7th term from the end of the arithmetic progression 7, 10, 13, ..... 184. (M'18)

Step 1: Find the total number of terms

First term (a) = 7

Common difference (d) = 10 - 7 = 3

Last term (l) = 184

l = a + (n-1)d

184 = 7 + (n-1)(3)

184 - 7 = 3(n-1)

177 = 3(n-1)

n-1 = 59

n = 60

Step 2: Find the 7th term from the end

7th term from end = (n-7+1)th term from beginning

= (60-7+1)th term = 54th term

Step 3: Calculate the 54th term

a₅₄ = a + (54-1)d

a₅₄ = 7 + 53(3)

a₅₄ = 7 + 159 = 166

∴ The 7th term from the end is 166.

In a rangoli design of 13 rows, every row increases its previous row by two dots and first row contains 5 dots, then how many total dots are in the design? (J'18)

Step 1: Identify the AP

First term (a) = 5

Common difference (d) = 2

Number of terms (n) = 13

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₁₃ = 13/2 [2(5) + (13-1)(2)]

S₁₃ = 13/2 [10 + 12(2)]

S₁₃ = 13/2 [10 + 24]

S₁₃ = 13/2 × 34

S₁₃ = 13 × 17 = 221

∴ The total number of dots in the design is 221.

Write the formula of nth term of G.P. and explain the terms in it. (M'19)

Step 1: Write the formula

aₙ = a · rⁿ⁻¹

Step 2: Explain each term

aₙ = nth term of the GP

a = First term of the GP

r = Common ratio of the GP

n = Position of the term

Step 3: Additional explanation

The formula shows that each term is obtained by multiplying the previous term by the common ratio r.

∴ aₙ = a·rⁿ⁻¹, where a is first term, r is common ratio, and n is term position.

Which term of the G.P.: √2, 2, 2√2, 4, ... is 32? (J'19)

Step 1: Identify values

First term (a) = √2

Common ratio (r) = 2/√2 = √2

nth term (aₙ) = 32

Step 2: Apply GP formula

aₙ = a · rⁿ⁻¹

32 = √2 · (√2)ⁿ⁻¹

32 = (√2)ⁿ

Step 3: Solve for n

32 = 2⁵

(√2)ⁿ = 2⁵

(2¹ᐟ²)ⁿ = 2⁵

2ⁿᐟ² = 2⁵

n/2 = 5

n = 10

∴ 32 is the 10th term of the GP.

5, 8, 11, 14, ........... is an arithmetic progression. Find the sum of first 20 terms of it. (May 22)

Step 1: Identify values

First term (a) = 5

Common difference (d) = 8 - 5 = 3

Number of terms (n) = 20

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₂₀ = 20/2 [2(5) + (20-1)(3)]

S₂₀ = 10 [10 + 19(3)]

S₂₀ = 10 [10 + 57]

S₂₀ = 10 × 67 = 670

∴ The sum of first 20 terms is 670.

3, 6, 9, 12, ............ is an arithmetic progression. Find the sum of first 20 terms of the progression. (Aug 22)

Step 1: Identify values

First term (a) = 3

Common difference (d) = 6 - 3 = 3

Number of terms (n) = 20

Step 2: Apply sum formula

Sₙ = n/2 [2a + (n-1)d]

S₂₀ = 20/2 [2(3) + (20-1)(3)]

S₂₀ = 10 [6 + 19(3)]

S₂₀ = 10 [6 + 57]

S₂₀ = 10 × 63 = 630

∴ The sum of first 20 terms is 630.

Which term of the A.P. 21, 18, 15, ... is –81? Also find the term which becomes zero. (Apr'23)

Step 1: Identify values

First term (a) = 21

Common difference (d) = 18 - 21 = -3

Step 2: Find n when aₙ = -81

aₙ = a + (n-1)d

-81 = 21 + (n-1)(-3)

-81 - 21 = (n-1)(-3)

-102 = (n-1)(-3)

n-1 = 102/3 = 34

n = 35

Step 3: Find n when aₙ = 0

0 = 21 + (n-1)(-3)

-21 = (n-1)(-3)

n-1 = 21/3 = 7

n = 8

∴ -81 is the 35th term and 0 is the 8th term.

If 6 times of 6th term of an arithmetic progression is equal to 9 times of 9th term of it, then show that 15th term of that A.P. is zero. (Jun'23)

Step 1: Write the given condition

6 × a₆ = 9 × a₉

Step 2: Express terms using AP formula

a₆ = a + 5d

a₉ = a + 8d

6(a + 5d) = 9(a + 8d)

Step 3: Solve for a in terms of d

6a + 30d = 9a + 72d

6a - 9a = 72d - 30d

-3a = 42d

a = -14d

Step 4: Find the 15th term

a₁₅ = a + 14d = -14d + 14d = 0

∴ The 15th term is 0, as required to show.

Progressions - 4 Mark Problems

Important Formulas:

If the sum of the first 7 terms of an Arithmetic Progression is 49 and that of first 17 terms is 289, then find the sum of first 'n' terms. (M'15)

Step 1: Write the given conditions

S₇ = 7/2 [2a + (7-1)d] = 49

7/2 [2a + 6d] = 49

7(a + 3d) = 49

a + 3d = 7 ...(1)

Step 2: Use second condition

S₁₇ = 17/2 [2a + (17-1)d] = 289

17/2 [2a + 16d] = 289

17(a + 8d) = 289

a + 8d = 17 ...(2)

Step 3: Solve equations (1) and (2)

Subtract (1) from (2):

(a + 8d) - (a + 3d) = 17 - 7

5d = 10

d = 2

Substitute in (1): a + 3(2) = 7

a + 6 = 7

a = 1

Step 4: Find sum of first n terms

Sₙ = n/2 [2a + (n-1)d]

Sₙ = n/2 [2(1) + (n-1)(2)]

Sₙ = n/2 [2 + 2n - 2]

Sₙ = n/2 [2n]

Sₙ = n²

∴ The sum of first n terms is n².

A manufacturer of TV sets produced 500 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production of TV sets in the 15th year and the total production of TV sets in the first 10 years. (J'15)

Step 1: Set up the AP

Let production in 1st year = a

Common difference = d

Production in 3rd year: a + 2d = 500 ...(1)

Production in 7th year: a + 6d = 700 ...(2)

Step 2: Solve for a and d

Subtract (1) from (2):

(a + 6d) - (a + 2d) = 700 - 500

4d = 200

d = 50

Substitute in (1): a + 2(50) = 500

a + 100 = 500

a = 400

Step 3: Find production in 15th year

a₁₅ = a + 14d = 400 + 14(50)

a₁₅ = 400 + 700 = 1100

Step 4: Find total production in first 10 years

S₁₀ = 10/2 [2a + (10-1)d]

S₁₀ = 5 [2(400) + 9(50)]

S₁₀ = 5 [800 + 450]

S₁₀ = 5 × 1250 = 6250

∴ Production in 15th year = 1100 sets, Total production in first 10 years = 6250 sets.

Find the sum of all the three digit numbers which are divisible by 4. (M'16)

Step 1: Identify the sequence

Three-digit numbers divisible by 4: 100, 104, 108, ..., 996

This is an AP with a = 100, d = 4

Step 2: Find number of terms

aₙ = a + (n-1)d

996 = 100 + (n-1)(4)

996 - 100 = 4(n-1)

896 = 4(n-1)

n-1 = 224

n = 225

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₂₂₅ = 225/2 [100 + 996]

S₂₂₅ = 225/2 × 1096

S₂₂₅ = 225 × 548

S₂₂₅ = 123,300

∴ The sum of all three-digit numbers divisible by 4 is 123,300.

The sum of the three terms which are in an arithmetic progression is 33. If the product of the first and the third terms exceeds the second term by 29, find the Arithmetic Progression. (J'16)

Step 1: Let the three terms be

a-d, a, a+d

Step 2: Apply sum condition

(a-d) + a + (a+d) = 33

3a = 33

a = 11

Step 3: Apply product condition

(a-d)(a+d) = a + 29

a² - d² = a + 29

11² - d² = 11 + 29

121 - d² = 40

d² = 121 - 40 = 81

d = ±9

Step 4: Find the AP

For d = 9: Terms are 11-9, 11, 11+9 = 2, 11, 20

For d = -9: Terms are 11-(-9), 11, 11+(-9) = 20, 11, 2

Both give the same AP: 2, 11, 20

∴ The Arithmetic Progression is 2, 11, 20, ...

Find the sum of all three-digit natural numbers, which are divisible by 3 and not divisible by 6. (M'17)

Step 1: Numbers divisible by 3 but not by 6

These are numbers divisible by 3 but not by 2 (odd multiples of 3)

Step 2: Find the sequence

Three-digit numbers divisible by 3: 102, 105, 108, ..., 999

Odd multiples of 3: 105, 111, 117, ..., 999

This is an AP with a = 105, d = 6

Step 3: Find number of terms

aₙ = a + (n-1)d

999 = 105 + (n-1)(6)

999 - 105 = 6(n-1)

894 = 6(n-1)

n-1 = 149

n = 150

Step 4: Find the sum

Sₙ = n/2 [a + l]

S₁₅₀ = 150/2 [105 + 999]

S₁₅₀ = 75 × 1104

S₁₅₀ = 82,800

∴ The sum of all three-digit numbers divisible by 3 but not by 6 is 82,800.

The sum of 5th and 9th terms of A.P. is 72 and the sum of 7th and 12th terms is 97. Find the A.P. (J'17)

Step 1: Write the given conditions

a₅ + a₉ = 72

(a + 4d) + (a + 8d) = 72

2a + 12d = 72

a + 6d = 36 ...(1)

Step 2: Second condition

a₇ + a₁₂ = 97

(a + 6d) + (a + 11d) = 97

2a + 17d = 97 ...(2)

Step 3: Solve equations (1) and (2)

From (1): a = 36 - 6d

Substitute in (2): 2(36 - 6d) + 17d = 97

72 - 12d + 17d = 97

72 + 5d = 97

5d = 25

d = 5

a = 36 - 6(5) = 36 - 30 = 6

Step 4: Write the AP

AP: a, a+d, a+2d, ...

AP: 6, 11, 16, 21, ...

∴ The Arithmetic Progression is 6, 11, 16, 21, ...

Which term of G.P.: 3, 9, 27, ... is 2187? (J'17)

Step 1: Identify values

First term (a) = 3

Common ratio (r) = 9/3 = 3

nth term (aₙ) = 2187

Step 2: Apply GP formula

aₙ = a · rⁿ⁻¹

2187 = 3 · 3ⁿ⁻¹

2187 = 3ⁿ

Step 3: Solve for n

3ⁿ = 2187

3ⁿ = 3⁷ (since 3⁷ = 2187)

n = 7

∴ 2187 is the 7th term of the GP.

Find the sum of all two-digit odd positive integers which are divisible by 3 but not by 2. (M'18)

Step 1: Understand the condition

Numbers divisible by 3 but not by 2 are odd multiples of 3

Step 2: Find the sequence

Two-digit odd multiples of 3: 15, 21, 27, ..., 99

This is an AP with a = 15, d = 6

Step 3: Find number of terms

aₙ = a + (n-1)d

99 = 15 + (n-1)(6)

99 - 15 = 6(n-1)

84 = 6(n-1)

n-1 = 14

n = 15

Step 4: Find the sum

Sₙ = n/2 [a + l]

S₁₅ = 15/2 [15 + 99]

S₁₅ = 15/2 × 114

S₁₅ = 15 × 57 = 855

∴ The sum of all two-digit odd positive integers divisible by 3 is 855.

Find the sum of the integers between 100 and 500 that are divisible by 9. (J'18)

Step 1: Identify the sequence

Numbers between 100 and 500 divisible by 9: 108, 117, 126, ..., 495

This is an AP with a = 108, d = 9

Step 2: Find number of terms

aₙ = a + (n-1)d

495 = 108 + (n-1)(9)

495 - 108 = 9(n-1)

387 = 9(n-1)

n-1 = 43

n = 44

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₄₄ = 44/2 [108 + 495]

S₄₄ = 22 × 603

S₄₄ = 13,266

∴ The sum of integers between 100 and 500 divisible by 9 is 13,266.

Find the sum of all two-digit odd multiples of 3. (M'19)

Step 1: Find the sequence

Two-digit odd multiples of 3: 15, 21, 27, ..., 99

This is an AP with a = 15, d = 6

Step 2: Find number of terms

aₙ = a + (n-1)d

99 = 15 + (n-1)(6)

99 - 15 = 6(n-1)

84 = 6(n-1)

n-1 = 14

n = 15

Step 3: Find the sum

Sₙ = n/2 [a + l]

S₁₅ = 15/2 [15 + 99]

S₁₅ = 15/2 × 114

S₁₅ = 15 × 57 = 855

∴ The sum of all two-digit odd multiples of 3 is 855.

Find the sum of all integers between 1 to 50 which are not divisible by 3. (J'19)

Step 1: Find sum of all integers from 1 to 50

S = 50/2 [1 + 50] = 25 × 51 = 1275

Step 2: Find sum of integers divisible by 3 from 1 to 50

Numbers: 3, 6, 9, ..., 48

This is an AP with a = 3, d = 3

aₙ = a + (n-1)d

48 = 3 + (n-1)(3)

48 - 3 = 3(n-1)

45 = 3(n-1)

n-1 = 15

n = 16

Sum = 16/2 [3 + 48] = 8 × 51 = 408

Step 3: Find required sum

Sum of numbers not divisible by 3 = Total sum - Sum of numbers divisible by 3

= 1275 - 408 = 867

∴ The sum of all integers from 1 to 50 not divisible by 3 is 867.

Author

zaheerpashamd

M.Sc,B.Ed: SDVR ZPHS B.Gangaram Sathupally(md) Khammam(dt)
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