Chapter 6: Exploring some more Progressions

1. Find the sum of the series 0.15 + 0.015 + 0.0015 + ... to 15 terms.

Solution:

The given series is

0.15 + 0.015 + 0.0015 + ...

This is a geometric progression (G.P.) in which

First term, a = 0.15

Common ratio, r = 0.015 / 0.15 = 0.1

Number of terms, n = 15

The sum of the first n terms of a G.P. is

S_n = a(1 - rⁿ) / (1 - r),   r ≠ 1

Substituting the values, we get

S₁₅ = 0.15(1 - (0.1)¹⁵) / (1 - 0.1)

= 0.15(1 - 10^-15) / 0.9

= (1 / 6)(1 - 10^-15)

Therefore, S₁₅ = 0.1666666666666665 (approximately).

2. Find the sum to n terms of the series 0.9 + 0.99 + 0.999 + ...

Solution:

The given series is

S_n = 0.9 + 0.99 + 0.999 + ...

Using the given hint, we can write

0.9 = 1 - 0.1

0.99 = 1 - 0.01

0.999 = 1 - 0.001

Therefore,

S_n = (1 - 0.1) + (1 - 0.01) + (1 - 0.001) + ... to n terms

Separating the terms, we get

S_n = n - (0.1 + 0.01 + 0.001 + ... to n terms)

The series inside the brackets is a G.P. with

First term, a = 0.1

Common ratio, r = 0.1

Number of terms = n

Its sum is

S = a(1 - rⁿ) / (1 - r)

= 0.1(1 - (0.1)ⁿ) / (1 - 0.1)

= (1 / 9)(1 - 10^-n)

Hence,

S_n = n - (1 / 9)(1 - 10^-n)

Therefore, S_n = n - (1 - 10^-n) / 9.

3. Find the sum to n terms of the series 5 + 55 + 555 + ...

Solution:

The given series is

5 + 55 + 555 + ...

We can express its terms as

5 = 5(1)

55 = 5(11)

555 = 5(111)

Thus,

S_n = 5 + 55 + 555 + ... to n terms

Each term can be written using powers of 10:

5 = 5(10 - 1) / 9

55 = 5(10² - 1) / 9

555 = 5(10³ - 1) / 9

In general, the k-th term is

a_k = 5(10^k - 1) / 9

Therefore,

S_n = (5 / 9)[(10 - 1) + (10² - 1) + (10³ - 1) + ... + (10ⁿ - 1)]

Separating the terms, we get

S_n = (5 / 9)[(10 + 10² + 10³ + ... + 10ⁿ) - n]

The sum of the G.P. 10 + 10² + ... + 10ⁿ is

= 10(10ⁿ - 1) / (10 - 1)

= 10(10ⁿ - 1) / 9

Substituting this value, we get

S_n = (5 / 9)[10(10ⁿ - 1) / 9 - n]

Therefore, S_n = 5[10(10ⁿ - 1) - 9n] / 81.

4. The sum of the first n terms of the sequence 3, 6, 12, ... is 381. Find n.

Solution:

The given sequence is

3, 6, 12, ...

This is a G.P. with

First term, a = 3

Common ratio, r = 6 / 3 = 2

Sum of the first n terms, S_n = 381

Using the sum formula for a G.P.,

S_n = a(rⁿ - 1) / (r - 1),   r ≠ 1

Substituting the values, we get

381 = 3(2ⁿ - 1) / (2 - 1)

381 = 3(2ⁿ - 1)

Dividing both sides by 3,

127 = 2ⁿ - 1

2ⁿ = 128

Since 128 = 2⁷,

2ⁿ = 2⁷

Therefore, n = 7.

Hence, the required number of terms is n = 7.

5. Find the sum to n terms of the series

x(x + y) + x²(x² + y²) + x³(x³ + y³) + ...

Solution:

The given series is

S_n = x(x + y) + x²(x² + y²) + x³(x³ + y³) + ...

Expanding the terms, we get

S_n = (x² + xy) + (x⁴ + x²y²) + (x⁶ + x³y³) + ...

Separating the terms,

S_n = (x² + x⁴ + x⁶ + ... + x²ⁿ) + (xy + x²y² + x³y³ + ... + xⁿyⁿ)

The first bracket is a G.P. with first term x² and common ratio x².

Its sum is

= x²(1 - x²ⁿ) / (1 - x²)

The second bracket is a G.P. with first term xy and common ratio xy.

Its sum is

= xy(1 - (xy)ⁿ) / (1 - xy)

Therefore, for x² ≠ 1 and xy ≠ 1,

S_n = x²(1 - x²ⁿ) / (1 - x²) + xy(1 - (xy)ⁿ) / (1 - xy).

6. Find the sum of the infinite terms of the series

1 + 1/2 + 1/3² + 1/2² + 1/3⁴ + 1/2³ + 1/3⁶ + ...

Solution:

The given series can be rearranged as

S = (1 + 1/3² + 1/3⁴ + ...) + (1/2 + 1/2² + 1/2³ + ...)

Both brackets are infinite G.P.s.

For the first G.P.,

First term, a = 1

Common ratio, r = 1/9

Since |r| < 1, its sum to infinity is

S₁ = a / (1 - r)

= 1 / (1 - 1/9)

= 9/8

For the second G.P.,

First term, a = 1/2

Common ratio, r = 1/2

Its sum to infinity is

S₂ = a / (1 - r)

= (1/2) / (1 - 1/2)

= 1

Therefore,

S = S₁ + S₂

= 9/8 + 1

S = 17/8.

7. The sum of an infinite series of G.P. is 6 and the sum of the squares of these terms is 12. Find the common ratio of the original G.P.

Solution:

Let the first term of the G.P. be a and its common ratio be r.

The sum to infinity of a G.P. is

S_∞ = a / (1 - r)

Given that S_∞ = 6,

a / (1 - r) = 6

Therefore,

a = 6(1 - r)   ... (1)

The squares of the terms form another G.P. with first term a² and common ratio r².

The sum of the squares to infinity is

a² / (1 - r²) = 12

Substituting a = 6(1 - r),

36(1 - r)² / (1 - r²) = 12

Since 1 - r² = (1 - r)(1 + r),

36(1 - r) / (1 + r) = 12

Dividing both sides by 12,

3(1 - r) = 1 + r

3 - 3r = 1 + r

2 = 4r

r = 1/2

Therefore, the common ratio is r = 1/2.

8. If the sum to infinity of the series 1 + r + r² + r³ + ... is S, given that |r| < 1, write r in terms of S.

Solution:

The given series is an infinite G.P. with

First term, a = 1

Common ratio = r

Sum to infinity = S

Using the formula,

S = a / (1 - r)

Substituting a = 1,

S = 1 / (1 - r)

Therefore,

S(1 - r) = 1

1 - r = 1/S

r = 1 - 1/S

Hence, r = (S - 1) / S.

9. Find the value of 4^1/2 · 4^1/4 · 4^1/8 · 4^1/16 · ... to infinity.

Solution:

Let

P = 4^1/2 · 4^1/4 · 4^1/8 · 4^1/16 · ...

Using the law of exponents, a^m · aⁿ = a^m+n, we get

P = 4^{(1/2 + 1/4 + 1/8 + 1/16 + ...)}

The exponents form an infinite G.P. with

First term, a = 1/2

Common ratio, r = 1/2

Since |r| < 1, the sum to infinity is

S_∞ = a / (1 - r)

= (1/2) / (1 - 1/2)

= 1

Therefore,

P = 4¹

Hence, the required value is 4.

10. The midpoints D, E and F of the sides of an equilateral triangle ABC are joined to form another smaller equilateral triangle. This process is repeated indefinitely. If each side of triangle ABC is 16 cm, find the sum of the perimeters of all the triangles so formed.

Solution:

Each side of the equilateral triangle ABC is 16 cm.

Therefore, the perimeter of triangle ABC is

P₁ = 3 × 16 = 48 cm

Since D, E and F are the midpoints of the sides of triangle ABC, each side of triangle DEF is half the side of triangle ABC.

Side of triangle DEF = 16/2 = 8 cm

Therefore, its perimeter is

P₂ = 3 × 8 = 24 cm

Similarly, the side of the next triangle is 4 cm, and its perimeter is 12 cm.

Thus, the perimeters of the triangles form the G.P.

48 + 24 + 12 + 6 + ...

Here,

First term, a = 48

Common ratio, r = 24/48 = 1/2

Since |r| < 1, the sum of the infinite G.P. is

S_∞ = a / (1 - r)

= 48 / (1 - 1/2)

= 48 / (1/2)

= 96 cm

Hence, the sum of the perimeters of all the triangles, including the original triangle, is 96 cm.

11. Let f(x) = 2x + 1. Find the number of values of x for which f(x), f(2x), f(4x) are in a G.P.

Solution:

Given,

f(x) = 2x + 1

Substituting 2x for x,

f(2x) = 2(2x) + 1 = 4x + 1

Similarly,

f(4x) = 2(4x) + 1 = 8x + 1

Therefore, the three terms are

2x + 1, 4x + 1, 8x + 1

For three numbers a, b and c to be in a G.P.,

b² = ac

Therefore,

(4x + 1)² = (2x + 1)(8x + 1)

Expanding both sides,

16x² + 8x + 1 = 16x² + 10x + 1

Cancelling the common terms,

8x = 10x

2x = 0

x = 0

When x = 0, the three terms are 1, 1, 1, which are in a G.P. with common ratio 1.

Hence, there is exactly one value of x, namely x = 0.

12. Let t₁, t₂, t₃, t₄, ... be the terms of a G.P. whose common ratio is r, where -1 < r < 1. Evaluate the following:

(t₁ - t₃ + t₅ - ...)/(t₂ - t₄ + t₆ - ...)

Solution:

Let the first term of the G.P. be a and its common ratio be r.

Then,

t₁ = a, t₂ = ar, t₃ = ar², t₄ = ar³, ...

The numerator is

N = t₁ - t₃ + t₅ - ...

= a - ar² + ar⁴ - ar⁶ + ...

= a(1 - r² + r⁴ - r⁶ + ...)

This is an infinite G.P. with first term a and common ratio -r².

Since |r| < 1, we have |r²| < 1. Therefore,

N = a / (1 - (-r²))

N = a / (1 + r²)

The denominator is

D = t₂ - t₄ + t₆ - ...

= ar - ar³ + ar⁵ - ar⁷ + ...

= ar(1 - r² + r⁴ - r⁶ + ...)

Therefore,

D = ar / (1 + r²)

Hence, the required expression is

N / D = [a / (1 + r²)] / [ar / (1 + r²)]

= 1 / r

Therefore, the required value is 1/r, provided r ≠ 0.

Note: If r = 0, the denominator of the given expression is zero, so the expression is undefined.

13. A particular ball rebounds to 3/5 of the height from which it falls whenever it strikes the floor. Find the total distance it covers before coming to rest, if it falls due to gravity from a height of 90 metres.

Solution:

The initial height from which the ball falls is 90 m.

After striking the floor for the first time, the ball rebounds to

90 × 3/5 = 54 m

After the second strike, it rebounds to

54 × 3/5 = 162/5 = 32.4 m

Similarly, the successive rebound heights are

54, 54(3/5), 54(3/5)², ...

These heights form an infinite G.P. with

First term, a = 54

Common ratio, r = 3/5

The sum of all rebound heights is

S_∞ = a / (1 - r)

= 54 / (1 - 3/5)

= 54 / (2/5)

= 54 × 5/2

= 135 m

For every rebound, the ball travels the rebound height twice: once while moving upwards and once while falling downwards.

Therefore, the total distance travelled is

D = Initial falling distance + 2 × (sum of rebound heights)

D = 90 + 2 × 135

D = 90 + 270

Hence, the total distance travelled by the ball before coming to rest is 360 metres.

1. Find the nth term and the sum of the first n terms of the series 1 + 9 + 24 + 46 + 75 + ...

Solution:

The given series is

1 + 9 + 24 + 46 + 75 + ...

Writing the terms and their differences, we get

Terms: 1, 9, 24, 46, 75, ...

First differences: 8, 15, 22, 29, ...

Second differences: 7, 7, 7, ...

Since the second differences are constant, the nth term is a quadratic expression.

Let the nth term be

t_n = an² + bn + c

For a quadratic sequence, the constant second difference is 2a.

Therefore,

2a = 7

a = 7/2

Thus,

t_n = (7/2)n² + bn + c

For n = 1, t₁ = 1:

7/2 + b + c = 1

b + c = -5/2 ... (1)

For n = 2, t₂ = 9:

14 + 2b + c = 9

2b + c = -5 ... (2)

Subtracting equation (1) from equation (2),

b = -5/2

Substituting in equation (1),

c = 0

Therefore,

t_n = n(7n - 5)/2.

Now, the sum of the first n terms is

S_n = Σ_k=1ⁿ (7k² - 5k)/2

= (1/2)[7Σk² - 5Σk]

Using the formulas

Σk = n(n + 1)/2

Σk² = n(n + 1)(2n + 1)/6

We get

S_n = (1/2)[7n(n + 1)(2n + 1)/6 - 5n(n + 1)/2]

= n(n + 1)(14n - 8)/24

Hence, S_n = n(n + 1)(7n - 4)/12.

2. Find the 10th term and the sum of the first 10 terms of the series 4 + 5 + 9 + 16 + 26 + ...

Solution:

The given series is

4 + 5 + 9 + 16 + 26 + ...

First differences: 1, 4, 7, 10, ...

Second differences: 3, 3, 3, ...

Since the second differences are constant, let

t_n = an² + bn + c

Since the second difference is 3,

2a = 3

a = 3/2

Therefore,

t_n = (3/2)n² + bn + c

For n = 1, t₁ = 4:

3/2 + b + c = 4

b + c = 5/2 ... (1)

For n = 2, t₂ = 5:

6 + 2b + c = 5

2b + c = -1 ... (2)

Subtracting equation (1) from equation (2),

b = -7/2

Substituting in equation (1),

c = 6

Thus,

t_n = (3/2)n² - (7/2)n + 6

Therefore,

t_n = (3n² - 7n + 12)/2

Finding the 10th term:

t₁₀ = (3 × 10² - 7 × 10 + 12)/2

= (300 - 70 + 12)/2

= 242/2

t₁₀ = 121.

Finding the sum of the first 10 terms:

S_n = Σ_k=1ⁿ (3k² - 7k + 12)/2

= (1/2)[3Σk² - 7Σk + 12n]

Using the formulas for Σk and Σk²,

S_n = (1/2)[n(n + 1)(2n + 1)/2 - 7n(n + 1)/2 + 12n]

= n(n² - 2n + 9)/2

Putting n = 10,

S₁₀ = 10(100 - 20 + 9)/2

= 10 × 89/2

S₁₀ = 445.

3. Find the nth term and the sum of the first 12 terms of the series 3 + 6 + 11 + 18 + 27 + ...

Solution:

The given series is

3 + 6 + 11 + 18 + 27 + ...

First differences: 3, 5, 7, 9, ...

Second differences: 2, 2, 2, ...

Let the nth term be

t_n = an² + bn + c

Since the second difference is 2,

2a = 2

a = 1

Therefore,

t_n = n² + bn + c

For n = 1, t₁ = 3:

1 + b + c = 3

b + c = 2 ... (1)

For n = 2, t₂ = 6:

4 + 2b + c = 6

2b + c = 2 ... (2)

Subtracting equation (1) from equation (2),

b = 0

Therefore, c = 2.

Hence,

t_n = n² + 2.

Finding the sum of the first 12 terms:

S₁₂ = Σ_k=1¹² (k² + 2)

= Σk² + 2 × 12

Using the formula

Σk² = n(n + 1)(2n + 1)/6

We get

S₁₂ = 12 × 13 × 25/6 + 24

= 650 + 24

S₁₂ = 674.

4. Find the 8th term and the sum of the first 8 terms of the series 4 + 13 + 28 + 49 + 76 + ...

Solution:

The given series is

4 + 13 + 28 + 49 + 76 + ...

First differences: 9, 15, 21, 27, ...

Second differences: 6, 6, 6, ...

Let the nth term be

t_n = an² + bn + c

Since the second difference is 6,

2a = 6

a = 3

Therefore,

t_n = 3n² + bn + c

For n = 1, t₁ = 4:

3 + b + c = 4

b + c = 1 ... (1)

For n = 2, t₂ = 13:

12 + 2b + c = 13

2b + c = 1 ... (2)

Subtracting equation (1) from equation (2),

b = 0

Therefore, c = 1.

Hence,

t_n = 3n² + 1.

Finding the 8th term:

t₈ = 3 × 8² + 1

= 3 × 64 + 1

t₈ = 193.

Finding the sum of the first 8 terms:

S_n = Σ_k=1ⁿ (3k² + 1)

= 3Σk² + n

Using the formula for the sum of squares,

S_n = 3n(n + 1)(2n + 1)/6 + n

= n(n + 1)(2n + 1)/2 + n

Putting n = 8,

S₈ = 8 × 9 × 17/2 + 8

= 612 + 8

S₈ = 620.