Sequence and series (Class- 9, Experience- 2)

[Exercise- Page 56]

E.2.1. Are the following sequences arithmetic, geometric, Fibonacci or none of the three? Why? Determine the common terms and explain.

(i) 2, 5, 10, 17,....

(ii) -2, 7, 12, 17,....

(iii) -12, 24, -48, 96,.....

(iv) 13, 21, 34, 55,....

(v) 5, -3, 95, -2725,.....

(vi) 13, 23, 43, 83,......

Solution:

E.2.2. Fill in the gaps in the following sequences:

(i) 2, 9, 16, ____,____, 37,____.

(ii) -35, ____, ____, -5, 5, ____.

(iii) ____,____, ____, 5, -4,____.

(iv) ____, 10x², 50x³,____, ____.

Solution:

E.2.3. Fill in the blank cells of the following table.

Solution:

E.2.4. You want to make a mosaic in the shape of an equilateral triangle on the floor of your study room, whose side length is 12 feet. The mosaic will have white and blue tiles. Each tile is also an equilateral triangle with a side length of 12 inches. The tiles have to be placed in opposite colors to complete the mosaic.

a) Make a model of triangular mosaic.

b) How many tiles of each color will you need?

c) How many tiles will you need in total?

Solution:

E.2.5. Fill in the blank cells of the following table:

Solution:

E.2.6.

a) Observe the sequence in Figure - 1 closely. Then construct the 10th figure and determine the number of coins.

b) Determine the number of coins in the nth figure based on the given information.

c) If n = 5, determine the numbers in the 2nd column of Figure- 2 and show that the sum of the numbers in the nth row supports the formula 2ⁿ.

d) Create a series with the sums of each row and determine the value of n when the sum of the first n terms of the series is 2046.

Solution:

E.2.7. Determine the value of n, where n ∈ N.

ii.

iii.

iv.

Solution:

Find the value of n, where n ∈ N

Given, k = 1, 2, 3,.... n

Here,
(20 – 4 ✕ 1) + (20 – 4 ✕ 2) + (20 – 4 ✕ 3) +....... (20 – 4 ✕ n) = -20

or, 20n – 4.(1+2+3+....n) = -20

or, 20n – 4 ✕ .12.n{2.1 + (n - 1)1} = -20 [Formula, S_n = 12n{2a + (n - 1)d}]

or, 20n – 2.n(2 + n – 1) = -20

or, 20n – 2n(n + 1) = -20

or, 20n – 2n² – 2n = -20

or, -2n² + 18n = -20

or, -2n² + 18n + 20 = 0

or, 2n² - 18n -20 = 0

or, 2(n² – 9n – 10) = 0

or, n² – 9n – 10 = 0

or, n² – 10n + n – 10 = 0

or, n(n - 10) + 1(n - 10) = 0

or, (n + 1)(n - 10) = 0

or, n + 1 = 0 and, n - 10 = 0

or, n = -1 or, n = 10

Here, negative value of n is not acceptable. So, n = 10

∴ n = 10 (Answer)

ii)

Given, k = 1, 2, 3, ….. n

Here,

(3.1 + 2) + (3.2 + 2) +(3.3 + 2) +.....+ (3.n + 2) = 1105

or, 3(1 + 2 + 3 +....n) + 2n = 1105

or, 3.12.n{2.1 + (n - 1).1} + 2n = 1105 [S_n = 12(n{2a + (n - 1)d} According to formula]

or, 3.12.n{2 + n - 1} + 2n = 1105

or, 3.12.n(n + 1) + 2n = 1105

or, 3.12.(n² + n) + 2n = 1105

or, 2 ✕ {3 .12.(n² + n) + 2n} = 2 ✕1105 [both side multiplied by 2]

or, 3.(n² + n) + 4n = 2210

or, 3n² + 3n + 4n = 2210

or, 3n² + 7n – 2210 = 0

or, 3n² - 78n + 85n – 2210 = 0

or, 3n(n - 26) + 85(n – 26) = 0

or, (n - 26)(3n + 85) = 0

or, n - 26 = 0 and, 3n + 85 = 0

or, n = 26 or, 3n = - 85

Here, negative value of n is not acceptable. So, n = 26

∴ n = 26 (Answer)

iii)

Given, k = 1, 2, 3, ….. n

(-8). (0.5)^1-1 + (-8). (0.5)^2-1 + (-8). (0.5)^3-1 +......+ (-8). (0.5)^n-1 = - 255 16

or, (-8). {(0.5)⁰ + (0.5)¹ + (0.5)² +......+ (0.5)^n-1} = - 255 16

or, {(0.5)⁰ + (0.5)¹ + (0.5)² +......+ (0.5)^n-1} = -255 -8 ✕ 16

or, (0.5)⁰ + (0.5)¹ + (0.5)² +......+ (0.5)^n-1 = 255 128

or, (0.5)⁰ ✕ (1 - 0.5ⁿ) (1 - 0.5) = 255 128 [Formula, S_n = a(1 - rⁿ) (1 - r)]

or, 1 ✕ (1 - 0.5ⁿ) 0.5 = 255 128

or, (1 - 0.5ⁿ) ✕ 2 0.5 ✕ 2 = 255128

or, (1 - 0.5ⁿ) ✕ 2 1 = 255128

or, (1 - 0.5ⁿ) ✕ 2 = 255 128

or, 1 - 0.5ⁿ = 255 128 ✕ 2

or, 1 - ( 1 2)ⁿ = 255 256

or, 1 - ( 1 2)ⁿ = 255 256 - 1

or, - ( 1 2)ⁿ = 255 - 256 256

or, - ( 1 2)ⁿ = -1 256

or, ( 1 2)ⁿ = 1 256

or, ( 1 2)ⁿ = ( 1 2)⁸

or, n = 8 [According to exponential rule]

∴ n = 8 (Answer)

iv)

Given, k = 1, 2, 3, ...... n

অতএব,

(3)^1-1 + (3)^2-1 + (3)^3-1 +......+(3)^n-1 = 3280

or, (3)⁰ + (3)¹ + (3)² +......+(3)^n-1 = 3280

or, 1 + 3 + 9 +......+(3)^n-1 = 3280

or, 1(1 - 3ⁿ) (1 - 3) = 3280 [Formula, S_n = a(1 - rⁿ) (1 - r) ]

or, (1 - 3ⁿ) (-2) = 3280

or, (1 - 3ⁿ) = 3280 ✕ (-2)

or, 1 - 3ⁿ = -6560

or, -3ⁿ = -6560 - 1

or, -3ⁿ = -6561

or, 3ⁿ = 6561

or, 3ⁿ = 3⁸

or, n = 8 [According to exponential rule]

∴ n = 8 (Answer)

E.2.8. The first, second and tenth terms of an arithmetic series are equal to the first, fourth and seventeenth terms of a geometric series respectively.

a) If the first term of the arithmetic series is ‘a’, the common difference is ‘d’ and the common ratio of the geometric series is ‘r’, then form two equations by combining the two series.

b) Determine the value of common ratio ‘r’.

c) If the tenth term of the geometric series is 5120, then determine the values of ‘a’ and ‘d’.

d) Determine the sum of the first 20 terms of the arithmetic series.

Solution:

The first, second and tenth terms of an arithmetic series are equal to the first, fourth and seventh terms of a geometric series respectively.

a) Form two equations by combining the two series-

Given,
1st term of arithmetic series a,
Common difference d
geometric series common ratio r
and The first, second and tenth terms of an arithmetic series are equal to the first, fourth and seven terms of a geometric series respectively.

According to formula, We know,
nth term of arithmetic series, a_n = a + (n − 1)d
nth term of geometric series, b_n = a⋅r^n-1

Here,
Arithmetic series, 1st term = a
Arithmetic series, 2nd term = a + d
Arithmetic series, 10th term = a + (10 - 1)d = a + 9d

Here, Geometric series,
Geometric series, 1st term = a
Geometric series, 4th term = ar^4-1 = ar³
Geometric series, 7th term = ar^7-1 = ar⁶

According to question,
a = a [1st term of arithmetic series = 1st term of geometric series]
a + d = ar³ [2nd term of arithmetic series = 4th term of geometric series]
a + 9d = ar⁶ [10th term of arithmetic series = 7th term of geometric series]

So, two equations are, a + d = ar³ and a + 9d = ar⁶

(Answer)

b) Find the value of common ratio ‘r’-

From (a) we get the equations,
a = a
a + d = ar³
a + 9d = ar⁶

From equation (i),
a + d = ar³
or, a + d - ar³ = 0
or, a - ar³ = -d
or, a(1 - r³) = -d
or, a = -   d 1 - r³.....(iii)

From equation (ii),
a + 9d = ar⁶
or, a - ar⁶ = -9d
or, a(1 - r⁶) = -9d
or, a = -   9d 1 - r⁶.......(iv)

From equation (iii) and (iv)-
-   d1 - r³ = -   9d1 - r⁶

or, 11 - r³ = 91 - r⁶

or, 11 - r⁶ = 9(1 - r³)

or, 1 - r⁶ = 9 - 9r³

or, 1 - r⁶ - 9 + 9r³ = 0

or, - r⁶ + 9r³ - 8 = 0

or, - r⁶ + r³ + 8r³ - 8 = 0

or, - r³(r³ -1) + 8(r³ – 8) = 0

or, (r³ -1)( -r³ + 8) = 0

Here,
(r³ -1) = 0
or, r³ = 1
or, r = ∛ 1
or, r = 1 [1 will not be the value of common ratio of geometric series]

or,
( -r³ + 8) = 1
or, -r³ = -8
or, r³ = 8
or, r³ = 8
or, r = ∛ 8
∴ r = 2 (Answer)

c) the tenth term of the geometric series is 5120, then determine the values of ‘a’ and ‘d

Given,
Geometric series, common ratio, r = 2 [From equation (b)]
Geometric series 10th term = 5120
Geometric series, 1st term, a =?
Common difference, d= ?

According to formula, We know,

ar^10-1 = 5120

or, a ✕ (2)⁹ = 5120

or, a ✕ 512 = 5120

or, a = 5120 512

∴ a = 10

Again,
a + d = ar³ [From (a), equations]

or, 10 + d = 10 ✕ 23

or, 10 + d = 10 ✕ 8

or, d = 80 – 10

∴ d = 70

d) Find the sum of the first 20 terms of the arithmetic series.

Given,
1st term of arithmetic series, a = 10
Common difference, d = 70
number of terms, n = 20

According to formula, We know,

S_n = 12n{2a + (n - 1)d}

or, S_n = 20 2{2 ✕ 10 + (20 - 1)70}

or, S_n = 10(2 ✕ 10 + 19 ✕ 70)

or, S_n = 10(20 + 1330)

or, S_n = 10 ✕ 1350

∴ S_n = 13500 (Answer)

E.2.9. Draw an equilateral triangle. Draw another equilateral triangle connecting the midpoints of its sides. Draw another equilateral triangle connecting the midpoints of the sides of that triangle. Draw 10 triangles in this manner. If each side of the innermost triangle is 64 mm, find the sum of the perimeters of all the triangles.

Solution:

E.2.10. Shahana plunted a sapling in his educational institute. After one year, the height of the sapling is 1.5 feet. The next year, its height increases by 0.75 foot. Every year, the height of the tree increases by 50% of the previous year’s increase. If it continues to grow like this, how high can the tree be after 20 years?

Solution:

E.2.11. Find out your family’s expenses for the last six months. Consider each month’s expenses as a term and try to convert them into a series if possible. Then try to solve the following problems.

a) Is it possible to create a series? If so, explain what kind of series you have got.

b) Express the sum of the series in an equation.

c) Determine how much can be the total expenses for the next six months.

d) Write down your observations about your family’s monthly/annual expenses.

Solution:

Active Math Class

Sequence and series (Class- 9, Experience- 2)

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