AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.4 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.4

### 10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.

In which of the following situations, does the list of numbers involved in the form a G.P.?

i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .

Answer:

Given: Sharmila’s yearly salary = Rs. 5,00,000.

Rate of annual increment = 10 %.

Here, a = a_{1} = 5,00,000

a_{2} = 5,00,000 × \(\frac{11}{10}\) = 5,50,000

a_{3} = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,05,000

a_{4} = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number \(\frac{11}{10}\).

∴ r = common ratio = \(\frac{11}{10}\)

Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.

Answer:

Given: Bricks needed for the bottom step = 100.

Each successive step needs 2 bricks less than the previous step.

∴ Second step from the bottom needs = 100 – 2 = 98 bricks.

Third step from the bottom needs = 98 – 2 = 96 bricks.

Fourth step from the bottom needs = 96 – 2 = 94 bricks.

Here the numbers are 100, 98, 96, 94, ….

Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:

Given: An equilateral triangle whose perimeter = 24 cm.

Side of the equilateral triangle = \(\frac{24}{3}\) = 8 cm.

[∵ All sides of equilateral are equal] ……. (1)

Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = \(\frac{8}{2}\) = 4 cm

[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]

Similarly, the side of third triangle = \(\frac{4}{2}\) = 2 cm

∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,

a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).

∴ The situation forms a G.P.

Question 2.

Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.

i) a = 4 ; r = 3.

Answer:

The terms are a, ar, ar^{2}, ar^{3}, ……..

∴ 4, 4 × 3, 4 × 3^{2} , 4 × 3^{2} , ……

⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = \(\frac{1}{5}\)

Answer:

The terms are a, ar, ar^{2}, ar^{3}, ……..

iii) a = 81 ; r = –\(\frac{1}{3}\)

Answer:

The terms of a G.P are:

a, ar, ar^{2}, ar^{3}, ……..

⇒ 81, -27, 9,

iv) a = \(\frac{1}{64}\); r = 2.

Answer:

Given: a = \(\frac{1}{64}\); r = 2.

∴ The G.P is \(\frac{1}{64}\), \(\frac{1}{32}\), \(\frac{1}{16}\), …….

Question 3.

Which of the following are G.P. ? If they are G.P, write three more terms,

i) 4, 8, 16, ……

Answer:

Given: 4, 8, 16, ……

where, a_{1} = 4; a_{2} = 8; a_{3} = 16, ……

\(\frac{a_{2}}{a_{1}}=\frac{8}{4}=2\)

\(\frac{a_{3}}{a_{2}}=\frac{16}{8}=2\)

∴ r = \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}\) = 2

Hence 4, 8, 16, … is a G.P.

where a = 4 and r = 2

a_{4} = a . r^{3} = 4 × 2^{3} = 4 × 8 = 32

a_{5} = a . r^{4} = 4 × 2^{4} = 4 × 16 = 64

a_{6} = a . r^{5} = 4 × 2^{5} = 4 × 32 = 128

ii) \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Answer:

Given: t_{1} = \(\frac{1}{3}\), t_{2} = –\(\frac{1}{6}\), t_{3} = \(\frac{1}{12}\), ….

\(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Hence the ratio is common between any two successive terms.

∴ \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), ……. is G.P.

where a = \(\frac{1}{3}\) and r = –\(\frac{1}{2}\)

iii) 5, 55, 555, ……..

Answer:

Given: t_{1} = 5, t_{2} = 55, t_{3} = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……

Given: t_{1} = -2, t_{2} = -6, t_{3} = -18

∴ -2, -6, -18, is a G.P.

where a = -2 and r = 3

a_{n} = a . r^{n-1} =

a_{4} = a . r^{3} = (-2) × 3^{3} = -2 × 27 = -54

a_{5} = a . r^{4} = (-2) × 3^{4} = -2 × 81 = -162

a_{6} = a . r^{5} = (-2) × 3^{5} = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …….

Answer:

i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), ….. is not a G.P.

vi) 3, -3^{2}, 3^{3}, ……

Given: t_{1} = 3, t_{2} = -3^{2}, t_{3} = 3^{3}, ……

i.e., every term is obtained by multiplying its preceding term by a fixed number -3.

3, -3^{2}, 3^{3}, …… forms a G.P,

where a = 3; r = -3

a_{n} = a . r^{n-1}

a_{4} = a . r^{3} = 3 × (-3)^{3} = 3 × (-27) = -81

a_{5} = a . r^{4} = 3 × (-3)^{4} = 3 × 81 = 243

a_{6} = a . r^{5} = 3 × (-3)^{5} = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\), …….

Answer:

Given: t_{1} = x, t_{2} = 1, t_{3} = \(\frac{1}{x}\), ……

Hence x, 1, \(\frac{1}{x}\), …. forms a G.P.

where a = x; r = \(\frac{1}{x}\)

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), …….

Answer:

Given: t_{1} = \(\frac{1}{\sqrt{2}}\), t_{2} = -2, t_{3} = \(\frac{8}{\sqrt{2}}\), ……

ix) 0.4, 0.04, 0.004, ……..

Answer:

Given: t_{1} = 0.4, t_{2} = 0.04, t_{3} = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.

where a = 0.4; r = \(\frac{1}{10}\)

Question 4.

Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.

Answer:

Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.

∴ r = \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) = \(\frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}\)

⇒ \(\frac{x+2}{x}\) = \(\frac{x+6}{x+2}\)

⇒(x + 2)^{2} = x(x + 6)

⇒ x^{2} + 4x + 4 = x^{2} + 6x

⇒ 4x – 6x = – 4 = -2x = -4

∴ x = 2