Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.4 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.4

Question 1.

In which of the following situations, does the list of numbers involved in 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%.

Solution:

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

Rate of annual increment = 10%

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

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

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

Hence the situation forms a G.R

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.

Solution:

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.R

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.

Solution:

Perimeter of the 1^{st} equilateral triangle = 3 × 24 = 72 cm

Perimeter of the 2^{nd} equilateral triangle = 3 × 12 = 36 cm

Perimeter of the 3^{rd} equilateral triangle = 3 × 6 = 18 cm

Successive terms are obtained by dividing with 2 the preceding term except first term.

∴ The above situation is 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.

Solution:

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

∴ 4, 4 × 3, 4 × 3^{2}, 4 × 3^{3}

∴ 4, 12, 36, 108,…………

ii) a = \(\sqrt{5}\); r = \(\frac{1}{5}\)

Solution:

The terms of a G.P are:

a, ar, ar^{2}, ar^{3},……….

iii) a = 81; r = \(-\frac{1}{3}\).

Solution:

The terms of a G.P are:

a, ar, ar^{2},…………

⇒ 81, 81 × \(\left(\frac{-1}{3}\right)\), 81 × \(\left(\frac{-1}{3}\right)^2\)

⇒ 81, -27, 9,…………

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

Solution:

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

a_{1} = a = \(\frac{1}{64}\)

a_{2} = ar = \(\frac{1}{64}\) × 2 = \(\frac{1}{32}\)

a_{3} = ar^{2} = \(\frac{1}{64}\) × 2^{2} = \(\frac{1}{64}\) × 4 = \(\frac{1}{16}\)

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

Question 3.

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

i) 4, 8, 16,………….

Solution:

Given : 4, 8, 16,………….

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

\(\frac{a_2}{a_1}\) = \(\frac{8}{4}\) = 2

\(\frac{\mathrm{a}_3}{\mathrm{a}_2}\) = \(\frac{16}{8}\) = 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) \(\frac{1}{3}\), \(\frac{-1}{6}\), \(\frac{1}{12}\)………, (x ≠ 0)

Solution:

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

Hence the ratio is common between any two successive terms.

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

iii) 5, 55, 555,………….

Solution:

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

\(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) = \(\frac{55}{5}\) = 11

\(\frac{\mathrm{t}_3}{\mathrm{t}_2}\) = \(\frac{555}{55}\) = \(\frac{111}{11}\)

∴ \(\frac{t_2}{t_1}\) ≠ \(\frac{t_3}{t_2}\)

iv) -2, -6, -18,…………..

Solution:

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

\(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) = \(\frac{-6}{-2}\) = 3; \(\frac{t_3}{t_2}\) = \(\frac{-18}{-6}\) = 3

∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\) = …….. = 3

∴ -2, -6, -18,……….. is a G.P.

where a = -2 and r = 3

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

a_{4} = ar^{3} = (-2) × 3^{3} = -2 × 27 = -54

a_{5} = ar^{4} = (-2) × 3^{4} = – 2 × 81 = -162

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

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

Solution:

Given: t_{1} = \(\frac{1}{2}\), t_{2} = \(\frac{1}{4}\), t_{3} = \(\frac{1}{6}\)

\(\frac{t_2}{t_1}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\frac{1}{4}\) × \(\frac{2}{1}\) = \(\frac{1}{2}\)

\(\frac{t_3}{t_2}\) = \(\frac{\frac{1}{6}}{\frac{1}{4}}\) = \(\frac{1}{6}\) × \(\frac{4}{1}\) = \(\frac{2}{3}\)

∴ \(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) ≠ \(\frac{t_3}{t_2}\)

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

vi) 3, -3^{2}, 3^{3},…………

Solution:

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

\(\frac{t_2}{t_1}\) = \(\frac{-3^2}{3}\) = -3

\(\frac{\mathrm{t}_3}{\mathrm{t}_2}\) = \(\frac{3^3}{-3^2}\) = -3

∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\) = …….. = -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} = 3 × (-3)^{4-1} = 3 × (-3)^{3} = -81

a_{5} = 3 × (-3)^{4} = 3 × 81 = 243

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

vii) x, 1, \(\frac{1}{x}\),……… (x ≠ 0)

Solution:

Given:

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

Solution:

Given :

Given terms are not in G.P.

ix) 0.4, 0.04, 0.004,………

Solution:

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

Question 4.

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

Solution:

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

∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\)

⇒ \(\frac{x+2}{x}\) = \(\frac{x+3}{x+2}\)

⇒ (x + 2)^{2} = x(x + 6)

⇒ x^{2} + 4x + 4 = x^{2} + 6x

⇒ 4x – 6x = -4 ⇒ -2x = -4

∴ x = 2