Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.5 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.5

Question 1.

For each geometric progression find the common ratio ‘r’, and then find a_{n}.

i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\),………..

Solution:

ii) 2, -6, 18, -54,…………

Solution:

Given

G.P. 2, -6, 18, -54,…………

a = 2, r = \(\frac{a_2}{a_1}\) = \(\frac{-6}{2}\) = -3

a_{n} = a. r^{n-1} = 2 × (-3)^{n-1}

∴ r = -3; a_{n} = 2(-3)^{n-1}

iii) -1, -3, -9, -27,………..

Solution:

Given G.P. = – 1, – 3, – 9, – 27, ….

a = -1, r = \(\frac{a_2}{a_1}\) = \(\frac{-3}{-1}\) = 3

a_{n} = a . r^{n-1} = (-1) × 3^{n-1}

∴ r = 3, a_{n} = (-1) × 3^{n-1}

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\),……….

Solution:

Given G.P: 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\),…………

a = 5, r = \(\frac{\mathrm{a}_2}{\mathrm{a}_1}\) = \(=\frac{2}{5}\)

a_{n} = a . r^{n-1} = 5 × \(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)

∴ r = \(\frac{2}{5}\) ; a_{n} = 5\(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)

Question 2.

Find the 10^{th} and n^{th} term of G.P.:

5, 25, 125,……..

Solution:

Given G.P. = 5, 25, 125,……….

a = 5, r = \(\frac{a_2}{a_1}\) = \(\frac{25}{5}\) = 5

a_{n} = a. r^{n-1} = 5 × 5^{n-1} = 5^{1+n-1} = 5^{n}

∴ a_{10} = a . r^{9} = 5 × 5^{9} = 5^{10}

∴ a_{10} = 5^{10}; a_{n} = 5^{n}

Question 3.

Find the indicated term of each geometric progression.

i) a_{1} = 9 ; r = \(\frac{1}{3}\) ; find a_{7}.

Solution:

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

a_{7} = 9 × \(\left(\frac{1}{3}\right)^{7-1}\) = 3^{2} \(\left(\frac{1}{3}\right)^6\) = \(\frac{3^2}{3^6}\) = \(\frac{1}{3^4}\)

∴ a_{7} = \(\frac{1}{3^4}\)

ii) a_{1} = -12 ; r = \(\frac{1}{3}\) ; find a_{6}.

Solution:

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

a_{6} = (-12) × \(\left(\frac{1}{3}\right)^{6-1}\)

= (-12) × \(\frac{1}{3^5}\) = \(\frac{-4 \times 3}{3^5}\) = \(\frac{-4}{3^4}\)

∴ a_{6} = \(\frac{-4}{3^4}\)

Question 4.

Which term of the G.P.

i) 2, 8, 32,………. is 512?

Solution:

Given GP. :2, 8, 32,………. is 512

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

Let the n^{th} term of G.P. be 512

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

512 = 2 × (4)^{n-1}

2^{9} = 2 × (2^{2})^{n-1}

2^{9} = 2^{1} × 2^{2(n-1)}

2^{9} = 2^{2n-2+1}

2^{9} = 2^{2n-1}

[∵ bases are equal, exponents are also equal]

∴ 2n = 9 + 1

n = \(\frac{10}{2}\) = 5

∴ 512 is the 5^{th} term of the given G.P

ii) \(\sqrt{3}\), 3, 3\(\sqrt{3}\),……. is 729?

Solution:

[∵ bases are equal, exponents are also equal]

\(\frac{1}{2}\)(n-1) = 6 – 1

⇒ n – 1 = 5 × 2

⇒ n = 10 + 1 = 11

∴ 729 is the 11^{th} term of the given G.P

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\),……….. is \(\frac{1}{2187}\) ?

Solution:

Given

[∵ bases are equal, exponents are also equal]

∴ 7^{th} term of GP is \(\frac{1}{2187}\)

Question 5.

Find the 12^{th} term of a G.P whose 8^{th} term is 192 and the common ratio is 2.

Solution:

Given a G.P such that a_{8} = 192 and

r = 2

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

∴ a_{8} = a(2)^{8-1} = 192

a.2^{7} = 192

⇒ a = \(\frac{192}{2^7}\) = \(\frac{192}{128}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)

∴ a_{12} = a. r^{11} = \(\frac{3}{2}\) × (2)^{11}

= 3 × 2^{10} = 3 × 1024 = 3072.

Question 6.

The 4^{th} term of a geometric progressions is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.

Now substituting r = \(\frac{2}{3}\) in equation (1) we get,

a\(\left(\frac{2}{3}\right)^3\) = \(\frac{2}{3}\) ⇒ a = \(\frac{2}{3}\) × \(\frac{3}{2}\) × \(\frac{3}{2}\) × \(\frac{3}{2}\) = \(\frac{9}{4}\) ×

∴ The G.P is a, ar, ar^{2}, ar^{3},…………..

\(\frac{9}{4}\), \(\frac{9}{4}\) × \(\frac{2}{3}\) ; \(\frac{9}{4}\) × \(\left(\frac{2}{3}\right)^2\) …….. = \(\frac{9}{4}\), \(\frac{3}{2}\), 1,……..

Question 7.

If the geometric progressions 162, 54, 18 …. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……….. have their n^{th} term equal, find the value of n.?

Solution:

Given G.P. :

162, 54, 18,…….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),………

Given that n^{th} terms are equal

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

⇒ 162 × \(\left(\frac{1}{3}\right)^{\mathrm{n}-1}\) = \(\frac{2}{81}\) × (3)^{n-1}

⇒ 3^{n-1} × 3^{n-1} = 162 × \(\frac{81}{2}\)

⇒ 3^{n-1+n-1} = 81 × 81

⇒ 3^{2n-2} = 3^{8} [a^{m} . a^{n} = a^{m+n}]

⇒ 2n – 2 = 8

[∵ bases are equal, exponents are also equal]

2n = 8 + 2

∴ n = \(\frac{10}{2}\) = 5

The 5^{th} terms of the two G.Ps are equal.