AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

### 10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.

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

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

Answer:

Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….

Answer:

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, ……

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

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

a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)

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

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

Question 2.

Find the 10^{th} and nth term of G.P.: 5, 25, 125,…..

Answer:

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}.

Answer:

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

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

Answer:

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

Question 4.

Which term of the G.P.

i) 2, 8, 32,….. is 512?

Answer:

Given G.P.: 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

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{9}

∴ 2n – 1 = 9

[∵ bases are equal, exponents are also equal]

∴ 2n = 9 + 1 = 10

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

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

ii) √3, 3, 3√3, …….. is 729?

Answer:

Given G.P.: √3, 3, 3√3, …….. is 729

a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3

now a_{n} = a . r^{n-1} = 729

⇒ (√3)(√3)^{n-1} = 729

⇒ (√3)^{n} = 3^{6} = (√3)^{12}

⇒ n = 12

So 12th term of GP √3, 3, 3√3, …….. is 729.

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

Answer:

Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^{th} term of the G.P., then

a_{n} = a . r^{n-1} = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]

7^{th} term of G.P is \(\frac{1}{2187}\).

Question 5.

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

Answer:

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

= 3 × 2^{10} = 3 × 1024 = 3072.

Question 6.

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

Answer:

Given: In a G.P.

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

we get,

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.

Answer:

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}

⇒ 3^{n-1+n-1} = 81 × 81

⇒ 3^{2n-2} = 3^{4} × 3^{4}

⇒ 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.P.s are equal.