AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

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 an.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 1

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
an = a . rn-1 = 2 × (-3)n-1
∴ r = -3; an = 2(-3)n-1

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
an = a . rn-1 = (-1) × 3n-1
∴ r = 3; an = (-1) × 3n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
an = a . rn-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); an = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10th 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
an = a . rn-1 = 5 × (5)n-1 = 51+n-1 = 5n
a10 = a . r9 = 5 × 59 = 510
∴ a10 = 510; an = 5n

Question 3.
Find the indicated term of each geometric progression.
i) a1 = 9; r = \(\frac{1}{3}\); find a7.
Answer:
an = a . rn-1
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 2

ii) a1 = -12; r = \(\frac{1}{3}\); find a6.
Answer:
an = a . rn-1
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 3

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

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 nth term of G.P. be 512
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 4
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 29
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5th 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 an = a . rn-1 = 729
⇒ (√3)(√3)n-1 = 729
⇒ (√3)n = 36 = (√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 \(\frac{1}{2187}\)
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 5
Let \(\frac{1}{2187}\) be the nth term of the G.P., then
an = a . rn-1 = \(\frac{1}{2187}\)
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 6
[∵ bases are equal, exponents are also equal]
7th term of G.P is \(\frac{1}{2187}\).

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

Question 5.
Find the 12th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a8 = 192 and r = 2
an = a . rn-1
a8 = a . (2)8-1 = 192
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 7
= 3 × 210 = 3 × 1024 = 3072.

Question 6.
The 4th 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.
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 8
Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 9

AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their nth term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 10
Given that nth terms are equal
an = a . rn-1
AP SSC 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 11
⇒ 3n-1+n-1 = 81 × 81
⇒ 32n-2 = 34 × 34
⇒ 32n-2 = 38 [∵ am . an = am+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5th terms of the two G.P.s are equal.