Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

 

Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Exercise 11(C)

Selina Class 10 ICSE solution Chapter 11 Geometric Progression will give strength to you and the board will also clear the exam, Mathematics Subject demands time and practice, then keep practicing and you will definitely get success.

Selina Publishers Solutions for Class 10 is designed by experienced teachers keeping in mind the syllabus of the examinations and the requirement of the students. The ICSE solution for class 10 presents a firm platform for students to prepare more and more questions and refer to the correct answers. That can provide complete solutions for students.

Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

Exercise 11(C) Page No: 156

Question 1. Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32

Solution:

Given series: √2, 2, 2√2, …… , 32

Here,

a = √2

r = 2/ √2 = √2

And, the last term (l) = 32

l = tn = arn – 1 = 32

(√2)( √2)n – 1 = 32

(√2)n = 32

(√2)n = (2)5 = (√2)10

Equating the exponents, we have

n = 10

So, the 7th term from the end is (10 – 7 + 1)th term.

i.e. 4th term of the G.P

Hence,

t4 = (√2)(√2)4 – 1 = (√2)(√2)3 = (√2) x 2√2 = 4

Question 2. Find the third term from the end of the G.P.

2/27, 2/9, 2/3, ……., 162

Solution:

Given series: 2/27, 2/9, 2/3, ……., 162

Here,

a = 2/27

r = (2/9) / (2/27)

r = 3

And, the last term (l) = 162

l = tn = arn – 1 = 162

(2/27) (3)n – 1 = 162

(3)n – 1 = 162 x (27/2)

(3)n – 1 = 2187

(3)n – 1 = (3)7

n – 1 = 7

n = 7+1

n = 8

So, the third term from the end is (8 – 3 + 1)th term

i.e 6th term of the G.P. = t6

Hence,

t6 = ar6-1

t6 = (2/27) (3)6-1

t6 = (2/27) (3)5

t6 = 2 x 32

t6 = 18

Question 3. Find the G.P. 1/27, 1/9, 1/3, ……, 81; find the product of fourth term from the beginning and the fourth term from the end.

Solution:

Given G.P. 1/27, 1/9, 1/3, ……, 81

Here, a = 1/27, common ratio (r) = (1/9)/ (1/27) = 3 and l = 81

We know that,

l = tn = arn – 1 = 81

(1/27)(3)n – 1 = 81

3n – 1 = 81 x 27 = 2187

3n – 1 = 37

n – 1 = 7

n = 8

Hence, there are 8 terms in the given G.P.

Now,

4th term from the beginning is t4 and the 4th term from the end is (8 – 4 + 1) = 5th term (t5)

Thus,

the product of t4 and t5 = ar4 – 1 x ar5 – 1 = ar3 x ar4 = a2r7 = (1/27)2(3)7 = 3

Question 4. If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that:

(q – r) log a + (r – p) log b + (p – q) log c = 0

Solution:

Let’s take the first term of the G.P. be A and its common ratio be R.

Then,

pth term = a ⇒ ARp – 1 = a

qth term = b ⇒ ARq – 1 = b

rth term = c ⇒ ARr – 1 = c

Now,

On taking log on both the sides, we get

log( aq-r x br-p x cp-q ) = log 1

⇒ (q – r)log a + (r – p)log b + (p – q)log c = 0

– Hence Proved

Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

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