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Class 11 Applied Maths Chapter 6 Solutions
Sequences and Series
EXERCISE- 6.6
Q.1 (i) Insert 3 geometric means between 1 and 256.
(ii) Insert 4 geometric means between 4/9 and 27/8.
(iii) Insert 6 geometric means between 27 and 1/81.
Ans. (i) a = 1, b = 256, no. of geometric mean = 3
a, G1, G2, G3, b……(GP)
with a = 1, b = 256
Therefore, b = A5
Using, An = ar^(n-1), we get,
b = ar^4
256 = 1 x r^4
256 = r^4
4^4 = r^4
r = 4
Therefore, three geometric mean = 4, 16, 64
(ii) a = 4/9, b = 27/8, no. of geometric mean = 4
Here, a, G1, G2, G3, G4b……(GP)
with a = 4/9, b = 27/8
Using, An = ar^(n-1), we get,
A6 = ar^5
b = ar^5
27/8 = 4/9 x r^5
(27×9)/(8×4) = r^5
(3^3×3^2)/(2^3×2^2) = r^5
3^5/2^5 = r^5
r = 3/2
Therefore, three geometric mean = 2/3, 1, 3/2, 9/4
(iii) a = 27, b = 1/81, no. of geometric mean = 6
Here, a, G1, G2, G3, G4, G5, G6, b……(GP)
with a = 27, b = 1/81
Using, An = ar^(n-1), we get,
A8 = ar^7
b = ar^7
1/81 = 27 x r^7
1/28×27 = r^7
1/(3^4×3^3) = r^4
1/3^7 = r^7
r^ 7 = 1/3
Therefore, three geometric mean = 9, 3, 1, 1/3, 1/9, 1/27
Q.2 If the fourth term of a G.P. is 2, then what is the product of its first 7 terms?
Ans. Here, A4 = 2, Product of 7th terms = ?
We know that,
Product of n terms = [middle term]^n (when n is odd)
P7 = {[(7+1)/2]^th term}^7
= [4th term]^7
= A4^7
= 2^7
Q.3 If the fourth term of a G.P. is 8 and fifth term is 16, then what is the product of first 8 terms?
Ans. Here, A4 = 8, A5 = 16, Product of 8 terms (even) = ?
Product of n terms = [GM of middle terms]^n (when n is even)
P8 = [GM of (8/2)th & (8/2 + 1)th term)^8
= [GM of A4 & A5]^8
= [GM of 8 & 16]^8
= [√ 8×16)^8
= [(128)^1/2]^8
= (2^7)^8
= 2^28
Q.4 If A1, A2 are two A.M.’s and G1, G2 are two G.M.’s between two positive numbers a and b, then find the value of (A1 + A2)/G1G2
Ans. A1 = a + (b-a)/(n+1)
= a + (b-a)/(2+1)
= a + (b-a)/3
= (3a + b – a)/3
= (2a + b)/3
A2 = a + 2(b-a)/3
= (3a + 2b – 2a)/3
= (a + 2b)/3
G1 = a (b/a)^1/(n+1)
= a (b/a)^(1/3)
G2 = a (b/a)^(2/3)
(A1+A2)/(G1G2) = [(2a+b)/3 + (a+2b)/3]/[a(b/a)^1/3 x a(b/a)^2/3]
= [(3a+3b)/3]/[a^2(b/a)^1]
= [3(a + b)/3]/[a^2 x b/a]
= (a + b)/(a + b)
Q.5 Insert 5 geometric means between 576 and 9. Show that their product is equal to the fifth power of the single G.M. between the given numbers.
Ans. 576, G1, G2, G3, G4, G5, 9
a = 576, a7 = 9
ar^6 = 9
576r^6 = 9
r^6 = 9/576
r^6 = (1/2)^6
r = 1/2
G1 = 576 x 1/2 = 288
G2 = 288 x 1/2 = 144
G3 = 144 x 1/2 = 72
G4 = 72 x 1/2 = 36
G5 = 36 x 1/2 = 18
(G3)^5 = (72)^5
Product = (middle term)^n
= (G3)^5
= (72)^5
Q.6 The arithmetic mean between two numbers is 34 and G.M. is 16. Find the numbers.
Ans. (a + b)/2 = 34
a + b = 68
a = 68 – b
√ab = 16
ab = 256
(68 – b)b = 256
68b – b^2 = 256
b^2 – 68 + 256 = 0
(b – 64) (b – 4) = 0
b = 64, b = 4
If b = 64, a = 68 – 64 = 4
If b = 4, a = 68 – 4 = 64
Q.7 The A.M. of two numbers a and b exceeds their G.M. by 2. The ratio of two numbers is 4. Find the numbers.
Ans. AM = GM + 2, a : b = 4
a/b = 4
a = 4b
(a + b)/2 = √ab + 2
(a + b)/2 – 2 = √ab
(4b + b)/2 – 2 = √4bxb
(5b – 4)/2 = 2b
5b – 4 = 4b
b = 4
a = 4 x 4 = 16
Q.8 If G is the geometric mean between a and b, show that 1/(G + a) + 1/(G + b) = 1/G.
Ans. G = √ab
G^2 = ab
1/(G + a) + 1/(G + b)
(G + b + G + a)/(G + a)(G + b)
(2G +a + b)/(G^2 + bG + aG + ab)
(2√ab + a + b)/(ab + (a+b)G + ab)
(√a + √b)^2/(2ab + (a+b)√ab)
(√a + √b)^2/(√ab[2√ab + a + b])
(√a + √b)^2/(G(√a + √b)^2)
1/G = RHS
Q.9 Find the minimum value of 4^x + 4^(1 – x), x ∊ R.
Ans. Let, 4^x + 4^(1-x) be two real no.
We know that, for two real no.
AM ≥ GM
Using, AM =(a +b)/2 & GM = √ab, we get,
[4^x + 4^(1 – x)]/2 ≥ √4^x+4^(1-x)
[4^x + 4^(1 – x)]/2 ≥ √4^x*4/4^x
4^x + 4^(1 – x) ≥ 2√4
4^x + 4^(1 – x) ≥ 2 x 2
4^x + 4^(1 – x) ≥ 4
Minimum value is 4.
Q.10 For any two positive real numbers a and b, prove that a/b + b/a ≥ 2.
Ans. Here, we have two positive real no. a & b
We know that, for a & b, AM(a, b) > GM(a, b)
(a + b)/2 ≥ √ab
(a + b)/√ab ≥ 2
Squaring both sides, we get,
(a + b)^2/ab ≥ 4
(a^2 + b^2 + 2ab)/ab ≥ 4
a^2/ab + b^2/ab + 2ab/ab ≥ 4
a/b + b/a + a ≥ 4
a/b + b/a ≥ 4 – 2
a/b + b/a ≥ 2
Q.11 The sum of two numbers is 6 times their geometric mean, show that the numbers are in the ratio (3 + 2√2) : (3 – 2√2).
Ans. a + b = 6√ab
(a + b)/2√ab = 3
Componendo & Dividendo
(a + b + 2√ab)/(a + b – 2√ab) = (3+1)/(3-1)
(√a + √b)^2/(√a – √b)^2 = 4/2
(√a + √b)/(√a – √b) = √2/1
Again by C & D
(√a + √b + √a – √b)/(√a + √b-(√a – √b) = (√2+1)/(√2-1)
(2√a)/(2√b) = (√2+1)/(√2-1)
Squaring both the sides
a/b = (√2+1)^2/(√2-1)^2
= (2 + 1 + 2√2)/(2 + 1 – 2√2)
= (3 + 2√2)/(3 – 2√2)
FAQ’s related to Class 11 Applied Maths Chapter 6 on Sequences and Series:
Q.1 What is a sequence?
Ans. A sequence is an ordered list of numbers following a particular pattern. Each number in a sequence is called a term.
Q.2 What is a series?
Ans. A series is the sum of the terms of a sequence. If we add the terms of a sequence, we get a series.
Q.3 What is the difference between an arithmetic sequence and a geometric sequence?
Ans.
- Geometric Sequence – A geometric sequence (or geometric progression) is a sequence in which each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio (r).
- Arithmetic Sequence – An arithmetic sequence (or arithmetic progression) is a sequence in which the difference between consecutive terms is constant. This difference is called the common difference (d).
These are a few Frequently Asked Questions relating to Class 11 Applied Maths Chapter 6
In Class 11 Applied Maths chapter 6, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 11 Applied Maths Chapter 6 promises an enriching learning experience that will set you on the path to success. Class 11 Applied Maths Chapter 6, we delve deep into advanced mathematical concepts that are crucial for understanding.
Class 11 Applied Maths Chapter 6 Exercise :
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