StudentAidUnit 6: Sequences and Series — Exercise 6 9
11th Class Mathematics · Unit 6: Sequences and Series
1.(i) Find the 9th term of the harmonic sequence 1/3, 1/5, 1/7,... (ii) Find the 9th term of the harmonic sequence -1/5, -1/3, -1,...
2.(i) Insert five harmonic means between -2/5 and 2/13. (ii) Insert five harmonic means between 1/4 and 1/24.
Given
(i) H.P. with five means to be inserted
-25, A1, A2, A3, A4, A5, 213
(i) Reciprocals in A.P.
-52, 1A1, 1A2, 1A3, 1A4, 1A5, 132 in A.P.
(i) First and last terms
a1 = -52, a7 = 132
Formula
(i) Common difference
a7 = a1 + 6d ⇒ 132 = -52 + 6d
Working
(i) Solving for d
6d = 9 ⇒ d = 32
(i) A.P. terms
-52, -1, 12, 2, 72, 5, 132
Result
(i) Five H.M.
-1, 2, 12, 27, 15
Given
(ii) H.P. with five means to be inserted
14, A1, A2, A3, A4, A5, 124
(ii) Reciprocals in A.P.
4, 1A1, 1A2, 1A3, 1A4, 1A5, 24 in A.P.
(ii) First and last terms
a1 = 4, a7 = 24
Working
(ii) Solving for d
24 = 4 + 6d ⇒ 6d = 20 ⇒ d = 103
(ii) A.P. terms
4, 223, 323, 14, 523, 623, 24
Result
(ii) Five H.M.
322, 332, 114, 352, 362
3.The first term of an H.P is -1/3 and the fifth term is 1/5. Find its 9th term.
Given
First and fifth terms
a1 = -13, a5 = 15
Reciprocals in A.P.
a1 = -3, a5 = 5 (in A.P.)
Working
Finding common difference
a5 = a1 + 4d ⇒ 5 = -3 + 4d ⇒ 4d = 8 ⇒ d = 2
Formula
9th term of A.P.
a9 = a1 + 8d = -3 + 8(2) = -3 + 16 = 13
Result
9th term of H.P.
113
4.If 5 is the harmonic mean between 2 and b, find b.
Given
Harmonic mean formula
H.M. = 2aba+b
Formula
Setting up equation
5 = 2(2)(b)2+b
Working
Simplifying
5 = 4b2+b
Cross-multiplying
5(2+b) = 4b ⇒ 10 + 5b = 4b
Result
Value of b
b = -10
5.If the numbers 1/k, 1/(2k+1), and 1/(4k-1) are in harmonic sequence, find k.
Condition for H.P.
If 1k, 12k+1, 14k-1 are in H.P., then k, 2k+1, 4k-1 are in A.P.
Formula
A.P. condition
(2k+1) - k = (4k-1) - (2k+1)
Working
Simplifying
k + 1 = 2k - 2
Solving for k
1 + 2 = 2k - k
Result
Value of k
k = 3
8.If the H.M and A.M between two numbers are 4 and 9/2 respectively, find the numbers.
Given
Arithmetic mean
A.M. = a+b2 = 92
From A.M.
a + b = 9
Harmonic mean
H.M. = 2aba+b = 4
Working
Substituting A.M.
2ab9 = 4 ⇒ 2ab = 36 ⇒ ab = 18
Formula
Quadratic with sum and product
x2 - 9x + 18 = 0
Factoring
(x-3)(x-6) = 0
Result
The numbers
3 or 6
9.If the (positive) G.M and H.M between two numbers are 4 and 16/5 respectively, find the numbers.
Given
Geometric mean
G.M. = sqrt{ab} = 4
From G.M.
ab = 16
Harmonic mean
H.M. = 2aba+b = 165
Working
Substituting ab = 16
2(16)a+b = 165 ⇒ 32a+b = 165
Solving for sum
a + b = 32 × 516 = 10
Formula
Quadratic equation
x2 - 10x + 16 = 0
Factoring
(x-2)(x-8) = 0
Result
The numbers
2 or 8
13.The H.M of two numbers is 4. The A.M, A and G.M, G satisfy the relation 2A + G² = 27. Find the numbers.
Given
Harmonic mean
H.M. = 2aba+b = 4
From H.M.
2ab = 4(a+b) ⇒ a+b = ab2
Arithmetic and geometric means
A = a+b2, quad G2 = ab
Formula
Given relation
2A + G2 = 27 ⇒ a+b + ab = 27
Working
Substituting a+b = ab/2
ab2 + ab = 27 ⇒ 3ab2 = 27 ⇒ ab = 18
Finding sum
a+b = 182 = 9
Formula
Quadratic equation
x2 - 9x + 18 = 0
Factoring
(x-3)(x-6) = 0
Result
The numbers
3 or 6
17.If 2+x, 5+x and 9+x are in H.P., find the value of x.
Condition for H.P.
If 2+x, 5+x, 9+x are in H.P., then 12+x, 15+x, 19+x are in A.P.
Formula
A.P. condition
2 · 15+x = 12+x + 19+x
Combining fractions (R.H.S.)
25+x = (9+x) + (2+x)(2+x)(9+x) = 11+2x(2+x)(9+x)
Working
Cross-multiplying
2(2+x)(9+x) = (5+x)(11+2x)
Expanding L.H.S.
2(18 + 2x + 9x + x2) = 2(18 + 11x + x2) = 36 + 22x + 2x2
Expanding R.H.S.
55 + 10x + 11x + 2x2 = 55 + 21x + 2x2
Equating and simplifying
36 + 22x + 2x2 = 55 + 21x + 2x2 ⇒ 22x - 21x = 55 - 36
Result
Value of x
x = 19