Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 4 Sequences and Series Ex 4.4 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.4
Question 1
Maharashtra Board Solution
Verify whether the following sequences are H.P. (i) (ii) (iii)
Solution & Step-by-Step Answer:
(i) Here, the reciprocal sequence is 3, 5, 7, 9, … ∴ t1 = 3, t2 = 5, t3 = 7, ….. ∵ t2 – t1 = t3 – t2 = t4 – t3 = 2, constant ∴ The reciprocal sequence is an A.P. ∴ the given sequence is H.P.
(ii)
Here, the reciprocal sequence is 3, 6, 9, 12 …
∴ t1= 3, t2= 6, t3= 9, t4= 12, …
∵ t2– t1= t3– t2= t4– t3= 3, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.
(iii)
Here, the reciprocal sequence is 7, 9, 11, 13, 15, ……
∴ t1= 7, t2= 9, t3= 11, t4= 13, …..
∵ t2– t1= t3– t2= t4– t3= 2, constant
∴ The reciprocal sequence is an A.P.
∴ The given sequence is H.P.
Question 2
Maharashtra Board Solution
Find the nth term and hence find the 8th term of the following H.P.s: (i) (ii) (iii)
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and .
Solution & Step-by-Step Answer:
G.M. = 4, H.M. = ∵ (G.M.)2 = (A.M.) (H.M.) ∴ 16 = A.M. × ∴ A.M. = 5
Question 4
Maharashtra Board Solution
Find H.M. of two positive numbers whose A.M. and G.M. are and 6.
Solution & Step-by-Step Answer:
A.M. = , G.M. = 6 Now, (G.M.)2 = (A.M.) (H.M.) ∴ 62 = × H.M. ∴ H.M. = 36 × ∴ H.M. =
Question 5
Maharashtra Board Solution
Find G.M. of two positive numbers whose A.M. and H.M. are 75 and 48.
Solution & Step-by-Step Answer:
A.M. = 75, H.M. = 48 (G.M.)2 = (A.M.) (H.M.) ∵ (G.M.)2 = 75 × 48 ∵ (G.M.)2 = 25 × 3 × 16 × 3 ∵ (G.M.)2 = 52 × 42 × 32 ∴ G.M. = 5 × 4 × 3 ∴ G.M. = 60
Question 6
Maharashtra Board Solution
Insert two numbers between and so that the resulting sequence is a H.P.
Solution & Step-by-Step Answer:
Let the required numbers be and . ∴ are in H.P. ∴ 7, H1, H2 and 13 are in A.P. ∴ t1 = a = 7 and t4 = a + 3d = 13 ∴ 7 + 3d = 13 ∴ 3d = 6 ∴ d = 2 ∴ H1 = t2 = a + d = 7 + 2 = 9 and H2 = t3 = a + 2d = 7 + 2(2) = 11 ∴ and are the required numbers to be inserted between and so that the resulting sequence is a H.P.
Question 7
Maharashtra Board Solution
Insert two numbers between 1 and -27 so that the resulting sequence is a G.P.
Solution & Step-by-Step Answer:
Let the required numbers be G1 and G2. ∴ 1, G1, G2, -27 are in G.P. ∴ t1 = 1, t2 = G1, t3 = G2, t4 = -27 ∴ t1 = a = 1 tn = arn-1 ∴ t4 = (1) r4-1 ∴ -27 = r3 ∴ r3 = (-3)3 ∴ r = -3 ∴ G1 = t2 = ar = 1(-3) = -3 ∴ G2 = t3 = ar = 1(-3)2 = 9 ∴ -3 and 9 are the required numbers to be inserted between 1 and -27 so that the resulting sequence is a G.P.
Question 8
Maharashtra Board Solution
Find two numbers whose A.M. exceeds their G.M. by and their H.M. by .
Solution & Step-by-Step Answer:
Let a, b be the two numbers. ∴ a + b = 13 ∴ b = 13 – a …….(iii) and ab = 36 ∴ a(13 – a) = 36 …… [From (iii)] ∴ a2 – 13a + 36 = 0 ∴ (a – 4)(a – 9) = 0 ∴ a = 4 or a = 9 When a = 4, b = 13 – 4 = 9 When a = 9, b = 13 – 9 = 4 ∴ the two numbers are 4 and 9.
Question 9
Maharashtra Board Solution
Find two numbers whose A.M. exceeds G.M. bv 7 and their H.M. by .
Solution & Step-by-Step Answer:
Let a, b be the two numbers. ∴ a + b = 70 ∴ b = 70 – a …..(ii) ∴ G = A – 7 = 35 – 7 = 28 …….[From (i)] ∴ √ab = 28 ∴ ab = 282 = 784 ∴ a(70 – a) = 784 ……[From (ii)] ∴ 70a – a2 = 784 ∴ a2 – 70a + 784 = 0 ∴ a2 – 56a – 14a + 784 = 0 ∴ (a – 56) (a – 14) = 0 ∴ a = 14 or a = 56 When a = 14, b = 70 – 14 = 56 When a = 56, b = 70 – 56 = 14 ∴ the two numbers are 14 and 56.
