Maharashtra State Board 11th Maths Solutions Chapter 2 Sequences and Series Ex 2.4
Question 1
Maharashtra Board Solution
Verify whether the following sequences are H.P. (i)
Solution & Step-by-Step Answer:
Here, the reciprocal sequence is 3, 5, 7, 9,… t1 = 3, t2 = 5, t3 = 7, t4 = 9, ….. t2 – t1 = t3 – t2 = t4 – t3 = 2 = constant ∴ The reciprocal sequence is an A.P. ∴ The given sequence is a H.P.
(ii)
Solution:
Here, the reciprocal sequence is 3, 6, 12, 24,…
t1= 3, t2= 6, t3= 12, ……
t2– t1= 3, t3– t2= 6
t2– t1≠ t3– t2
∴ The reciprocal sequence is not an A.P.
∴ The given sequence is not a H.P.
(iii)
Solution:
∴ The reciprocal sequence is an A.P.
∴ The given sequence is a H.P.
Question 2
Maharashtra Board Solution
Find the nth term and hence find the 8th term of the following HPs. (i)
Solution & Step-by-Step Answer:
are in H.P. ∴ 2, 5, 8, 11,… are in A.P. ∴ a = 2, d = 3 tn = a + (n – 1)d = 2 + (n – 1)(3) = 3n – 1 ∴ nth term of H.P. = ∴ 8th term of H.P. = =
(ii)
Solution:
are in H.P.
∴ 4, 6, 8, 10, … are in A.P.
∴ a = 4, d = 2
tn= a + (n – 1)d
= 4 + (n – 1) (2)
= 2n + 2
∴ nth term of H.P. =
∴ 8th term of H.P. = =
(iii)
Solution:
are in H.P.
∴ 5, 10, 15, 20, … are in A.P.
∴ a = 5, d = 5
tn= a + (n – 1)d
= 5 + (n – 1) (5)
= 5n
∴ nth term of H.P. =
∴ 8th term of H.P. = =
Question 3
Maharashtra Board Solution
Find A.M. of two positive numbers whose G.M. and H.M. are 4 and respectively.
Solution & Step-by-Step Answer:
G.M. = 4, H.M. = Now, (G.M.)2 = (A.M.) (H.M.) ∴ 42 = A.M. × ∴ A.M. = 16 × ∴ 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 Now, (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. ∴ 4, H1, H2, 3 are in A.P. t1 = 4, t2 = H1, t3 = H2, t4 = 3 ∴ t1 = a = 4, t4 = 3 tn = a + (n – 1)d t4 = 4 + (4 – 1)d 3 = 4 + 3d 3d = -1 ∴ d = H1 = t2 = a + d = 4 – = H2 = t3 = a + 2d = 4 – = ∴ For resulting sequence to be H.P. we need to insert numbers and .
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 = ar2 = 1(-3)2 = 9 ∴ For resulting sequence to be G.P. we need to insert numbers -3 and 9.
Question 8
Maharashtra Board Solution
If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by , find the numbers.
Solution & Step-by-Step Answer:
Let a and b be the two numbers. A = , G = , H = According to the given conditions, Consider, G = A – 2 = 10 – 2 = 8 = 8 ab = 64 a(20 – a) = 64 …..[From (i)] a2 – 20a + 64 = 0 (a – 4)(a – 16) = 0 ∴ a = 4 or a = 16 When a = 4, b = 20 – 4 = 16 When a = 16, b = 20 – 16 = 4 ∴ The two numbers are 4 and 16.
Question 9
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 and b be the two numbers. A = , G = , H = According to the given conditions, = 6 ab = 36 a(13 – a) = 36 ……[From (i)] 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.
