Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 4 Sequences and Series Ex 4.2 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Sequences and Series Ex 4.2
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
(i) a = 2, r = 3, Sn = 242 Sn = , for r > 1
Solution & Step-by-Step Answer:
(i) t3 = 20, t6 = 160 tn = arn-1 ∴ t3 = ar3-1 = ar2 ∴ ar2 = 20 ∴ a = ……(i)
Solution & Step-by-Step Answer:
(i) Sn = 3 + 33 + 333 +….. upto n terms = 3(1 + 11 + 111 +….. upto n terms) = (9 + 99 + 999 + … upto n terms) = [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms] = [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)] But 10, 100, 1000, … n terms are in G.P. with a = 10, r = = 10
(ii) Sn= 8 + 88 + 888 + … upto n terms
= 8(1 + 11 + 111 + … upto n terms)
= (9 + 99 + 999 + … upto n terms)
= [(10 – 1) + (100 – 1) + (1000 – 1) +… upto n terms]
= [(10 + 100 + 1000 + … upto n terms) – (1 + 1 + 1 + … n times)]
But 10, 100, 1000, … n terms are in G.P. with
a = 10, r = = 10
Solution & Step-by-Step Answer:
(i) Sn = 0.4 + 0.44 + 0.444 + ….. upto n terms = 4(0.1 + 0.11 + 0.111 + …. upto n terms) = (0.9 + 0.99 + 0.999 + … upto n terms) = [(i – 0.1) + (1 – 0.01) + (1 – 0.001) … upto n terms] = [(1 + 1 + 1 + …n times) – (0.1 + 0.01 + 0.001 +… upto n terms)] But 0.1, 0.01, 0.001, … n terms are in G.P. with a = 0.1, r = = 0.1 ∴ Sn =
(ii) Sn= 0.7 + 0.77 + 0.777 + … upto n terms
= 7(0.1 + 0.11 + 0.111 + … upto n terms)
= (0.9 + 0.99 + 0.999 + … upto n terms)
= [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) +… upto n terms]
= [(1 + 1 + 1 +… n times) – (0.1 + 0.01 + 0.001 +… upto n terms)]
But 0.1, 0.01, 0.001, … n terms are in G.P.
with a = 0.1, r = = 0.1
Solution & Step-by-Step Answer:
(i) Let t1 = 0.5, t2 = 0.55, t3 = 0.555 and so on. t1 = 0.5 t2 = 0.55 = 0.5 + 0.05 t3 = 0.555 = 0.5 + 0.05 + 0.005 ∴ tn = 0.5 + 0.05 + 0.005 + … upto n terms But 0.5, 0.05, 0.005, … upto n terms are in G.P. with a = 0.5 and r = 0.1 ∴ tn = the sum of first n terms of the G.P.
(ii) Let t1= 0.2, t2= 0.22, t3= 0.222 and so on
t1= 0.2
t2= 0.22 = 0.2 + 0.02
t3= 0.222 = 0.2 + 0.02 + 0.002
∴ tn= 0.2 + 0.02 + 0.002 + … upto n terms
But 0.2, 0.02, 0.002, … upto n terms are in G.P. with a = 0.2 and r = 0.1
∴ tn= the sum of first n terms of the G.P.
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
Let a be the 1st term and r be the common ratio of the G.P. ∴ the G.P. is a, ar, ar2, ar3, …, arn-1
Solution & Step-by-Step Answer:
Let a and r be the 1st term and common ratio of the G.P. respectively.
