Question

NCERT Solutions Class 11 Maths Chapter 9 Sequence and Series is one of the important chapters of class 11. Our NCERT Solutions are made in very concise form.

One must refer to our NCERT Solutions Class 11 for complete understanding of all the concepts. We have covered all the important concepts such as

• Sequence – sequence is orderly arrangement of grouped numbers which is dictated by some specific rule. Example of sequence 1, 3, 5, 7, 9, 11.  
• Series – things or the events of similar class which comes one after one another in spatial or temporal succession is known as series. It is also the sum of the elements in the sequence.

• Arithmetic Progression (A.P.) – an arithmetic progression is a sequence in which the difference between every successive term is same. In arithmetic progression every next term can be obtained by adding some fixed number to the previous term. Example: 1, 3, 5, 7, 9, 11, 13

• Arithmetic Mean (A.M.) – arithmetic mean is the average sum of set of given numbers. In Arithmetic Progression the arithmetic mean is the middle term of the arithmetic progression. In a given arithmetic progression when there are three terms a, b, and c then the arithmetic mean is the middle term, i.e. ‘b’, hence the b is the arithmetic mean of the given arithmetic progression.

• Geometric Progression (G.P.) – in Geometric Progression each new term is obtained by multiplying a fixed number to the last term. The number which is being multiplied is also called the common ratio. Example: 1,2,4,8,16,32,64.

• Sum of Geometric series of infinite terms – in infinite geometric series there is no particular last term. There are two such cases of infinite geometric series. In one case the ratio of geometric series is greater than 1 and other where the ratio of geometric series is less than 1.
• Geometric mean (G.M.) – the geometric mean is a peculiar type of average where we take the reciprocal of number of number of term and put it over the product of numbers. Example: geometric mean of 4 and 16 is 8. \(4\times 16\) = 64, \(\sqrt[]{64} \) = 8.

This NCERT Solutions Class 11 Maths provides a holistic approach to provide solution of all queries of students.

Answer 1

 98. Find the sum of the following series up to n terms:

\(\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}\) 

Answer:

The n^th term of the given series is

Here, 1, 3, 5,... (2n - 1) is an A.P. with first term a, last term (2n - 1) and number of terms as n 

99. Show that

Answer:

n^th term of the numerator = n(n + 1)² = n³ + 2n² + n 

n^th term of the denominator = n²(n + 1) = n³ + n²

From (1), (2), and (3), we obtain

Thus, the given result is proved.

100. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?

Answer:

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000 

According to the given condition, the interest paid annually is 

12% of 6000, 12% of 5500, 12% of 5000… 12% of 500

Thus, total interest to be paid 

= 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500 

= 12% of (6000 + 5500 + 5000 + … + 500) 

= 12% of (500 + 1000 + 1500 + … + 6000) 

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common difference equal to 500. 

Let the number of terms of the A.P. be n.

∴ 6000 = 500 + (n – 1) 500 

⇒ 1 + (n – 1) = 12 

⇒ n = 12 

∴ Sum of the A.P

Thus, total interest to be paid 

= 12% of (500 + 1000 + 1500 + … + 6000) 

= 12% of 39000 = Rs 4680 

Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

101. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

Answer:

It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash. ∴

 Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000 

According to the given condition, the interest paid annually is 

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000 

Thus, total interest to be paid 

= 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000 

= 10% of (18000 + 17000 + 16000 + … + 1000) 

= 10% of (1000 + 2000 + 3000 + … + 18000) 

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000. 

Let the number of terms be n. 

∴ 18000 = 1000 + (n – 1) (1000) 

⇒ n = 18

∴ Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000) 

= 10% of Rs 171000 = Rs 17100 

∴ Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

102. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8^th set of letter is mailed.

Answer:

The numbers of letters mailed forms a G.P.: 4, 42 , … 48 

First term = 4 

Common ratio = 4 

Number of terms = 8 

It is known that the sum of n terms of a G.P. is given by

It is given that the cost to mail one letter is 50 paisa. 

∴ Cost of mailing 87380 letters = Rs 87830 x \(\frac{50}{100}\) = Rs 43690

Thus, the amount spent when 8^th set of letter is mailed is Rs 43690.

103. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15^th year since he deposited the amount and also calculate the total amount after 20 years.

Answer:

It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

∴ Interest in first year 

 

∴ Amount in 15^th year 

= Rs 10000 + 14 × Rs 500 

= Rs 10000 + Rs 7000 

= Rs 17000

Amount after 20 years = 

= Rs 10000 + 20 × Rs 500 

= Rs 10000 + Rs 10000 

= Rs 20000

104. A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

Answer:

Cost of machine = Rs 15625 

Machine depreciates by 20% every year. 

Therefore, its value after every year is 80% of the original cost i.e., \(\frac45\) of the original cost.

∴ Value at the end of 5 years = 

= 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

105. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Answer:

Let x be the number of days in which 150 workers finish the work. 

According to the given information, 

150x = 150 + 146 + 142 + …. (x + 8) terms 

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common difference –4 and number of terms as (x + 8)

However, x cannot be negative. x = 17 

Therefore, originally, the number of days in which the work was completed is 17. 

Thus, required number of days = (17 + 8) = 25

106. Find the sum to infinity in each of the following Geometric Progression.

\(1,\frac13,\frac19,...\)

Answer:

Let S = \(1,\frac13,\frac19,...\) 

Here, a = 1 and r = \(\frac13\)

So,

Answer 2

107. Find the sum to infinity in each of the following Geometric Progression.

6, 1.2, 0.24 …

Answer:

Let S = 6 + 1.2 + 0.24 + ⋯ 

Here, a = 6 and r = 0.2 

So,

108. Find the sum to infinity in each of the following Geometric Progression.

\(5,\frac{20}{7},\frac{80}{49},...\)

Answer:

Let S = \(5,\frac{20}{7},\frac{80}{49},...\) 

Here, a = 5 and r = \(\frac43\) 

So,

109. Find the sum to infinity in each of the following Geometric Progression.

\(\frac{-3}{4},\frac3{16},\frac{-3}{64},...\)

Answer:

Let S = \(\frac{-3}{4},\frac3{16},\frac{-3}{64},...\) 

Here, a = \(\frac{-3}{4}\) and r = \(\frac{-1}4\)

So,

110. Prove that:

\(3^{\frac12}\times3^{\frac14}\times3^{\frac18}... = 3\)

Answer:

LHS = \(3^{\frac12}\times3^{\frac14}\times3^{\frac18}... \) 

111. Let x = 1 + a + a² + ⋯ and y = 1 + b + b² + ⋯, where |a| < 1 and |b| < 1. Prove that: 1 + ab + a²b² + ⋯ = \(\frac{xy}{x+y-1}\)

Answer:

Here,

And

From (1) and (2),