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#### Chapters

Chapter 2: Relations and Functions

Chapter 3: Trigonometric Functions

Chapter 4: Principle of Mathematical Induction

Chapter 5: Complex Numbers and Quadratic Equations

Chapter 6: Linear Inequalities

Chapter 7: Permutations and Combinations

Chapter 8: Binomial Theorem

▶ Chapter 9: Sequences and Series

Chapter 10: Straight Lines

Chapter 11: Conic Sections

Chapter 12: Introduction to Three Dimensional Geometry

Chapter 13: Limits and Derivatives

Chapter 14: Mathematical Reasoning

Chapter 15: Statistics

Chapter 16: Probability

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## Solutions for Chapter 9: Sequences and Series

Below listed, you can find solutions for Chapter 9 of CBSE, Karnataka Board PUC NCERT for Mathematics Class 11.

### NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.1 [Pages 180 - 181]

Write the first five terms of the sequences whose n^{th} term is `a_n = n(n+2)`

Write the first five terms of the sequences whose n^{th} term is `a_n = n/(n + 1)`

Write the first five terms of the sequences whose n^{th} term is *a*_{n} = 2^{n}

Write the first five terms of the sequences whose *n*^{th} term is `a_n = (2n -3)/6`

Write the first five terms of the sequences whose *n*^{th} term is `a_n = (-1)^(n-1) 5^(n+1)`

Write the first five terms of the sequences whose *n*^{th} term is `a_n = n (n^2 + 5)/4`

Find the 17^{th} term in the following sequence whose *n*^{th} term is a_{n} = 4n – 3; `a_17 , a_24`

Find the indicated terms in each of the sequences whose n^{th }terms are `a_n = n^2/2^n`; `a_7`

Find the 9^{th} term in the following sequence whose *n*^{th} term is `a_n = (–1)^(n – 1) n^3; a_9`

Find the 20^{th} term in the following sequence whose *n*^{th} term is `a_n = (n(n-2))/(n+3)` ;`a_20`

Write the first five terms of the following sequence and obtain the corresponding series:

`a_1 = 3, a_n = 3a_( n-1) + 2 for all n > 1`

Write the first five terms of the following sequence and obtain the corresponding series: `a_1 = -1, a_n = (a_(n-1))/n , n >= 2`

Write the first five terms of the following sequence and obtain the corresponding series:

`a_1 = a_2 = 2, a_n = a_(n-1) -1, n > 2`

The Fibonacci sequence is defined by 1 = a_{1} = a_{2} and an = a_{n – 1} + a_{n – 2} , n > 2.

Find `a_(n+1)/a_n`, for n = 1, 2, 3, 4, 5

### NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.2 [Pages 185 - 186]

Find the sum of odd integers from 1 to 2001.

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20^{th} term is –112.

How many terms of the A.P. -6 , `-11/2` , -5... are needed to give the sum –25?

In an A.P., if *p*^{th} term is 1/q and qth term is 1/p, prove that the sum of first *pq *terms is 1/2 (pq + 1) where `p != q`

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term

Find the sum to *n* terms of the A.P., whose *k*^{th }term is 5*k* + 1.

If the sum of *n* terms of an A.P. is (*pn *+ *qn*^{2}), where *p* and *q* are constants, find the common difference.

The sums of *n* terms of two arithmetic progressions are in the ratio 5*n* + 4: 9*n* + 6. Find the ratio of their 18^{th} terms

If the sum of first *p* terms of an A.P. is equal to the sum of the first *q* terms, then find the sum of the first (*p* + *q*) terms.

Sum of the first *p, q* and *r* terms of an A.P. are *a, b* and *c*, respectively.

Prove that `a/p (q - r) + b/q (r- p) + c/r (p - q) = 0`

The ratio of the sums of *m* and *n* terms of an A.P. is *m*^{2}: *n*^{2}. Show that the ratio of *m*^{th }and *n*^{th}term is (2*m* – 1): (2*n* – 1)

If the sum of *n* terms of an A.P. is 3n^{2} + 5n and its* **m*^{th}^{ }term is 164, find the value of *m*.

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

if `(a^n + b^n)/(a^(n-1) + b^(n-1))` is the A.M. between *a* and *b*, then find the value of *n*.

Between 1 and 31, *m* numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^{th} and (*m* – 1)^{th} numbers is 5:9. Find the value of *m*.

A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30^{th} installment?

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

### NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.3 [Pages 192 - 193]

Find the 20^{th} and *n*^{th}terms of the G.P. `5/2, 5/4 , 5/8,...`

Find the 12^{th} term of a G.P. whose 8^{th} term is 192 and the common ratio is 2.

The 5^{th}, 8^{th} and 11^{th} terms of a G.P. are *p*, *q* and *s*, respectively. Show that *q*^{2} = *ps*.

The 4^{th} term of a G.P. is square of its second term, and the first term is –3. Determine its 7^{th}term.

Which term of the following sequences:

`2, 2sqrt2, 4,.... is 128`

Which term of the following sequences:

`sqrt3, 3, 3sqrt3, .... is 729`?

Which term of the following sequences:

`1/3, 1/9, 1/27, .... is 1/19683` ?

For what values of *x,* the numbers `2/7, x, -7/2` are in G.P?

Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Find the sum to *n* terms in the geometric progression `sqrt7, sqrt21,3sqrt7...`

Find the sum to *n* terms in the geometric progression 1, – a, a^{2}, – a^{3}, ... n terms (if a ≠ – 1).

Find the sum to *n* terms in the geometric progression x^{3}, x^{5}, x^{7}, ... n terms (if x ≠ ± 1).

Evaluate `sum_(k=1)^11 (2+3^k )`

The sum of first three terms of a G.P. is `39/10` and their product is 1. Find the common ratio and the terms.

How many terms of G.P. 3, 3^{2}, 3^{3}, … are needed to give the sum 120?

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to* n* terms of the G.P.

Given a G.P. with *a* = 729 and 7^{th} term 64, determine S_{7}.

Find a G.P. for which sum of the first two terms is – 4 and the fifth term is4 times the third term.

If the 4^{th}, 10^{th} and 16^{th} terms of a G.P. are *x, y *and *z*, respectively. Prove that *x*,* y*,* z *are in G.P.

Find the sum to* n* terms of the sequence, 8, 88, 888, 8888…

Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2

Show that the products of the corresponding terms of the sequences a, ar, ar^{2}, …ar^{n – 1} and A, AR, AR^{2}, … `AR^(n-1)`^{ }form a G.P, and find the common ratio

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4^{th} by 18.

If the `p^(th), q^(th) and r^(th)` terms of a G.P. are a, b and c, respectively. Prove that `a^(q - r) b^(r-p) c^(p-q) = 1`

If the first and the *n*^{th} term of a G.P. are *a* ad *b*, respectively, and if *P* is the product of *n *terms, prove that *P*^{2} = (*ab*)^{n}.

Show that the ratio of the sum of first *n* terms of a G.P. to the sum of terms from `(n + 1)^(th) " to "(2n)^(th) " term is " 1/r^n`.

If a, b, c and d are in G.P. show that (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2} .

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Find the value of* n* so that `(a^(n+1) + b^(n+1))/(a^n + b^n)` may be the geometric mean between *a* and *b*.

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio `3 + 2sqrt2) ":" (3 - 2sqrt2)`

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are `A+- sqrt((A+G)(A-G))`

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2^{nd} hour, 4^{th} hour and *n*^{th}hour?

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

### NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Exercise 9.4 [Page 196]

Find the sum to *n* terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Find the sum to *n* terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Find the sum to *n* terms of the series 3 × 1^{2} + 5 × 2^{2} + 7 × 3^{2} + …

Find the sum to *n* terms of the series `1/(1xx2) + 1/(2xx3)+1/(3xx4)+ ...`

Find the sum to *n* terms of the series 5^{2} + 6^{2} + 7^{2} + ... + 20^{2}

Find the sum to *n* terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Find the sum to *n* terms of the series 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + …

Find the sum to *n* terms of the series whose *n*^{th} term is given by *n* (*n* + 1) (*n* + 4).

Find the sum to n terms of the series whose n^{th} terms is given by n^{2} + 2^{n}

Find the sum to *n* terms of the series whose *n*^{th} terms is given by (2*n* – 1)^{2}

### NCERT solutions for Mathematics Class 11 Chapter 9 Sequences and Series Miscellaneous Exercise [Pages 199 - 200]

Show that the sum of (*m* + *n*)^{th} and (*m* – *n*)^{th} terms of an A.P. is equal to twice the *m*^{t}^{h} term.

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the sum of *n*, 2*n*, 3*n* terms of an A.P. be S_{1}, S_{2} and S_{3},_{ }respectively, show that S_{3} = 3 (S_{2}– S_{1})

Find the sum of all numbers between 200 and 400 which are divisible by 7.

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

If f is a function satisfying f (x +y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and `sum_(x = 1)^n` f(x) = 120, find the value of n.

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

if `(a+ bx)/(a - bx) = (b +cx)/(b - cx) = (c + dx)/(c- dx) (x != 0)` then show that *a*, *b*, *c* and *d* are in G.P.

Let S be the sum, P the product and R the sum of reciprocals of *n* terms in a G.P. Prove that P^{2}R^{n}^{ }= S^{n}

The *p*^{th}, *q*^{th} and *r*^{th} terms of an A.P. are *a, b, c* respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0

if `a(1/b + 1/c), b(1/c+1/a), c(1/a+1/b)` are in A.P., prove that *a, b, c* are in A.P.

If a, b, c, d are in G.P, prove that (a^{n} + b^{n}), (b^{n} + c^{n}), (c^{n} + d^{n}) are in G.P.

If *a* and *b *are the roots of are roots of x^{2} – 3x + p = 0 , and c, d are roots of x^{2} – 12x + q = 0, where *a*,* b*,* c*, *d*, form a G.P. Prove that (*q* + *p*): (*q* – *p*) = 17:15.

The ratio of the A.M and G.M. of two positive numbers *a* and *b*, is *m*:* n*. Show that `a:b = (m + sqrt(n^2 - n^2)):(m - sqrt(m^2 - n^2))`.

If *a, b, c* are in A.P*,*;* b, c, d* are in G.P and ` 1/c, 1/d,1/e` are in A.P. prove that *a*,* c*,* e* are in G.P.

Find the sum of the following series up to *n* terms:

5 + 55 + 555 + …

Find the sum of the following series up to *n* terms:

.6 +.66 +. 666 +…

Find the 20^{th} term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + *n* terms.

Find the sum of the first *n* terms of the series: 3 + 7 + 13 + 21 + 31 + …

If S_{1}, S_{2}, S_{3} are the sum of first *n* natural numbers, their squares and their cubes, respectively, show that `9S_2^2 = S_3(1 + 8S_1)`

Find the sum of the following series up to *n* terms `1^3/1 + (1^3 + 2^3)/(1+3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) +...`

Show that `(1xx2^2 + 2xx3^2 + ...+nxx(n+1)^2)/(1^2 xx 2 + 2^2 xx3 + ... + n^2xx (n+1))` = `(3n + 5)/(3n + 1)`

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?

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?

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.

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.

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.

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.

## Solutions for Chapter 9: Sequences and Series

## NCERT solutions for Mathematics Class 11 chapter 9 - Sequences and Series

Shaalaa.com has the CBSE, Karnataka Board PUC Mathematics Mathematics Class 11 CBSE, Karnataka Board PUC solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Mathematics Class 11 CBSE, Karnataka Board PUC 9 (Sequences and Series) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

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Concepts covered in Mathematics Class 11 chapter 9 Sequences and Series are Sum to N Terms of Special Series, Introduction of Sequence and Series, Concept of Sequences, Concept of Series, Arithmetic Progression (A.P.), Geometric Progression (G. P.), Relationship Between A.M. and G.M..

Using NCERT Mathematics Class 11 solutions Sequences and Series exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Mathematics Class 11 students prefer NCERT Textbook Solutions to score more in exams.

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