Important Questions for Class 11 Maths Chapter 9: Sequences and Series

Important Questions for Class 11 Maths Chapter 9: Sequences and Series are provided in the article. A sequence is an ordered list of items in mathematics. It has members, just like a set (also called elements or terms). The length of the sequence is defined as the number of ordered elements (potentially infinite). Unlike a set, order matters, and a phrase might appear many times in the sequence at different points. 

The video below explains this:

Sequence and Series Detailed Video Explanation:

Also check: NCERT Solutions for class 11 mathematics Chapter 9: Sequences and Series

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Very Short Answer Type [1 Mark]

Ques. If a, b, and c are all in G.P., then ax² + 2bx + c = 0 and dx² + 2ex + f = 0 share a common root if d/a, e/b, and f/c are all in G.P.

1. AP
2. GP
3. HP
4. none of these

Ans. (a) A.P

Solution:

Given that a, b, c are in GP (series)

⇒ b² = ac

⇒ b² – ac = 0

Hence, ax² + 2bx + c = 0 have equal roots.

Now the D = 4b² – 4ac

and the root of the above equation is -2b/2a = -b/a

So -b/a is the common root of the above quadratic equation.

Now,

dx² + 2ex + f = 0

⇒ d(-b/a)² + 2e×(-b/a) + f = 0

⇒ db² /a² – 2be/a + f = 0

⇒ d×ac /a² – 2be/a + f = 0

⇒ dc/a – 2be/a + f = 0

⇒ d/a – 2be/ac + f/c = 0

⇒ d/a + f/c = 2be/ac

⇒ d/a + f/c = 2be/b²

⇒ d/a + f/c = 2e/b

⇒ d/a, e/b, f/c are in Arithmetic Progression (AP).

Ques. If a, b, c are in AP , then

1. b = a + c
2. 2b = a + c
3. b² = a + c
4. 2b² = a + c

Ans. (b) 2b = a + c

Solution:

Given that the a, b, c are in AP, then

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

Ques. An increasing GP is formed by three numbers. The new numbers are in Ap if the middle term is doubled. The common ratio of GP is

Ans. (a) 2 + √3

Solution:

Let’s assume the three numbers be a/r, a, ar

Since the numbers form is increasing GP, So r > 1

Now, it is given that, the numbers a/r, 2a, ar are in AP

Hence,

⇒ 4a = a/r + ar

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± √3

⇒ r = 2 + √3 {Since r > 1}

Ques. The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is

1. n/(n+1)
2. 1/(n+1)
3. 1/n
4. None of these

Ans. (a) n/(n+1)

Solution: From the given question, the series can be defined as:

S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)

⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))

⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)

⇒ S = 1 – 1/(n+1)

⇒ S = (n + 1 – 1)/(n+1)

⇒ S = n/(n+1)

Ques. If the numbers 1/(b + c), 1/(c + a), 1/(a + b) are in AP Series, then

1. a, b, c are in AP
2. a², b², c² are in AP
3. 1/1, 1/b, 1/c are in AP
4. None of these

Ans. (b) a², b², c² are in AP

Solution:

From the given question, the numbers are , 1/(b + c), 1/(c + a), 1/(a + b)

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP (or AP Series).

Ques. The sum of the series 1/2! + 1/4! + 1/6! + ….. Is

1. e² – 1 / 2
2. (e – 1)² /2 e
3. e² – 1 / 2 e
4. e² – 2 / e

Ans. (b) (e – 1)² /2 e

Solution:

From the given series, we already know that,

e^x = 1 + x/1! + x² /2! + x³ /3! + x⁴ /4! + ………..

Now,

e^1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..

e^-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..

e^1 + e^-1 = 2(1 + 1/2! + 1/4! + ………..)

⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..

⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..

⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..

⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Also check:

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Short Answer Type [2 Marks]

Ques. A geometric progression's third term is 4. The result of multiplying the first five terms is

Ans: From the given question, it is given that T_3 = 4.

⇒ ar² = 4

Now, the product of the first five terms can be defined as = a.ar.ar².ar³.ar4, then

= a5r10

= (ar²)5

= 4^5

Hence, the product of the first five terms is 4^5

Ques. Let the Tr be the r th term of an A.P Series., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n is equivalent to

Ans: Let’s assume the first term is a has the common difference of d in the AP

Then, Tm = 1/n

⇒ a + (m-1)d = 1/n ………… (1)

and Tn = 1/m

⇒ a + (n-1)d = 1/m ………. (2)

From equation 2 – 1, we get

(m-1)d – (n-1)d = 1/n – 1/m

⇒ (m-n)d = (m-n)/mn

⇒ d = 1/mn

From equation (1), we get

a + (m-1)/mn = 1/n

⇒ a = 1/n – (m-1)/mn

⇒ a = {m – (m-1)}/mn

⇒ a = {m – m + 1)}/mn

⇒ a = 1/mn

Now, the Tmn = 1/mn + (mn-1)/mn

⇒ Tmn = 1/mn + 1 – 1/mn

⇒ Tmn = 1

Ques. The sum of two numbers is 13/6. Between them, an even number of mathematical means are placed, and their sum exceeds their number by one. The number of means inserted is then multiplied by two, then

Ans: Let’s take a and b are two numbers such that

a + b = 13/6

Let’s take A1, A2, A3, ………A2n be 2n arithmetic means between a and b

Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that, the series A1 + A2 + A3 + ………+ A2n = 2n + 1

⇒ 13n/6 = 2n + 1, then

⇒ n = 6

Ques. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in

Ans: From the given question, the equation is

ax² + bx + c = 0

Let p and q are the roots of the above equation.

Now p+q = -b/a

and pq = c/a

Given that, the p + q = 1/p² + 1/q², then

⇒ p + q = (p² + q²)/(p² ×q²)

⇒ p + q = {(p + q)² – 2pq}/(pq)²

⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²

⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}

⇒ -bc²/a³ = {b² – 2ca}/a²

⇒ -bc²/a = b² – 2ca

Divide both sides by bc, we get the following

⇒ -c /a = b/c – 2a/b

⇒ 2a/b = b/c + c/a

⇒ b/c, a/b, c/a are in AP

⇒ c/a, a/b, b/c are in AP

⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP

⇒ a/c, b/a, c/b are in Harmonic Progression

Ques. If the numbers 1/(b + c), 1/(c + a), 1/(a + b) are in AP then

Ans: From the given question, the numbers are 1/(b + c), 1/(c + a), 1/(a + b), then

⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)

⇒ 2b² = a² + c²

⇒ a², b², c² are in AP

Also check:

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Long Answer Type [3 Marks]

Ques. A GP's 5th, 8th, and 11th words are p, q, and s, respectively. Prove that (q^2) = ps.

Ans: As a result from the given question:

P is the 5th term.

q is the 8th term.

s = 11th term

To demonstrate this, ps = q2

We may express the equation as a5 = ar5-1 = ar4 = p using the information above.... —-(1)

a8 = ar8-1 = ar7 = q..... - (2)

a11 = ar11-1 = ar10 = s.... – (3)

When we divide (2) by (1), we get r3 = q/p... — (4)

When we divide (3) by (2), we get r3 = s/q… — (5)

When we add the equations (4) and (5) together, we get q/p = s/q.

q2 = ps is the result.

As a result, it has been proven.

Ques. Place five integers between 8 and 26 in such a way that the result is an A.P.

Ans: Let's assume A1, A2, A3, A4, and A5 are five numbers between 8 and 26, resulting in an A.P sequence of 8, A1, A2, A3, A4, A5, and 26.

Here, a=8, l=26, and n=5 are the values.

As a result, 26= 8+(7-1)d

It thus becomes,

26 = 8+6d

6d = 26-8 = 18

6d= 18

d = 3

A1= a+d = 8+ 3 =11

A2= a+2d = 8+ 2(3) =8+6 = 14

A3= a+3d = 8+ 3(3) =8+9 = 17

A4= a+4d = 8+ 4(3) =8+12 = 20

A5= a+5d = 8+ 5(3) =8+15 = 23

As a result, the required five digits between 8 and 26 are 11, 14, 17, 20, 23, and so on.

Ques. A total of 150 workers were hired to complete a job in a particular number of days. On the second day, four workers dropped out, and on the third day, four more workers dropped out, and so on. Find the number of days it took to finish the task and the number of days it took to finish the work.

Ans:  A = 150, d = -4

Sn= n2[2×150+(n−1)(−4)]

If the total number of workers who would have worked for the complete n days was 150, (n - 8)

∴n2[300 + (n - 1)( - 4)] = 150(n - 8)

⇒n = 25

Ques. Prove that the sum of n terms of the series 11 + 10^3 + 100^5 + .....is10/9(10^n - 1) + n^2.

Ans: Here, using the formula,

Sn = 11 + 10^3 + 100^5 + ...... + n terms

Sn = (10 + 1) + (10^2 + 3) + (10^3 + 5) + .... + (10^n+(2^n - 1))

Sn = 10(10^n - 1)10 - 1 + n^2(1 + 2^n - 1)

=10/9(10^n - 1) + n^2

Ques. Find the sum to n terms of the series 1² + (1² + (1² +2²) + (1² + 2² + 3²) + ------ 

Ans: 

From the formula, a_n = 1² + 2² + 3² + --- +n²

\(a_n = \frac{n(n+1)(2n+1)}{6}\)

\(S_n = \frac{1}{6} [ 2 \sum^n_{k-1} k^3 + 3 \sum^n_{k-1} k^2 + \sum k\)

\(= \frac{1}{6} [ 2. \frac{n^2(n+1)^2}{4} + \frac{3.n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}]\)

\(= \frac{n(n+1)}{12} [n(n+1 + (2n+1) + 1]\)

\(= \frac{n(n+1)}{12}(n^2 + n + 2n + 1 + 1)\)

\(=\frac{n(n+1)(n^2 + 3n + 2)}{12}\)

\(=\frac{n(n+1)^2(n+2)}{12}\)

This is the required solution.

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Very Long Answer Type [5 Marks]

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

Ans: Let the initial term and the common difference of the first and second arithmetic progressions be a_1, a_2 and d_1, d_2 correspondingly. Then

(5n+4)/(9n+6) = [(n/2)[2a1+ (n-1)d1] = (Sum of n terms of the first A.P)/(Sum of n terms of the second A.P)

[(n/2)[2a2+ (n-1)d2]] [(n/2)[2a2+ (n-1)d2] (5n + 4)/(9n + 6)

On L.H.S [2a1+ (n-1)d1], cancel out (n/2) both the numerator and denominator.

[2a2+ (n-1)d2]= (5n+4)/(9n+6) /[2a2+ (n-1)d2]= (5n+4)/(9n+6) /[2a2+ (n-1)] —-(1)

Substitute 35 for n in equation (1), because (n-1)/2 = 17.

[2a1+ 34d1]/[2a2+ 34d2]= (5(35)+4)/(9(35+6) [a1+ 17d1]/[a2+ 17d2]= 179/321...(2)

At this point, we can state that [a1+ 17d1]/[a2+ 17d2] = 18th term of first AP/18th term of second AP —-(3)

Using (2) and (3), we can deduce that the 18th term of the first AP divided by the 18th term of the second AP equals 179/321.

Ques. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Ans: Let a and d represent the A.P.'s initial term and common difference, respectively. It is well known that an A.P.'s kth term is given by ak = a +(k -1)d.

As a result, am+n = a +(m+n -1)d.

a + a + a + a + a + a + a + (m-n -1)

d am = a +(m-1)d...

As a result, the sum of an A.P's

(m + n)th and (m – n)th terms is expressed as am+n+ am-n = a +(m+n -1).

(m-n -1)d + d + a

= 2a +(m + n -1+ m – n – n – n – n – n – n – n – n – d \s

=2a+(2m-2) 2a + 2 = d (m-1) d

= two [a + (m-1)d] [a + (m-1)d] [a + (m-1)

2 am. [since am is equal to a +(m-1)d]

As a result, the sum of an A.P.'s (m + n)thand (m – n)th terms equals twice the mth term.

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

Ans: The divisible by 2 integers from 1 to 100 are 2, 4, 6,.... 100.

This results in an A.P. in which the first term and the common difference are both equal to 2.

⇒ 100=2+(n-1)2

⇒ n= 50

So, the sum of integers from 1 to 100 which are divisible by 2 is given as:

2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]

= (50/2)(4+98)

= 25(102)

= 2550

Divisible by 5, 10,.... 100 integers from 1 to 100.

This generates an A.P. with a common difference of 5 and a first term of 5.

Then, 100= 5+(n-1)5

⇒5n = 100

⇒ n= 100/5

⇒ n= 20

As a result, the total of all divisible by 2 integers from 1 to 100 is:

5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]

= (20/2)(10+95)

= 10(105)

= 1050

As a result, the divisible by both 2 and 5 integers from 1 to 100 are 10, 20,.... 100.

This also creates an Arithmetic Progression. because the first term and the common difference are both equal to ten.

Then, 100= 10+(n-1)10

⇒10n = 100

⇒ n= 100/10

⇒ n= 10

10+20+…+100= (10/2)[2(10)+(10-1)(10)]

= (10/2)(20+90)

= 5(110)

= 550

So, the required sum is:

= 2550+ 1050 – 550

= 3050

As a result, the sum of the divisible by 2 or 5 integers from 1 to 100 is 3050.

Ques. Between 1 and 31, m numbers have been introduced in such a way that the resulting sequence is an A.P., with a 7:(m – 1) ratio of 5:9. Calculate the value of m.

Ans: Let assume 1, A1, A2, …., Am, 31 are in A.P.

a = 1, an = 31

am+2 = 314

an = a + (n - 1)d

31 = a + (m + 2 - 1)d

d = 30m + 1

A7Am - 1 = 59(given)

⇒1 + 7(30m + 1)1 + (m - 1)(30m + 1) = 59

⇒m = 1

Ques. The ratio of A M and G. M of two positive numbers. a and b is m : n show that a:b = (m + \(\sqrt{m^2 - n^2}\) ) : (m – \(\sqrt{m^2 - n^2}\)

)

Ans: \(\frac{\frac{a+b}{2}}{\sqrt{ab}} = \frac{m}{n}\)

\(\frac{a+b}{2\sqrt{ab}} = \frac{m}{n}\) 

by C and D 

\(\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}\) = \(\frac{m+n}{m-n}\)

\(\frac{(\sqrt{a} + \sqrt{b})^2}{(\sqrt{a} - \sqrt{b})^2} = \frac{m+n}{m-n}\)

\(\frac{(\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{m+n}}{\sqrt{m-n}}\)

by C and D  \(\frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{m+n} + \sqrt{m-n}}{\sqrt{m+n} - \sqrt{m-n}}\)

Sq both side 

\(\frac{a}{b} = \frac{m+ \sqrt{m^2 -n^2}}{m- \sqrt{m^2 -n^2}}\)

Hence, this is the required solution.

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