NCERT Solutions of Class 11 Maths Exercise 9.3 -Sequences and Series

NCERT Solutions of Class 11 Maths Exercise 9.3 of the chapter 9-Sequences and Series are created here by a step by step method by an expert teacher of maths.These NCERT solutions of of exercise 9.3 are the solutions of chapter 9-Sequences and series of class 11 NCERT text book. All questions of exercise are based on Geometric Progression.If you Study these NCERT solutions till the end you will clear your concept that enable you in solving all questions based on the Geometric Progressions in maths question paper of class 11 CBSE board exam of 2020-21.

Class 11 maths NCERT Solutions of chapter 9-Sequences and Series

exercise 9.1 -Sequence and Series

Exercise 9.2- Sequences and Series

Exercise 9.4-Sequence and Series

NCERT Solutions of Class 11 Maths Exercise 9.3 of the chapter 9-Sequences and Series

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Q1. Find the 20^thand n^thterms of the G.P. 5/2, 5/4,5/8…..

Ans. The given G.P is 5/2, 5/4, 5/8…..

In the G.P. first term, a = 5/2

Common ratio, r

$\boldsymbol{r=\frac{5}{4}\div \frac{5}{2}=\frac{1}{2}}$

n^thterms of the G.P,a_n =ar^n-1

$\boldsymbol{a_{20}=\frac{5}{2}\left ( \frac{1}{2} \right )^{20-1}}$

$\boldsymbol{=\frac{5}{2^{20}}=5\times 2^{-20}}$

$\boldsymbol{a_{n}=\frac{5}{2}\left ( \frac{1}{2} \right )^{n-1}}$

$\boldsymbol{=\frac{5}{2^{n}}}$

Hence 20^thand n^thterms of the G.P are 5 × 2^-20 and 5/2ⁿrespectively

Q2. Find the 12^thterms of a G.P. whose 8^thterm is 192 and the common ratio is 2.

Ans. We are given 8^thterm(a₈) is 192 and the common ratio,r is 2

n^thterms of the G.P,a_n =ar^n-1

$\boldsymbol{a_{8}=a\left ( 2 \right )^{8-1}}$

$\boldsymbol{a\left ( 2 \right )^{8-1}=192}$

$\boldsymbol{a=\frac{192}{2^{7}}}$

$\boldsymbol{a=\frac{3}{2}}$

$\boldsymbol{a_{12}=\frac{3}{2}\left ( 2 \right )^{12-1}}$

$\boldsymbol{=\frac{3}{2}\left ( 2 \right )^{11}}$

$\boldsymbol{=3\times 2^{10}}$

= 3×1024 =3072

Q3.The 5^thterm, 8^thterm, and 11^thterms of a G.P are p, q, and s, respectively. Show that q² = ps.

Ans. According to question

5^thterm = p, 8^thterm = q and 11^thterms = s

The n^thterm of G.P

$\boldsymbol{a_{n}=ar^{n-1}}$

$\boldsymbol{p=ar^{4}....\left ( i \right )}\boldsymbol{,\: q=ar^{7}....\left ( ii \right ),\: s=ar^{10....\left ( iii \right )}}$

Squaring both sides of equation (ii)

$\boldsymbol{q^{2}=\left ( ar^{7} \right )^{2}}$

$\boldsymbol{q^{2}=a^{2}r^{14}}$

$\boldsymbol{q^{2}=\left ( ar^{4} \right )\left ( ar^{10} \right )}$

$\boldsymbol{q^{2}=ps}$

Hence proved

Q4. The 4^thterm of a G.P is square of its second term , and the first term is -3. Determine its 7^thterm.

Ans. We are given the first term, of G.P, a = -3

n^thterm of G.P,a_n = ar^n-1

The fourth term and and second term of the G.P is given by the (i) and (ii) equation

$\boldsymbol{a_{4}= -3r^{3}...\left ( i \right ),\: a_{2}=-3r...\left ( ii \right )}$

According to question

$\boldsymbol{a_{4}=\left ( a_{2} \right )^{2}}$

$\boldsymbol{-3r^{3}=\left ( -3r \right )^{2}}$

$\boldsymbol{-3r^{3}=9r^{2}}$

r = -3

The seventh term of G.P is given by

$\boldsymbol{a_{7}=-3\left ( -3 \right )^{6}}$

$\boldsymbol{=\left ( -3 \right )^{7}}$

= -2187

Therefore the 7^thterm of the series is -2187

Q5.Which term of the following sequences :

(a) 2, 2√2, 4……. is 128? (b) √3, 3, 3√3…..is 729

$\boldsymbol{\left ( c \right )\frac{1}{3},\frac{1}{9},\frac{1}{27}....is\frac{1}{19683}?}$

Ans.(a) The given G.P is 2, 2√2, 4……. is 128

Common ratio ,r of the G.P is = 2√2/2 = √2 and the first term is ,a = 2

n^thterm of G.P,a_n = ar^n-1

$\boldsymbol{2\left ( \sqrt{2} \right )^{n-1}=128}$

$\boldsymbol{2\times 2^{\left ( n-1 \right )/2}=128}$

$\boldsymbol{128=2^{1+\left ( n-1 \right )/2}}$

$\boldsymbol{2^{\left ( n-1 \right )/2}=128}$

$\boldsymbol{2^{\left ( n-1 \right )/2}=2^{7}}$

$\boldsymbol{\frac{n-1}{2}=7}$

n = 15

Therefore 15^thterm of G.P is 1s 128

(b) The given G.P is √3, 3, 3√3…..is 729

Common ratio ,r of the G.P is = 3/√3 = √3 and the first term is ,a = √3

n^thterm of G.P,a_n = ar^n-1

$\boldsymbol{\sqrt{3}\left ( \sqrt{3} \right )^{n-1}=729}$

$\boldsymbol{3^{1/2}.3^{\left ( n-1 \right )/2}=729}$

$\boldsymbol{3^{n/2}=3^{6}}$

$\boldsymbol{\frac{n}{2}=6}$

n = 12

Therefore 12^thterm of G.P is 729

$\boldsymbol{\left ( c \right )\frac{1}{3},\frac{1}{9},\frac{1}{27}....is\frac{1}{19683}?}$

The first term of the G.P

$\boldsymbol{a=\frac{1}{3}}$

Common ratio, r

$\boldsymbol{r=\frac{1}{9}\div \frac{1}{3}=\frac{1}{3}}$

n^thterm of G.P,a_n = ar^n-1

$\boldsymbol{\frac{1}{3}\left ( \frac{1}{3} \right )^{n-1}=\frac{1}{19683}}$

$\boldsymbol{\left ( \frac{1}{3} \right )^{n-1}=\left ( \frac{1}{3} \right )^{9}}$

n – 1 = 9

n = 10

There for 10^thterm of G.P is $\boldsymbol{\frac{1}{19683}}$

Q6. For what values of x, the numbers $\boldsymbol{-\frac{2}{7},x,-\frac{7}{2}}$ are in G.P ?

Ans. Since numbers are in G.P, therefore

$\boldsymbol{x\div -\frac{2}{7}=-\frac{7}{2}\div x}$

$\boldsymbol{-\frac{7x}{2}=-\frac{7}{2x}}$

14x² = 14

x = ±1

Hence for x = ±1, the given numbers are in G.P

Find the sum to indicated number of terms in each of the geometric progression in exercises 7 to 10.

Q7.0.15, 0.015, 0.0015…. 20 terms.

Ans. We are given the G.P 0.15, 0.015, 0.0015…. 20 terms

In the given G.P ,a = 0.15

The common ratio, r

$\boldsymbol{r=\frac{0.15}{0.015}=\frac{1}{10}}$

In case r< 1,the sum of the G.P is given by

$\boldsymbol{S_{n}=\frac{a\left ( 1-r^{n} \right )}{1-r}}$

$\boldsymbol{S_{20}=\frac{0.15\left [ 1-\left ( 1/10 \right )^{20} \right ]}{1-1/10}}$

$\boldsymbol{=\frac{1.5}{9}\left ( 1-0.1^{20} \right )}$

$\boldsymbol{=\frac{1}{6}\left ( 1-0.1^{20} \right )}$

Hence the required sum of the G.P is $\boldsymbol{\frac{1}{6}\left ( 1-0.1^{20} \right )}$

Q8.√7, √21, 3√7………n terms.

Ans. The given G.P is √7, √21, 3√7………n terms.

In the G.P ,a = √7 and common ratio, r = √21÷√7 = √3

Since, r > 1 so, the sum of the G.P is given by

$\boldsymbol{S_{n}=\frac{a\left [ r^{n} -1\right ]}{r-1}}$

$\boldsymbol{=\frac{\sqrt{7}\left ( (\sqrt{3})^{n}-1 \right )}{\sqrt{3}-1}}$

$\boldsymbol{=\frac{\sqrt{7}\left ( 3^{n/2}-1 \right )\left ( \sqrt{3} +1\right )}{\left ( \sqrt{3} -1\right )\left ( \sqrt{3} +1\right )}}$

$\boldsymbol{=\frac{\sqrt{7}}{2}\left ( \sqrt{3} +1\right )\left ( 3^{n/2} -1\right )}$

Q9.Find the sum to n terms in the geometric progression 1, -a, a², -a³ …. (if a ≠ -1)

Ans.The given series is 1, -a, a², -a³ …. (if a ≠ -1)

Common ratio,r = -a/1=-a and first term,A=1

Since Since r < 1, therefore the sum of n terms is given as

$\boldsymbol{S_{n}=\frac{A\left ( 1-r^{n} \right )}{1-r}}$

$\boldsymbol{=\frac{1}{1+a}\left [ 1-\left ( -a \right )^{n} \right ]}$

Q10.Find the sum to n terms in the geometric progression x³, x⁵, x⁷, … (if x ≠ ±1 )

Ans.First-term, a =x³, common ratio,r = x⁵/x³=x²

Since r <1 therefore the sum of n terms is given as

$\boldsymbol{S_{n}=\frac{a\left ( 1-r^{n} \right )}{1-r}}$

$\boldsymbol{S_{n}=\frac{x^{3}\left [ 1-\left ( x^{2} \right )^{n} \right ]}{1-x^{2}}}$

$\boldsymbol{=\frac{x^{3}}{1-x^{2}}\left ( 1-x^{2n} \right )}$

$\boldsymbol{Q11.Evaluate\: \sum_{k=1}^{11}\left ( 2+3^{k} \right )}$

Ans. The sum given to us defined as follows

$\boldsymbol{ \sum_{k=1}^{11}\left ( 2+3^{k} \right )}$

Expanding the sum by putting k=1.2.3…..11

⇒(2 +3) + (2 + 3²) +(2+ 3³) +…….(2 + 3¹¹)

=(2+2+2….up to 11 terms)+(3 +3² +3³ +……3¹¹)

= 2×11 + (3 +3² +3³ +……3¹¹)

=22+ (3 +3² +3³ +……3¹¹)….(i)

3 +3² +3³ +……3¹¹ is a G.P

Where a = 3, r= 3²/3 = 3

Since r > 1, therefore the sum of n terms is given as

$\boldsymbol{S_{n}=\frac{a\left ( r^{n}-1 \right )}{r-1}}$

$\boldsymbol{S_{11}=\frac{3\left ( 3^{11}-1 \right )}{3-1}}$

$\boldsymbol{=\frac{3}{2}\left ( 3^{11}-1 \right )}$

Substituting this sum in (i), we get the total required sum

$\boldsymbol{S_{11}=22+\frac{3}{2}\left ( 3^{11} -1\right )}$

Q12. The sum of the first three terms of a G.P is 39/10 and their product is 1. Find the common ratio and the terms.

Ans. Let the first term of the G.P is a and the common ratio of the sequence is r

Therefore the first three terms are = a, ar and ar²

According to first condition of the question

$\boldsymbol{a+ar+ar^{2}=\frac{39}{10}}$

$\boldsymbol{a(1+r+r^{2})=\frac{39}{10}....\left ( i \right )}$

According to second condition of the question

$\boldsymbol{a\times ar\times ar^{2}=1}$

(ar)³ = 1

ar =1

$\boldsymbol{a=\frac{1}{r}}$

Substituting the value of a in equation (i)

$\boldsymbol{\frac{1}{r}\left ( 1+r+r^{2} \right )=\frac{39}{10}}$

39r = 10 +10r +10r²

10r² – 29r +10 = 0

10r² – 25 r-4r +10 =0

5r(2r -5) -2(2r – 5) = 0

(2r -5)(5r -2) = 0

r = 5/2, r = 2/5

If r = 5/2 then a = 2/5 ( a = 1/r)

ar = (2/5)(5/2) =1

ar² = (2/5)(5/2)²= 5/2

When r =2/5, then a = 5/2

ar =(5/2)(2/5) =1²

ar² = (5/2)(2/5)²= 2/5

Hence the required terms are 2/5, 1 and 5/2 or 5/2,1 and 2/5.

Q13.How many terms of G.P. 3, 3^², 3^³, … are needed to give the sum 120?

Ans. The given series is 3, 3^², 3^³, …

Where, a = 2, r= 3²/3 = 3

Let the sum of n terms is = 120

Since r > 1, therefore the sum of n terms is given as

$\boldsymbol{S_{n}=\frac{a\left ( r^{n} -1\right )}{r-1}}$

$\boldsymbol{120=\frac{3\left ( 3^{n}-1\right )}{3-1}}$

$\boldsymbol{120=\frac{3\left ( 3^{n}-1\right )}{2}}$

$\boldsymbol{240=3^{n+1}-3}$

$\boldsymbol{3^{n+1}=243}$

$\boldsymbol{3^{n+1}=3^{5}}$

Comparing the exponents of both sides

n + 1 = 5

n = 5-1 = 4

Therefore 4 terms of the given G.P give the sum 120.

Q14.The sum of first three terms of a G.P is 16 and the sum of next three terms is 128. Determine the first term, the common ratio, and the sum to n terms of the G.P.

Ans.Let the first term of the G.P is a and the common ratio is r

Therefore G.P is a, ar,ar²,ar³……..

According to first condition of question

Sum of first three terms = 16

a+ ar+ ar² =16

a(1 + r +r²) = 16…….(i)

According to second condition of question

The sum of next three terms(4^th, 5^th, 6^thterms)

ar³+ar⁴+ar⁵= 128

ar³( 1 + r + r²) = 128……(ii)

Dividing equation (i) by equation (ii)

$\boldsymbol{\frac{a\left ( 1+r+r^{2} \right )}{ar^{3}\left ( 1+r+r^{2} \right )}=\frac{16}{128}}$

r³ = 8

r = 2

Putting the value of r =2 in equation (i)

a(1 + r +r²) = 16

a(1 + 2 + 2²) = 16

a(7) = 16

a = 16/7

$\boldsymbol{S_{n}=\frac{a\left ( r^{n} -1\right )}{r-1}}$

$\boldsymbol{=\frac{16/7\left ( 2^{n} -1\right )}{2-1}}$

$\boldsymbol{=\frac{16}{7}\left ( 2^{n} -1\right )}$

Therefore the first term of the sequence is 16/7, common ratio is 2 and sum of n terms is 16/7(2ⁿ-1).

Q15.Given a G.P. with a = 729 and 7^th term 64, determine S₇

Ans.We are given a G.P in which a = 729, 7^thterm =a₇=64

a_n=ar^n-1

a₇=ar^7-1

64= (729)r⁶

$\boldsymbol{r^{6}=\frac{64}{729}}$

$\boldsymbol{r^{6}=\left ( \frac{2}{3} \right )^{6}}$

$\boldsymbol{r=\frac{2}{3}}$

$\boldsymbol{S_{n}=\frac{a\left ( 1-r^{n} \right )}{1-r}}$

Putting the value n = 7,a=729 and r= 2/3

$\boldsymbol{S_{7}=\frac{729\left [ 1-\left ( 2/3 \right )^{7} \right ]}{1-2/3}}$

$\boldsymbol{=\frac{729\times 3\left ( 3^{7} -2^{7}\right )}{3^{7}}}$

$\boldsymbol{=\frac{3^{7}\left ( 3^{7} -2^{7}\right )}{3^{7}}}$

$\boldsymbol{=3^{7}-2^{7}}$

S₇=2187 – 128 = 2059

Q16. Find a GP for which the sum of the first two terms is -4 and fifth term is 4 times the third term.

Ans. Let a , r be the first term and common ratio of the GP respectively

First term of GP = a and second term is = ar

according to first condition of the question

a + ar = -4

a(1 + r) = -4……(i)

According to second condition of the question

fifth term = 4× the third term

ar^5-1 = 4× ar^3-1)

ar⁴ = 4ar²

r² = 4

r = ±2

If r = 2, then putting the value of r=2 in equation (i)

a(1 + 2) = -4

a = -4/3

So, the GP is -4/3, (-4/3) × 2, (-4/3)×2²…..= -4/3, -8/3, -16/3……

If r = -2, then putting the value of r= -2 in equation (i)

a(1 -2) = -4

-a = -4

a = 4

So, the GP is 4, 4×-2, -4×(-2)²…..,= 4, -8, -16……..

Hence the required GP is either -4/3, -8/3, -16/3……or 4, -8, -16……..

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