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Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.3 | Set 1
  • Last Updated : 19 Apr, 2021

Question 1.  Find the 20th and nth terms of the G.P 5/2, 5/4, 5/8, …

Solution:

According to the question

G.P: 5/2, 5/4, 5/8, …

So, first term(a) = 5/2

So, the common ratio(r) = \frac{a_1}{a}=\frac{\frac54}{\frac52}= \frac12    



Find: 20th and nth terms of the given G.P

So, the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio.

Now we find the 20th terms of the given G.P:

a20 = (5/2)(1/2)20-1 = (5/2)(1/2)19\frac{5}{2^{20}}

Find the nth terms of the given G.P:

an \frac52\frac1{2^{n-1}}=\frac5{2^{n}}



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

Solution:

According to the question

Common ration(r) = 2

and 8th term is 192

So let us considered a be the first term

So, 

a8 = ar7

ar7 = 192

a(2)7 = 192

a = 3/2

Find: 12th term of a G.P.

As we know that the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio

So, we find 12th term of a G.P. 

 a12 = ar12 – 1

 a12 = ar11

 a12 = (3/2)(2)11

a12 = 3072

Question 3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Solution:

According to the question

The 5th, 8th and 11th terms of a G.P. are p, q and s

Prove: q2 = ps

Proof:

 Let a G.P. with first term a and common ratio r,

So, a5 = ar4 = p  ….(1)

a8 = ar7= q  ….(2)

a11 = ar10 = s  ….(3)

Now divide eq(2) by (1), we get

ar7/ar4 = q/p

r3 = q/p  ….(4)

Now divide eq(3) by (2), we get

ar10/ar7= s/q

r3 = s/q  ….(5)

From eq(4) and (5), we get

q/p = s/q

q2 = ps

Hence Proved

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

Solution:

According to the question

First term(a) = – 3

and the 4th term of a G.P. is square of its second term

Find: 7th term

Let us considered r be the common ratio 

As we know that the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio

So, a4 = ar3

It is given that the 4th term of a G.P. is square of its second term

a4 = (a2)2

ar3 = (a2)2

ar3 = (ar)2

ar3 = a2r2

r = a

Now put the value of a, we get

r = -3

Now, we find the 7th term

a7 = ar7 – 1  

a7 = ar6  

= −3 x (−3)6 =−2187

Question 5. Which term of the following sequences:

(a) 2, 2√2, 4, … is 128 ?

Solution:

According to the question

G.P.: 2, 2√2, 4, …

So, first term(a) = 2

So, the common ratio(r) = √2

As we know that, the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio.

128 = 2(√2)n – 1

(2)7 = 2(√2)n – 1 

(2)7 = 2((2)1/2)n – 1 

(2)7 = 2(2)(n – 1)/2

(2)6 = (2)(n – 1)/2

6 = (n – 1)/2

12 = n – 1

12 + 1 = n

n = 13

Hence, the 13th term of the G.P. is 128

(b) √3, 3, 3√3, …. is 729 ?

Solution:

According to the question

G.P.: √3, 3, 3√3, …. 

So, first term(a) = √3

So, the common ratio(r) = √3

As we know that, the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio.

729 = √3(√3)n-1

(3)6 = √3(√3)n-1

(3)6 = (3)1/2((3)1/2)n-1

(3)6 = (3)1/2(3)n-1/2

(3)6 = (3)1/2+(n-1)/2

6 = 1/2 + (n – 1)/2

6 – 1/2 = (n – 1)/2

(12 – 1)/2 = (n – 1)/2

11 = n – 1

n = 12

Hence, the 12th term of the G.P. is 729

(c) \frac13,\frac19,\frac1{27},... is \frac1{19683} ?

Solution:

According to the question

G.P.:  \frac13,\frac19,\frac1{27},...

So, first term(a) = 1/3

So, the common ratio(r) = 1/3

As we know that, the nth term of G.P can be expressed using formula:

an = arn – 1  

Where a is 1st term and r is the common ratio.

\frac1{19683}=(\frac13)(\frac13)^{n-1}

(1/3)9 = (1/3)n

n = 9

Hence, the 9th term of the G.P. is 1/19683

Question 6.   For what values of x, the numbers -2/7, x. -7/2… are in G.P.? 

Solution:

According to the question

Numbers are -2/7, x. -7/2…

The common ration is (r) = \frac{x}{\frac{-2}{7}} = -7x/2

Again the common ration(r) = = \frac{\frac{-7}{2}}{x} = -7/2x

So, 

-7x/2 = -7/2x

7x/2 = 2/7x

14x2 = 14

x2 = 14/14

x = ±1

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10: 

Question 7. 0.15, 0.015, 0.0015, … 20 terms.

Solution:

According to the question

G.P.: 0.15, 0.015, 0.0015, … 

So, first term(a) = 0.15

So, the common ratio(r) = 0.1

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

So, S20 = (0.15)[(1 – (0.1)20)/(1 – 0.1)]

= (0.15/0.9)(1 – (0.1)20)

= 1/6(1 – (0.1)20)

Question 8. √7​, 21​, 37​,…  n terms.

Solution:

According to the question

G.P.: √7​, 21​, 37​,…

So, first term(a) = √7​

So, the common ratio(r) = √3

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

So, Sn = (√7​)[(1 – (√3)n)/(1 – √3)]

= (√7​)[(1 – (√3)n)/(1 – √3)] x [(1 + √3)/(1 + √3)]

= -(√7​)(1 + √3)/2(1 – (3)n/2)

= (√7​)(1 + √3)/2((3)n/2 – 1)

Question 9.  1, -a, a2, -a3,… n terms (if a ≠ – 1).

Solution:

According to the question

G.P.: 1, -a, a2, -a3,… 

So, first term(a) = 1

So, the common ratio(r) = -a

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

So, S_n=\frac{(1)(1-(-a)^n)}{1 -(-a)}=\frac{1-(-a)^n}{1+a}

Question 10.  x3, x5, x7, … n terms (if x ≠ ± 1). 

Solution:

According to the question

G.P.: x3, x5, x7, …

So, first term(a) = x3

So, the common ratio(r) = x2

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

So,

S_n=\frac{(x^3)(1-(x^2)^n)}{1-x^2}

S_n=\frac{(x^3)(1-x^{2n})}{1-x^2}

Question 11. Evaluate \sum_{k=1}^{11} (2+3^k)

Solution:

\sum_{k=1}^{11} (2+3^k) = (2 + 31) + (2 + 32) + (2 + 33) + …. + (2 + 311)

= (2 + 2 + …11 terms) + (3 + 32 + 33+…11 terms)

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

So, Sn = 2 x 11 + 3(311 – 1)/3 – 1 

Sn = 22 + 3/2((311 – 1))

Question 12.  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.

Solution:

Let us considered the three terms a/r, a, ar of the G.P. 

According to the question 

\frac{a}r\times a\times ar = 1 

a3 = 1

a = 1

Also. 

\frac{a}{r}+a+ar=\frac{39}{10}     

Now put the value of a, we get

\frac1r+1+r=\frac{39}{10}     

 \frac{r^2+r+1}r=\frac{39}{10}

10r2 + 10r + 10 = 39r

10r2 – 29r + 10 = 0

On solving the equation, we get 

r = 2/5, 5/2

So, the G.P. is 5/2, 1, 2/5.

Question 13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Solution:

According to the question

G.P.: 3, 32, 33, …

So, first term(a) = 3

So, the common ratio(r) = 3

Sn = ​120

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(1 - r^n)}{1-r}

Sn = (3)(1 – (3)n)(1-3)

120 = (3)(1 – (3)n)

-240 = (3)(1 – (3)n)

-80 = 1 – (3)n

-80 – 1 = – (3)n

-81 = – (3)n

(3)4 = (3)n

n = 4   

Hence, 4 terms of G.P. are needed to give the sum 120

Question 14. 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. 

Solution:

Let us considered the first three terms of the G.P. are a, ar, ar2 and

first term of G.P. be a and common ratio r.

Find: the first term, the common ratio, and the sum to n terms of the G.P. 

According to the question 

 a + ar + ar2 = 16   

a(1 + r + r2) = 16 …….(1)

 ar3 + ar4 + ar5 = 128 

ar3(1 + r + r2) = 128  …….(2)

Now divide eq(2) by (1), we get

ar3(1 + r + r2)/ a(1 + r + r2) = 128/16

r3 = 8

r = 2

Now put the value of r in eq(1), we get

a(1 + (2) + (2)2) = 16

a(1 + 2 + 4) = 16

a(7) = 16

a = 16/7

Now we find the sum to n terms of the G.P. 

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(r^n-1)}{r-1}

S_n=\frac{16(2^n-1)}{7(2-1)}=\frac{16(2^n-1)}7

Question 15. Given a G.P. with a = 729 and 7th term 64, determine S7

Solution:

According to the question 

The first term(a) = 729

and 7th term 64

Find: S7

Let the common ratio be r.

a7 = ar6 = 64

r6 = 64/729

r = ±2/3

Now we find the sum to 7th terms of the G.P. 

As we know that, the sum of n terms of G.P. with 1st term a & common ratio r is given by  

S_n=\frac{a(r^n-1)}{r-1}

On taking r = +2/3   

S_7=\frac{729((\frac23)^7-1)}{\frac23-1} = 2059

On taking r = -2/3  

S_7=\frac{729((-\frac23)^7-1)}{(-\frac23-1)} = 463

Question 16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Solution:

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

According to the question 

a + ar = -4   

a5 = 4(a3)

So, ar4 = 4(ar2)  

r2 = 4 

r = ±2 

When r = +2 then 

a + ar = a + 2a = 3a = −4

 a =  -4/3

Then the G.P. is \frac{-4}3,\frac{-8}3,\frac{16}3,...

When r = -2 then 

a + ar = a − 2a = −a = −4 

a = 4

Then the G.P. is 4, -8, 16, -32,…

 

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