A sequence, in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity, is called a Geometric Progression. So, this chapter is going to be completely about G.P., its general term, properties and the sum of terms in a G.P. Students who find it difficult to solve problems in this chapter can refer toÂ Selina Solutions for Class 10 Mathematics prepared by our experienced faculty at BYJUâ€™S. The solutions are prepared with an aim to boost confidence among students for taking up their Class 10 final exams fearlessly. This mainly improves problem-solving skills of the students, which are vital from an examination point of view. Selina Solutions for Class 10 Mathematics Chapter 11 Geometric Progression PDF are available right below.

## Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression Download PDF

## Access Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

### Exercise 11(A) Page No: 152

**1. Find which of the following sequence form a G.P.:**

**(i) 8, 24, 72, 216, â€¦â€¦â€¦**

**(ii) 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦**

**(iii) 9, 12, 16, 24, â€¦â€¦â€¦**

**Solution: **

(i) Given sequence: 8, 24, 72, 216, â€¦â€¦â€¦

Since,

24/8 = 3, 72/24 = 3, 216/72 = 3

â‡’ 24/8 = 72/24 = 216/72 = â€¦â€¦â€¦.. = 3

Therefore 8, 24, 72, 216, â€¦â€¦â€¦ is a G.P. with a common ratio 3.

(ii) Given sequence: 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦

Since,

(1/24)/ (1/8) = 1/3, (1/72)/ (1/24) = 1/3, (1/216)/ (1/72) = 1/3

â‡’ (1/24)/ (1/8) = (1/72)/ (1/24) = (1/216)/ (1/72) = â€¦â€¦â€¦.. = 1/3

Therefore 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦ is a G.P. with a common ratio 1/3.

(iii) Given sequence: 9, 12, 16, 24, â€¦â€¦â€¦

Since,

12/9 = 4/3; 16/12 = 4/3; 24/16 = 3/2

12/9 = 16/12 â‰ 24/16

Therefore, 9, 12, 16, 24 â€¦â€¦ is not a G.P.

**2. Find the 9 ^{th} term of the series: 1, 4, 16, 64, â€¦..**

**Solution: **

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = 4/1 = 4

We know that, the general term is

t_{n} = ar^{n â€“ 1}

Thus,

t_{9} = (1)(4)^{9 â€“ 1} = 4^{8 }= 65536

**3. Find the seventh term of the G.P: 1, âˆš3, 3, 3 âˆš3, â€¦..**

**Solution: **

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = âˆš3/1 = âˆš3

We know that, the general term is

t_{n} = ar^{n â€“ 1}

Thus,

t_{7} = (1)(âˆš3)^{7 â€“ 1} = (âˆš3)^{6 }= 27

**4. Find the 8 ^{th} term of the sequence: **

**Solution: **

The given sequence can be rewritten as,

3/4, 3/2, 3, â€¦..

Itâ€™s seen that, the first term is (a) = 3/4

And, common ratio(r) = (3/2)/ (3/4) = 2

We know that, the general term is

t_{n} = ar^{n â€“ 1}

Thus,

t_{8} = (3/4)(2)^{8 â€“ 1} = (3/4)(2)^{7 }= 3 x 2^{5} = 3 x 32 = 96

**5. Find the 10 ^{th} term of the G.P. : **

**Solution: **

The given sequence can be rewritten as,

12, 4, 4/3, â€¦..

Itâ€™s seen that, the first term is (a) = 12

And, common ratio(r) = (4)/ (12) = 1/3

We know that, the general term is

t_{n} = ar^{n â€“ 1}

Thus,

t_{10} = (12)(1/3)^{10 â€“ 1} = (12)(1/3)^{9 }= 12 x 1/(19683) = 4/ 6561

**6. Find the nth term of the series:**

**1, 2, 4,Â 8, â€¦â€¦..**

**Solution: **

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = 2/ 1 = 2

We know that, the general term is

t_{n} = ar^{n â€“ 1}

Thus,

t_{n} = (1)(2)^{n â€“ 1} = 2^{n – 1}

### Exercise 11(B) Page No: 154

**1. Which term of the G.P. :**

**Solution: **

In the given G.P.

First term, a = -10

Common ratio, r = (5/âˆš3)/ (-10) = 1/(-2âˆš3)

We know that, the general term is

t_{n} = ar^{n â€“ 1}

So,

t_{n} = (-10)( 1/(-2âˆš3))^{n â€“ 1} = -5/72

Now, equating the exponents we have

n â€“ 1 = 4

n = 5

Thus, the 5^{th} of the given G.P. is -5/72

**2. The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression. **

**Solution:**

Given,

t_{5} = 81 and t_{2} = 24

We know that, the general term is

t_{n} = ar^{n â€“ 1}

So,

t_{5} = ar^{5 â€“ 1} = ar^{4} = 81 â€¦. (1)

And,

t_{2} = ar^{2 â€“ 1} = ar^{1} = 24 â€¦. (2)

Dividing (1) by (2), we have

ar^{4}/ ar = 81/ 24

r^{3} = 27/ 8

r = 3/2

Using r in (2), we get

a(3/2) = 24

a = 16

Hence, the G.P. is

G.P. = a, ar, ar^{2}, ar^{3}â€¦â€¦

= 16, 16 x (3/2), 16 x (3/2)^{2}, 16 x (3/2)^{3}

= 16, 24, 36, 54, â€¦â€¦

**3. Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P. **

**Solution: **

Given,

t_{4} = 1/18 and t_{7} = -1/486

We know that, the general term is

t_{n} = ar^{n â€“ 1}

So,

t_{4} = ar^{4 â€“ 1} = ar^{3} = 1/18 â€¦. (1)

And,

t_{7} = ar^{7 â€“ 1} = ar^{6} = -1/486 â€¦. (2)

Dividing (2) by (1), we have

ar^{6}/ ar^{3} = (-1/486)/ (1/18)

r^{3} = -1/27

r = -1/3

Using r in (1), we get

a(-1/3)^{3} = 1/18

a = -27/ 18 = -3/2

Hence, the G.P. is

G.P. = a, ar, ar^{2}, ar^{3}â€¦â€¦

= -3/2, -3/2(-1/3), -3/2(-1/3)^{2}, -3/2(-1/3)^{3}, â€¦â€¦

= -3/2, 1/2, -1/6, 1/18, â€¦..

**4. If the first and the third terms of a G.P are 2 and 8 respectively, find its second term. **

**Solution:**

Given,

t_{1} = 2 and t_{3} = 8

We know that, the general term is

t_{n} = ar^{n â€“ 1}

So,

t_{1} = ar^{1 â€“ 1} = a = 2 â€¦. (1)

And,

t_{3} = ar^{3 â€“ 1} = ar^{2} = 8 â€¦. (2)

Dividing (2) by (1), we have

ar^{2}/ a = 8/ 2

r^{2} = 4

r = Â± 2

Hence, the 2^{nd} term of the G.P. is

When a = 2 and r = 2 is 2(2) = 4

Or when a = 2 and r = -2 is 2(-2) = -4

**5. The product of 3 ^{rd}Â and 8^{th}Â terms of a G.P. is 243. If its 4^{th}Â term is 3, find its 7^{th}Â term**

**Solution: **

Given,

Product of 3^{rd}Â and 8^{th}Â terms of a G.P. is 243

The general term of a G.P. with first term a and common ratio r is given by,

t_{n} = ar^{n â€“ 1}

So,

t_{3} x t_{8 }= ar^{3 â€“ 1} x ar^{8 â€“ 1} = ar^{2} x ar^{7 }= a^{2}r^{9} = 243

Also given,

t_{4} = ar^{4 â€“ 1} = ar^{3} = 3

Now,

a^{2}r^{9} = (ar^{3}) ar^{6} = 243

Substituting the value of ar^{3}in the above equation, we get,

(3) ar^{6} = 243

ar^{6} = 81

ar^{7 â€“ 1} = 81 = t_{7}

Thus, the 7^{th} term of the G.P is 81.

### Exercise 11(C) Page No: 156

**1. Find the seventh term from the end of the series: âˆš2, 2, 2âˆš2, â€¦â€¦ , 32**

**Solution: **

Given series: âˆš2, 2, 2âˆš2, â€¦â€¦ , 32

Here,

a = âˆš2

r = 2/ âˆš2 = âˆš2

And, the last term (l) = 32

l = t_{n} = ar^{n â€“ 1} = 32

(âˆš2)( âˆš2)^{n â€“ 1} = 32

(âˆš2)^{n } = 32

(âˆš2)^{n } = (2)^{5} = (âˆš2)^{10}

Equating the exponents, we have

n = 10

So, the 7^{th} term from the end is (10 â€“ 7 + 1)^{th }term.

i.e. 4^{th} term of the G.P

Hence,

t_{4} = (âˆš2)(âˆš2)^{4 â€“ 1} = (âˆš2)(âˆš2)^{3} = (âˆš2) x 2âˆš2 = 4

**2. Find the third term from the end of the G.P. **

**2/27, 2/9, 2/3, â€¦â€¦., 162**

**Solution: **

Given series: 2/27, 2/9, 2/3, â€¦â€¦., 162

Here,

a = 2/27

r = (2/9) / (2/27)

r = 3

And, the last term (l) = 162

l = t_{n} = ar^{n â€“ 1} = 162

(2/27) (3)^{n – 1} = 162

(3)^{n – 1} = 162 x (27/2)

(3)^{n – 1} = 2187

(3)^{n – 1} = (3)^{7}

n – 1 = 7

n = 7+1

n = 8

So, the third term from the end is (8 – 3 + 1)^{th} term

i.e 6^{th }term of the G.P. = t_{6}

Hence,

t_{6} = ar^{6-1}

t_{6} = (2/27) (3)^{6-1}

t_{6} = (2/27) (3)^{5}

t_{6} = 2 x 3^{2}

t_{6} = 18

**3. Find the G.P. 1/27, 1/9, 1/3, â€¦â€¦, 81; find the product of fourth term from the beginning and the fourth term from the end. **

**Solution: **

Given G.P. 1/27, 1/9, 1/3, â€¦â€¦, 81

Here, a = 1/27, common ratio (r) = (1/9)/ (1/27) = 3 and l = 81

We know that,

l = t_{n} = ar^{n â€“ 1} = 81

(1/27)(3)^{n â€“ 1} = 81

3^{n â€“ 1} = 81 x 27 = 2187

3^{n â€“ 1} = 3^{7}

n â€“ 1 = 7

n = 8

Hence, there are 8 terms in the given G.P.

Now,

4^{th} term from the beginning is t_{4} and the 4^{th} term from the end is (8 â€“ 4 + 1) = 5^{th} term (t_{5})

Thus,

the product of t_{4 }and t_{5} = ar^{4 â€“ 1} x ar^{5 â€“ 1} = ar^{3} x ar^{4} = a^{2}r^{7} = (1/27)^{2}(3)^{7} = 3

**4. If for a G.P.,Â p ^{th},Â q^{th}Â andÂ r^{th}Â terms are a, b and c respectively; prove that:**

**(q – r) log a + (r – p) log b + (p – q) log c = 0**

**Solution: **

Letâ€™s take the first term of the G.P. be A and its common ratio be R.

Then,

p^{th} term = a â‡’ AR^{p â€“ 1} = a

q^{th} term = b â‡’ AR^{q â€“ 1} = b

r^{th} term = c â‡’ AR^{r â€“ 1} = c

Now,

On taking log on both the sides, we get

log( a^{q-r} x b^{r-p} x c^{p-q} ) = log 1

â‡’ (q – r)log a + (r – p)log b + (p – q)log c = 0

– Hence Proved

### Exercise 11(D) Page No: 156

**1. Find the sum of G.P.:**

**(i) 1 + 3 + 9 + 27 + â€¦â€¦â€¦. to 12 terms**

**(ii) 0.3 + 0.03 + 0.003 + 0.0003 +â€¦..Â toÂ 8 terms.**

**(iii) 1 â€“ 1/2 + 1/4 â€“ 1/8 + â€¦â€¦.. to 9 terms**

**(iv) 1 – 1/3 + 1/3 ^{2} â€“ 1/3^{3} + â€¦â€¦â€¦ to n terms**

**(v) **

**(vi) **

**Solution: **

(i) Given G.P: 1 + 3 + 9 + 27 + â€¦â€¦â€¦. to 12 terms

Here,

a = 1 and r = 3/1 = 3 (r > 1)

Number of terms, n = 12

Hence,

S_{n} = a(r^{n }– 1)/ r â€“ 1

â‡’ S_{12} = (1)((3)^{12 }– 1)/ 3 â€“ 1

= (3^{12} â€“ 1)/ 2

= (531441 â€“ 1)/ 2

= 531440/2

= 265720

(ii) Given G.P: 0.3 + 0.03 + 0.003 + 0.0003 +â€¦..**Â **to 8 terms

Here,

a = 0.3 and r = 0.03/0.3 = 0.1 (r < 1)

Number of terms, n = 8

Hence,

S_{n} = a(1 – r^{n })/ 1 – r

â‡’ S_{8} = (0.3)(1 – 0.1^{8 })/ (1 â€“ 0.1)

= 0.3(1 â€“ 0.1^{8})/ 0.9

= (1 â€“ 0.1^{8})/ 3

= 1/3(1 â€“ (1/10)^{8})

(iii) Given G.P: 1 â€“ 1/2 + 1/4 â€“ 1/8 + â€¦â€¦.. to 9 terms

Here,

a = 1 and r = (-1/2)/ 1 = -1/2 (| r | < 1)

Number of terms, n = 9

Hence,

S_{n} = a(1 – r^{n })/ 1 – r

â‡’ S_{9} = (1)(1 – (-1/2)^{9 })/ (1 â€“ (-1/2))

= (1 + (1/2)^{9})/ (3/2)

= 2/3 x ( 1 + 1/512 )

= 2/3 x (513/512)

= 171/ 256

(iv) Given G.P: 1 – 1/3 + 1/3^{2} â€“ 1/3^{3} + â€¦â€¦â€¦ to n terms

Here,

a = 1 and r = (-1/3)/ 1 = -1/3 (| r | < 1)

Number of terms is n

Hence,

S_{n} = a(1 – r^{n })/ 1 – r

(v) Given G.P:

Here,

a = (x + y)/ (x â€“ y) and r = 1/[(x + y)/ (x â€“ y)] = (x – y)/ (x + y) (| r | < 1)

Number of terms = n

Hence,

S_{n} = a(1 – r^{n })/ 1 – r

(vi) Given G.P:

Here,

a = âˆš3 and r = 1/âˆš3/ âˆš3 = 1/3 (| r | < 1)

Number of terms = n

Hence,

S_{n} = a(1 – r^{n })/ 1 – r

**2. How many terms of the geometric progression 1 + 4 + 16 + 64 + â€¦â€¦..Â mustÂ be added to get sum equal to 5461?**

**Solution: **

Given G.P: 1 + 4 + 16 + 64 + â€¦â€¦..

Here,

a = 1 and r = 4/1 = 4 (r > 1)

And,

S_{n} = 5461

We know that,

S_{n} = a(r^{n }– 1)/ r â€“ 1

â‡’ S_{n} = (1)((4)^{n }– 1)/ 4 â€“ 1

= (4^{n} â€“ 1)/3

5461 = (4^{n} â€“ 1)/3

16383 = 4^{n} â€“ 1

4^{n} = 16384

4^{n} = 4^{7}

n = 7

Therefore, 7 terms of the G.P must be added to get a sum of 5461.

**3. The first term of a G.P. is 27 and its 8 ^{th} term is 1/81. Find the sum of its first 10 terms. **

**Solution:**

Given,

First term (a) of a G.P = 27

And, 8^{th} term = t_{8} = ar^{8 – 1} = 1/81

(27)r^{7} = 1/81

r^{7} = 1/(81 x 27)

r^{7} = (1/3)^{7}

r = 1/3 (r <1)

S_{n} = a(1 – r^{n})/ 1 – r

Now,

Sum of first 10 terms = S_{10}

**4. A boy spendsÂ Rs.10 on first day,Â Rs.20 on second day,Â Rs.40 on third day and so on. Find how much, in all, will he spend in 12 days?**

**Solution: **

Given,

Amount spent on 1^{st} day = Rs 10

Amount spent on 2^{nd} day = Rs 20

And amount spent on 3^{rd} day = Rs 40

Itâ€™s seen that,

10, 20, 40, â€¦â€¦ forms a G.P with first term, a = 10 and common ratio, r = 20/10 = 2 (r > 1)

The number of days, n = 12

Hence, the sum of money spend in 12 days is the sum of 12 terms of the G.P.

S_{n} = a(r^{n }– 1)/ r â€“ 1

S_{12} = (10)(2^{12 }– 1)/ 2 â€“ 1 = 10 (2^{12} – 1) = 10 (4096 – 1) = 10 x 4095 = 40950

Therefore, the amount spent by him in 12 days is Rs 40950

**5. The 4 ^{th} term and the 7^{th} term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P. **

**Solution:**

Given,

t_{4} = 1/27 and t_{7 }= 1/729

We know that,

t_{n} = ar^{n â€“ 1}

So,

t_{4 }= ar^{4 â€“ 1} = ar^{3} = 1/27 â€¦. (1)

t_{7 }= ar^{7 â€“ 1} = ar^{6} = 1/729 â€¦. (2)

Dividing (2) by (1) we get,

ar^{6}/ ar^{3} = (1/729)/ (1/27)

r^{3} = (1/3)^{3}

r = 1/3 (r < 1)

In (1)

a x 1/27 = 1/27

a = 1

Hence,

S_{n} = a(1 – r^{n})/ 1 – r

S_{n} = (1 â€“ (1/3)^{n })/ 1 â€“ (1/3)

= (1 â€“ (1/3)^{n })/ (2/3)

= 3/2 (1 â€“ (1/3)^{n })

**6. A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term. **

**Solution: **

Given,

For a G.P.,

r = 3, l = 486 and S_{n} = 728

1458 â€“ a = 728 x 2 = 1456

Thus, a = 2

**7. Find the sum of G.P.: 3, 6, 12, â€¦., 1536.**

**Solution: **

Given G.P: 3, 6, 12, â€¦., 1536

Here,

a = 3, l = 1536 and r = 6/3 = 2

So,

The sum of terms = (lr â€“ a)/ (r – 1)

= (1536 x 2 â€“ 3)/ (2 – 1)

= 3072 â€“ 3

= 3069

**8. How many terms of the series 2 + 6 + 18 + â€¦.. must be taken to make the sum equal to 728?**

**Solution: **

Given G.P: 2 + 6 + 18 + â€¦..

Here,

a = 2 and r = 6/2 = 3

Also given,

S_{n} = 728

S_{n} = a(r^{n }– 1)/ r â€“ 1

728 = (2)(3^{n }– 1)/ 3 â€“ 1 = 3^{n }â€“ 1

729 = 3^{n}

3^{6} = 3^{n}

n = 6

Therefore, 6 terms must be taken to make the sum equal to 728.

**9. In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. **

**Find its common ratio. **

**Solution: **

Given,

Therefore, the common ratio is 3/5.

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