# Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.3 | Set 1

### Question 1. Find the 20^{th} and n^{th} 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) =

^{ }Find: 20

^{th}and n^{th}terms of the given G.PSo, the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratio.Now we find the 20

^{th}terms of the given G.P:a

_{20 }= (5/2)(1/2)^{20-1}= (5/2)(1/2)^{19}=Find the n

^{th}terms of the given G.P:a

_{n }=

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

**Solution:**

According to the question

Common ration(r) = 2

and 8

^{th}term is 192So let us considered a be the first term

So,

a

_{8 }= ar^{7}ar

^{7}= 192a(2)

^{7}= 192a = 3/2

Find: 12

^{th}term of a G.P.As we know that the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratioSo, we find 12

^{th}term of a G.P.a

_{12 }= ar^{12 – 1}a

_{12 }= ar^{11}a

_{12 }= (3/2)(2)^{11}a

_{12}= 3072

### Question 3. The 5^{th}, 8^{th} and 11^{th} terms of a G.P. are p, q and s, respectively. Show that q^{2 }= ps.

**Solution:**

According to the question

The 5

^{th}, 8^{th}and 11^{th}terms of a G.P. are p, q and sProve: q

^{2 }= psProof:

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

So, a

_{5 }= ar^{4 }= p ….(1)a

_{8 }= ar^{7}= q ….(2)a

_{11 }= ar^{10 }= s ….(3)Now divide eq(2) by (1), we get

ar

^{7}/ar^{4 }= q/pr

^{3 }= q/p ….(4)Now divide eq(3) by (2), we get

ar

^{10}/ar^{7}= s/qr

^{3 }= s/q ….(5)From eq(4) and (5), we get

q/p = s/q

q

^{2 }= psHence Proved

### Question 4. The 4^{th} term of a G.P. is square of its second term, and the first term is â€“ 3. Determine its 7^{th} term.

**Solution:**

According to the question

First term(a) = â€“ 3

and the 4

^{th}term of a G.P. is square of its second termFind: 7

^{th}termLet us considered r be the common ratio

As we know that the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratioSo, a

_{4 }= ar^{3}It is given that the 4

^{th}term of a G.P. is square of its second terma

_{4 }= (a_{2})^{2}ar

^{3 }= (a_{2})^{2}ar

^{3 }= (ar)^{2}ar

^{3 }= a^{2}r^{2}r

^{ }= aNow put the value of a, we get

r

^{ }= -3Now, we find the 7

^{th}terma

_{7 }= ar^{7 – 1}a

_{7 }= ar^{6}= âˆ’3 x (âˆ’3)

^{6}=âˆ’2187

### Question 5. Which term of the following sequences:

### (a) 2, 2âˆš2, 4, … is 128 ?

**Solution:**

According to the question

G.P.: 2, 2âˆš2, 4, …

So, first term(a) = 2

So, the common ratio(r) = âˆš2

As we know that, the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratio.128 = 2(âˆš2)

^{n – 1}(2)

^{7}= 2(âˆš2)^{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) âˆš3, 3, 3âˆš3, …. is 729 ?

**Solution:**

According to the question

G.P.: âˆš3, 3, 3âˆš3, ….

So, first term(a) = âˆš3

So, the common ratio(r) = âˆš3

As we know that, the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratio.729 = âˆš3(âˆš3)

^{n-1}(3)

^{6}= âˆš3(âˆš3)^{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) is ?

**Solution:**

According to the question

G.P.:

So, first term(a) = 1/3

So, the common ratio(r) = 1/3

As we know that, the n

^{th}term of G.P can be expressed using formula:a

_{n }= ar^{n – 1}Where a is 1

^{st}term and r is the common ratio.(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) = = -7x/2

Again the common ration(r) = = = -7/2x

So,

-7x/2 = -7/2x

7x/2 = 2/7x

14x

^{2}= 14x

^{2}= 14/14x = Â±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 1

^{st}term a & common ratio r is given bySo, S

_{20}= (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. âˆš7â€‹, âˆš21â€‹, 3âˆš7â€‹,… n terms.

**Solution:**

According to the question

G.P.: âˆš7â€‹, âˆš21â€‹, 3âˆš7â€‹,…

So, first term(a) = âˆš7â€‹

So, the common ratio(r) = âˆš3â€‹

As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given bySo, S

_{n}= (âˆš7â€‹)[(1 – (âˆš3)^{n})/(1 – âˆš3)]= (âˆš7â€‹)[(1 – (âˆš3)

^{n})/(1 – âˆš3)] x [(1 + âˆš3)/(1 + âˆš3)]= -(âˆš7â€‹)(1 + âˆš3)/2(1 – (3)

^{n/2})= (âˆš7â€‹)(1 + âˆš3)/2((3)

^{n/2 }– 1)

### Question 9. 1, -a, a^{2}, -a^{3},… n terms (if a â‰ â€“ 1).

**Solution:**

According to the question

G.P.: 1, -a, a

^{2}, -a^{3},…So, first term(a) = 1

So, the common ratio(r) = -aâ€‹

As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given bySo,

### Question 10. x^{3}, x^{5}, x^{7}, … n terms (if x â‰ Â± 1).

**Solution:**

According to the question

G.P.: x

^{3}, x^{5}, x^{7}, …So, first term(a) = x

^{3}So, the common ratio(r) = x

^{2}â€‹As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given bySo,

### Question 11. Evaluate

**Solution:**

= (2 + 3

^{1}) + (2 + 3^{2}) + (2 + 3^{3}) + …. + (2 + 3^{11})= (2 + 2 + …11 terms) + (3 + 3

^{2 }+ 3^{3}+…11 terms)As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given bySo, S

_{n}= 2 x 11 + 3(3^{11}– 1)/3 – 1S

_{n }= 22 + 3/2((3^{11}– 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

= 1

a

^{3 }= 1a = 1

Also.

Now put the value of a, we get

10r

^{2 }+ 10r + 10 = 39r10r

^{2 }– 29r + 10 = 0On 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, 3^{2}, 3^{3}, â€¦ are needed to give the sum 120?

**Solution:**

According to the question

G.P.: 3, 3

^{2}, 3^{3}, â€¦So, first term(a) = 3

So, the common ratio(r) = 3

S

_{n }= â€‹120As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given byS

_{n }= (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, ar

^{2 }andfirst 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 + ar

^{2 }= 16a(1 + r + r

^{2}) = 16 …….(1)ar

^{3 }+ ar^{4 }+ ar^{5 }= 128ar

^{3}(1 + r + r^{2}) = 128 …….(2)Now divide eq(2) by (1), we get

ar

^{3}(1 + r + r^{2})/ a(1 + r + r^{2}) = 128/16r

^{3 }= 8r = 2

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

a(1 + (2) + (2)

^{2}) = 16a(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 1

^{st}term a & common ratio r is given by

### Question 15. Given a G.P. with a = 729 and 7^{th} term 64, determine S_{7}.

**Solution:**

According to the question

The first term(a) = 729

and 7

^{th}term 64Find: S

_{7}Let the common ratio be r.

a

_{7 }= ar^{6 }= 64r

^{6 }= 64/729r

^{ }= Â±2/3Now we find the sum to 7th terms of the G.P.

As we know that, the sum of n terms of G.P. with 1

^{st}term a & common ratio r is given byOn taking r = +2/3

= 2059

On taking r = -2/3

= 463

### Question 16. Find a G.P. for which sum of the first two terms is â€“ 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

a

_{5}= 4(a_{3})So, ar

^{4 }= 4(ar^{2})r

^{2 }= 4r

^{ }= Â±2When r = +2 then

a + ar = a + 2a = 3a = âˆ’4

a = -4/3

Then the G.P. is

When r = -2 then

a + ar = a âˆ’ 2a = âˆ’a = âˆ’4

a = 4

Then the G.P. is 4, -8, 16, -32,…

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