# Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.1 | Set 1

**Question 1.** Show that each one of following progression is a GP. Also, find the common ratio in each case:

**(i) 4, -2, 1, -1/2,…….**

**(ii) -2/3, -6, -54,……….**

**(iii) a, 3a ^{2}/4, 9a^{3}/16,…….**

**(iv) 1/2, 1/3, 2/9, 4/27,…….**

**Solution:**

(i)Given a1 = 4, a2 = -2, a3 = 1a2/a1 = -2/4 = -1/2

a3/a2 = -1/2

a4/a1 = (-1/2)/1 =-1/2

Now, a2/a1 = a3/a2 = a4/a3 = -1/2

Thus, a1, a2, a3, a4,… are in GP and common ratio(r) is -1/2.

(ii)Given a1 = -2/3, a2 = -6, a3 = -54a2/a1 = -6/(-2/3) = 9

a3/a2 = -54/-6 = 9

Now, a2/a1 = a3/a2 = 9

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 9.

(iii)Given a1= a, a2= 3a^{2}/4, a3 = 9a^{3}/16a2/a1 = (3a

^{2}/4)/a = 3a/4a3/a2 = (9a

^{3}/16)/(3a^{2}/4) = 3a/4Now, a2/a1 = a3/a2 = 9

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 3a/4.

(iv)Given a1=1/2, a2=1/3, a3=2/9a2/a1 = (1/3)/(1/2) = 2/3

a3/a2 = (2/9)/(1/3) = 2/3

Now, a2/a1 = a3/a2 = 2/3

Thus, a1, a2, a3,…… are in GP and common ratio(r) is 2/3.

**Question 2.** Show that the sequence <an>, defined by an = 2/(3^{n}), n∈N is a GP.

**Solution:**

Given an = 2/(3

^{n})We put n=1,2,3,….., n∈N

The sequence is 2/3, 2/(3

^{2}), 2/(3^{3}), ………..a2/a1 = (2/(3

^{2}))/(2/3)=1/3a3/a2 = (2/(3

^{3}))/(2/(3^{2})) = 1/3Now, a2/a1 = a3/a2 = 1/3

Thus, a1,a2,a3,…… are in GP and common ratio(r) is 1/3.

Hence, the given sequence is in GP.

**Question 3. Find:**

**(i) The 9th term of the GP 1,4,16,64,…….**

**(ii) The 10th term of the GP -3/4, 1/2, -1/3,-2/9, ………….**

**(iii) The 8th term of GP 0.3, 0.06, 0.012, …………..**

**(iv) The 12th term of GP 1/(a ^{3}x^{3}), ax, a5x^{5}, ………..**

**(v) The nth term of GP √3,1/√3,1/3√3,…………..**

**Solution:**

(i)Given a2/a1 = 4/1 = 4a3/a2 = 16/4 = 4

Hence r = a3/a2 = a2/a1 = 4 and a1 = 1,

a9 = a1*r

^{8}a9 = 1*(4)

^{8 }= (4)

^{8}

(ii)Given a2/a1 = (1/2)/(-3/4) = -2/3a3/a2 = (-1/3)/(1/2) = -2/3

Hence, r = a3/a2 = a2/a1 = -2/3 and a1 = -3/4

a10 = a1*r

^{9}a10 = (-3/4)*(-2/3)

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

^{8}

(iii)Given a2/a1 = 0.06/0.3 = 0.2a3/a2 = 0.012/0.06 = 0.2

Hence r = a3/a2 = a2/a1 = 0.2 and a1 = 0.3

a8 = a1*r

^{7}a8 = 0.3*(0.2)

^{7}

(iv)Given a2/a1 = (ax)/(1/(a^{3}x^{3})) = a^{4}x^{4}a3/a2 = (a

^{5}x^{5})/(ax) = a^{4}x^{4}Hence r = a3/a2 = a2/a1= a4x4 and a1 =1/(a

^{3}x^{3})a12 = a1*r

^{11}a8 = (1/(a

^{3}x^{3}))*((a^{4}x^{4})^{11})= a

^{41}x^{41}

(v)Given a2/a1 = (1/√3)/√3 = 1/3a3/a2 = (1/3√3)/(1/√3) = 1/3

Hence r = a3/a2 = a2/a1= 1/3 and a1 = √3

an = a1*r

^{(n-1)}= √3*(1/3)

^{(n-1) }

**Question 4.** Find the 4th term from the end of GP 2/27, 2/9, 2/3,………,162.

**Solution:**

Given a2/a1 = (2/9)/(2/27) = 3

a3/a2 = (2/3)/(2/9) = 3

Hence r = a3/a2 = a2/a1= 3 and a1 = 2/27

After reversing the GP it becomes 162, ……….,, 2/3, 2/9, 2/27.

a1’=162 and r’= 1/r =1/3

a4’= a1′ * r’

^{3}= 162 * (1/3)

^{3}= 6.

**Question 5.** Which term of the progression 0.004,0.02,0.1,….is 12.5?

**Solution:**

Given a2/a1= 0.02/0.004= 5

a3/a2 = 0.1/0.02 = 5

Hence, r=a3/a2 = a2/a1= 5 and a1=0.004

nth term of sequence is given by

an = a1*(r)

^{n-1}an=(0.004)*(5)

^{n-1}Let the nth term be 12.5.

Hence, (0.004)*(5)

^{n-1}=12.5(5)

^{n-1}= 12.5/0.004(5)

^{n-1}= 3125(5)

^{n-1}= (5)^{5}Hence, n-1=5

n=5+1= 6

**Question 6.** Which term of the GP :

**(i) √2, 1/√2, 1/2√2, 1/4√2,…………….is 1/(512√2)?**

**(ii) 2, 2√2, 4, …………………..is 128?**

**(iii) √3, 3, 3√3, …………………..is 729?**

**(iv) 1/3, 1/9, 1/27, ………….is 1/19683?**

**Solution:**

(i)In given sequence a1=√2, a2= 1/√2Common ratio(r) = a2/a1 = (1/√2)/√2 =1/2 and a1 = √2

Let the nth term be 1/(512√2)

a1*r

^{n-1 }= 1/(512√2)√2*(1/2)

^{n-1 }= 1/(512√2)(1/2)

^{n-1}= 1/1024(1/2)

^{n-1}= (1/2)10Hence, n-1= 10

n=10+1 =11.

(ii)In given sequence a1=2 ,a2= 2√2Common ratio(r)=a2/a1= 2√2/2 = √2 and a1=2

Let the nth term be 1/(512√2)

a1*r

^{n-1}= 1282*(√2)

^{n-1}= 128(√2)

^{n-1}=64(√2)

^{n-1}= (√2)12Hence n-1= 12

n= 12+1 =13.

(iii)In given sequence a1= √3, a2= 3Common ratio(r)=a2/a1 = 3/√3 = √3 and a1 = √3

Let the nth term be 729

a1*r

^{n-1}= 729√3*(√3)

^{n-1}= 729(√3)

^{n }= 729(√3)

^{n}= (√3)^{11}Hence n-1= 11

n= 11+1 =12.

(iv)In given sequence a1= 1/3, a2= 1/9Common ratio(r)=a2/a1 = 1/3 and a1 = 1/3

Let the nth term be 1/19683

a1*r

^{n-1}= 1/196831/3*(1/3)

^{n-1 }= 1/19683(1/3)

^{n}= 1/19683(1/3)

^{n}= (1/3)^{8}Hence n = 8.

**Question 7. **Which term of progression 18, -12, 8,…………..is 512/729?

**Solution:**

In given sequence a1=18 ,a2 =-12

Common ratio(r)= -12/18 = -2/3 and a1=18

Let the nth term be 512/729

a1*r

^{n-1}= 512/729(18)*(-2/3)

^{n-1 }= 512/729(-2/3

^{)n-1}= 256/6561(-2/3)

^{n-1}= (-2/3)^{8}Hence n-1 = 8

n = 8+1 = 9

**Question 8.** Find the 4th term from end of GP 1/2 1/6, 1/18, 1/54,…………….1/4373?

**Solution:**

In given sequence a1 = 1/2, a2 = 1/6

Common ratio(r)= (1/6)/(1/2) =1/3 and a1 = 1/2

After reversing the GP it becomes 1/4374,……..1/54, 1/18, 1/6, 1/2

Common ratio(r’)= 3 and a1 = 1/4374

4th term from end of original GP becomes 4th from beginning of reversed GP

Hence required term is given by

(a1′)*(r’

^{4-1}) = (a1′)*(r’)^{3 }= (1/4374)*(3)

^{3}= (27/4374)

= 1/162

**Question 9.** The 4th term of a GP is 27 and the 7th term is 729, find the GP?

**Solution:**

Let a be first term of required GP and r be common ratio.

Given, a4= 27 and a7 = 729

(a1*r

^{6}) / (a1*r^{3}) = 729/27r

^{3}= 729/27 = 27r

^{3}= 3^{3}r=3

Now, put r=3 in a1*r

^{3}=27a1*(3)

^{3}= 27a1 = 1

Thus, the given GP is 1,3, 9, 27,……….

### **Question 10. **The 7th term of a GP is 8 times the 4th term and 5th term is 48. Find the GP ?

**Solution:**

Let a be first term of required GP and r be common ratio.

Given, a7 = 8*a4 and a5 = 48

a7/a4 = 8

(a1*r

^{6})/(a1*r^{3}) = 8r

^{3}= 8r=2

Now put r = 2 in a5 = 48

a1*r

^{4}= 48a1* (2)

^{4}= 48a1* 16 = 48

a1 = 3

Thus, the given GP is 3, 6, 12, 24,…………..