Class 11 RD Sharma Solutions – Chapter 20 Geometric Progressions- Exercise 20.5 | Set 1

Question 1. If a, b, c are in G.P., prove that log a, log b, log c are in A.P.

Solution:

Given: a, b and c are in G.P.

Using property of geometric mean, we get

b² = ac 

(b²)ⁿ = (ac)ⁿ

b²n = aⁿ cⁿ

Now, use log on both the sides, we get,

log b²ⁿ = log (aⁿcⁿ)

log (bⁿ)² = log aⁿ + log cⁿ

2 log bⁿ = log aⁿ + log cⁿ

Hence, proved log aⁿ, log bⁿ, log cⁿ are in A.P

Question 2. If a, b, c are in G.P., prove that 1/log_a m, 1/log_b m, 1/log_c m are in A.P.

Solution:

Given: a, b and c are in GP

Using the property of geometric mean

b² = ac 

On applying log on both sides with base m, we get

log_m b² = log_m ac

Using property of log

log_m b² = log_m a + log_m c 

2log_m b = log_m a + log_m c

2/log_b m = 1/log_a m + 1/log_c m

Hence, proved 1/log_a m, 1/log_b m, 1/log_c m are in A.P.

Question 3. Find k such that k + 9, k – 6, and 4 forms three consecutive terms of a G.P.

Solution:

Let us considered  

a = k + 9

b = k − 6 

c = 4

AS we know that a, b and c are in GP, then

By using property of geometric mean, we get

b² = ac 

(k − 6)² = 4(k + 9)

k² – 12k + 36 = 4k + 36

k² – 16k = 0

k = 0 or k = 16

Question 4. Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they form a G.P. find the numbers.

Solution:

Let us considered the first term of an A.P. = a  

Common difference = d

a₁ + a₂ + a₃ = 15

Here, the three number are: a, a + d, and a + 2d

So,

a + a + d + a + 2d = 15

3a + 3d = 15 or a + d = 5

d = 5 – a          -(1)

a + 1, a + d + 3, and a + 2d + 9

They are in GP, that is:

(a + d + 3)/(a + 1) = (a + 2d + 9)/(a + d + 3)

(a + d + 3)² = (a + 2d + 9)(a + 1)

a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9

(5 – a)² – 4a + 4(5 – a) = 0

25 + a² – 10a – 4a + 20 – 4a = 0

a² – 18a + 45 = 0

a² – 15a – 3a + 45 = 0

a(a – 15) – 3(a – 15) = 0

a = 3 or a = 15

d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

Then,

For a = 3 and d = 2, A.P is 3, 5, 7

For a = 15 and d = -10, A.P is 15, 5, -5

Hence, the numbers are 3, 5, 7 or 15, 5, – 5

Question 5. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a G.P. Find the numbers.

Solution:

Let us considered the first term of an A.P. be = a

Common difference = d

a₁ + a₂ + a₃ = 21

Here, the three number are: a, a + d, and a + 2d

So,

3a + 3d = 21 or

a + d = 7.

d = 7 – a          -(1)

a, a + d – 1, and a + 2d + 1

They are now in GP, that is:

(a + d – 1)/a = (a + 2d + 1)/(a + d – 1)

(a + d – 1)² = a(a + 2d + 1)

a² + d² + 1 + 2ad – 2d – 2a = a² + a + 2da

(7 – a)² – 3a + 1 – 2(7 – a) = 0

49 + a² – 14a – 3a + 1 – 14 + 2a = 0

a² – 15a + 36 = 0

a² – 12a – 3a + 36 = 0

a(a – 12) – 3(a – 12) = 0

a = 3 or a = 12

d = 7 – a

d = 7 – 3 or d = 7 – 12

d = 4 or – 5

Then,

For a = 3 and d = 4, A.P is 3, 7, 11

For a = 12 and d = -5, A.P is 12, 7, 2

Hence, the numbers are 3, 7, 11 or 12, 7, 2

Question 6. The sum of three numbers a, b, c in A.P. is 18. If a and b are each increased by 4 and c is increased by 36, the new numbers form a G.P. Find a, b, c.

Solution:

Let us considered the first term of an A.P. = a 

Common difference = d

b = a + d; c = a + 2d.

Given:

a + b + c = 18

3a + 3d = 18 or a + d = 6.

d = 6 – a          -(1)

a + 4, a + d + 4, and a + 2d + 36

They are now in GP, that is:

(a + d + 4)/(a + 4) = (a + 2d + 36)/(a + d + 4)

(a + d + 4)² = (a + 2d + 36)(a + 4)

a² + d² + 16 + 8a + 2ad + 8d = a² + 4a + 2da + 36a + 144 + 8d

d² – 32a – 128

(6 – a)² – 32a – 128 = 0

36 + a² – 12a – 32a – 128 = 0

a² – 44a – 92 = 0

a² – 46a + 2a – 92 = 0

a(a – 46) + 2(a – 46) = 0

a = – 2 or a = 46

d = 6 –a

d = 6 – (– 2) or d = 6 – 46

d = 8 or – 40

Then,

For a = -2 and d = 8, A.P is -2, 6, 14

For a = 46 and d = -40, A.P is 46, 6, -34

Hence, the numbers are – 2, 6, 14 or 46, 6, – 34

Question 7. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. Find the numbers.

Solution:

Let us considered the three numbers = a, ar, ar²

a + ar + ar² = 56          -(1)

Now, subtract 1, 7, 21 from the numbers, we get,

(a – 1), (ar – 7), (ar² – 21)

The above numbers are in AP.

If three numbers are in AP, 

So, according to the arithmetic mean, we can write as 2b = a + c

2 (ar – 7) = a – 1 + ar² – 21

= (ar² + a) – 22

2ar – 14 = (56 – ar) – 22

2ar – 14 = 34 – ar

3ar = 48

ar = 48/3

ar = 16

a = 16/r          -(2)

Now, substitute the value of a in eq(1) we get,

(16 + 16r + 16r²)/r = 56

16 + 16r + 16r² = 56r

16r² – 40r + 16 = 0

2r² – 5r + 2 = 0

2r² – 4r – r + 2 = 0

2r(r – 2) – 1(r – 2) = 0

(r – 2) (2r – 1) = 0

r = 2 or 1/2

Substitute the value of r in eq(2) we get,

a = 16/r

= 16/2 or 16/(1/2)

= 8 or 32

Hence, the three numbers are (a, ar, ar²) is (8, 16, 32)

Question 8. if a, b, c are in G.P., prove that:

(i) a(b2 + c²) = c(a² + b²)

(ii) a²b²c² [1/a³ + 1/b³ + 1/c³] = a³ + b³ + c³

(iii) (a+b+c)² / (a² + b² + c²) = (a+b+c) / (a-b+c)

(iv) 1/(a² – b²) + 1/b² = 1/(b² – c²)

(v) (a + 2b + 2c) (a – 2b + 2c) = a² + 4c²

Solution:

(i) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

Let LHS: a(b² + c²)

Now, on substituting b² = ac, we get

a(ac + c²)

a²c + ac²

c(a² + ac)

On Substituting ac = b² we get,

c(a² + b²) = RHS

LHS = RHS

Hence, proved.

(ii) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

Let LHS: a²b²c² [1/a³ + 1/b³ + 1/c³]

a²b²c²/a³ + a²b²c²/b³ + a²b²c²/c³

b²c²/a + a²c²/b + a²b²/c

(ac)c²/a + (b²)²/b + a²(ac)/c         -(∵ b² = ac)

ac³/a + b⁴/b + a³c/c

c³ + b³ + a³ = RHS

LHS = RHS

Hence, proved.

(iii) Given: a, b, c are in GP.

 Using the property of geometric mean,

b² = ac

Let LHS: (a + b + c)² / (a² + b² + c²)

(a + b + c)² / (a² + b² + c²) = (a + b + c)²/(a² – b² + c² + 2b²)

= (a + b + c)² / (a² – b² + c² + 2ac)      -(∵ b² = ac)

= (a + b + c)² / (a + b + c)(a – b + c)       -(∵ (a + b + c)(a – b + c) = a² – b² + c² + 2ac)

= (a + b + c) / (a – b + c)

= RHS

LHS = RHS

Hence, proved.

(iv) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

Let LHS: 1/(a² – b²) + 1/b²

On taking LCM, we get

1/(a² – b²) + 1/b² = (b² + a² – b²)/(a² – b²)b²

= a² / (a²b² – b⁴)

= a² / (a²b² – (b²)²)

= a² / (a²b² – (ac)²)        -(∵ b² = ac)

= a² / (a²b² – a²c²)

= a² / a²(b² – c²)

= 1/ (b² – c²)

= RHS

LHS = RHS

Hence, proved.

(v) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

Let LHS: (a + 2b + 2c) (a – 2b + 2c)

Now, on expanding, we get

(a + 2b + 2c) (a – 2b + 2c) = a² – 2ab + 2ac + 2ab – 4b² + 4bc + 2ac – 4bc + 4c²

= a² + 4ac – 4b² + 4c²

= a² + 4ac – 4(ac) + 4c²         -(∵ b² = ac)

= a² + 4c²

= RHS

LHS = RHS

Hence, proved.

Question 9. If a, b, c, d are in G.P., prove that:

(i) (ab – cd) / (b² – c²) = (a + c) / b

(ii) (a + b + c + d)² = (a + b)² + 2(b + c)² + (c + d)²

(iii) (b + c) (b + d) = (c + a) (c + d)

Solution:

(i) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

bc = ad

c² = bd

Let LHS: (ab – cd) / (b² – c²)

(ab – cd) / (b² – c²) = (ab – cd) / (ac – bd)

= (ab – cd)b / (ac – bd)b

= (ab² – bcd) / (ac – bd)b

= [a(ac) – c(c²)] / (ac – bd)b

= (a²c – c³) / (ac – bd)b

= [c(a² – c²)] / (ac – bd)b

= [(a + c) (ac – c²)] / (ac – bd)b

= [(a + c) (ac – bd)] / (ac – bd)b

= (a + c) / b

= RHS

LHS = RHS

Hence, proved.

(ii) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

bc = ad

c² = bd

Let RHS: (a + b)² + 2(b + c)² + (c + d)²

Now, on expanding, we get

(a + b)² + 2(b + c)² + (c + d)² = (a + b)² + 2 (a + b) (c + d) + (c + d)²

= a² + b² + 2ab + 2(c² + b² + 2cb) + c² + d² + 2cd

= a² + b² + c² + d² + 2ab + 2(c² + b² + 2cb) + 2cd

= a² + b² + c² + d² + 2(ab + bd + ac + cb +cd)         -(∵ c² = bd, b² = ac)

(a + b + c)² + d² + 2d(a + b + c) = {(a + b + c) + d}²

RHS = LHS

Hence, proved.

(iii) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

bc = ad

c² = bd

Let LHS: (b + c) (b + d)

Now, on expanding, we get

(b + c) (b + d) = b² + bd + cb + cd

= ac + c² + ad + cd

= c (a + c) + d (a + c)

= (a + c) (c + d)

= RHS

LHS = RHS

Hence, proved.

Question 10. If a, b, c are in G.P., prove that the following are also in G.P.:

(i) a², b², c²

(ii) a³, b³, c³

(iii) a² + b², ab + bc, b² + c²

Solution:

(i) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

On squaring both the sides we get,

(b²)² = (ac)²

(b²)² = a²c²

Hence, proved a², b², c² are in G.P.

(ii) Given: a, b, c are in GP.

By using the property of geometric mean,

b² = ac

On squaring both the sides, we get

(b²)³ = (ac)³

(b²)³ = a³c³

(b³)² = a³c³

Hence, proved a³, b³, c³ are in G.P.

(iii) Given: a, b, c are in GP.

Using the property of geometric mean,

b² = ac

a² + b², ab + bc, b² + c² or (ab + bc)² = (a² + b²) (b² + c²) 

Let LHS: (ab + bc)²

Now, on expanding, we get

(ab + bc)² = a²b² + 2ab²c + b²c²

= a²b² + 2b²(b²) + b²c²            -(∵ ac = b²⁾

= a²b² + 2b⁴ + b²c²

= a²b² + b⁴ + a²c² + b²c²         -(∵ b² = ac)

= b²(b² + a²) + c²(a² + b²)

= (a² + b²)(b² + c²)

= RHS

LHS = RHS

Hence, a² + b², ab + bc, b² + c² are in GP.

Question 11. If a, b, c, d are in G.P. prove that;

(i) (a² + b²), (b² + c²), (c² + d²) are in G.P.

(ii) (a² – b²), (b² – c²), (c² – d²) are in G.P.

(iii)  are in G.P.

(iv) (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.

Solution:

(i) Given: a, b, c, d are in G.P.

So, a, b = ar, c = ar², d = ar³

Now,

(b² + c²)² = (a² + b²)(c² + d²)

(a²r² + a²r⁴)² = (a² + a²r²)(a²r⁴ + a²r⁶)

a⁴(r² + r⁴) = a²(1 + r²)a²r⁴(1 + r²)

a⁴r⁴(1 + r²)² = a⁴r⁴(1 + r²)²

L.H.S = R.H.S

⇒ (b² + c²)² = (a² + b²)(c² + d²)

Hence, proved (a² + b²), (b² + c²), (c² + d²) are in G.P.

(ii)  Given: a, b, c, d are in G.P.

So, a, b = ar, c = ar², d = ar³

Now,

(b² – c²)² = (a² – b²)(c² – d²)

(a²r² – a²r⁴) = (a² – a²r²)(a²r⁴ – a²r⁶)

a⁴(r² – r⁴)² = a²(1 – r²) a²r⁴ (1 – r²)

a⁴r⁴ (1 – r²)² = a⁴r⁴ (1 – r²)²

L.H.S = R.H.S

⇒ (b² – c²)² = (a² – b²)(c² – d²)

Hence, proved (a² – b²), (b² – c²), (c² – d²) are in G.P.

(iii) Given: a, b, c, d are in G.P.

So, a, b = ar, c = ar², d = ar³

Now,

L.H.S = R.H.S

Hence, proved  are in G.P.

(iv) Given: a, b, c, d are in G.P.

So, a, b = ar, c = ar², d = ar³

Now,

(ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²)

(a²r + a²r³ + a²r⁵)² = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶)

a⁴(r + r³ + r⁵)² = a²(1 + r² + r⁴) a²r² ( 1 + r² + r⁴)

a⁴r²(1 + r² + r⁴)² = a⁴r²(1 + r² + r⁴)²

L.H.S = R.H.S

⇒ (ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²)

Hence, proved (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.