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

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  • Last Updated : 21 Feb, 2021

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/loga m, 1/logb m, 1/logc 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

logm b2 = logm ac

Using property of log

logm b2 = logm a + logm

2logm b = logm a + logm c

2/logb m = 1/loga m + 1/logc m

Hence, proved 1/loga m, 1/logb m, 1/logc 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) \frac{1}{a^2+b^2},\ \frac{1}{b^2+c^2},\ \frac{1}{c^2+a^2}   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,

\left(\frac{1}{b^2+c^2}\right)^2=\left(\frac{1}{a^2+b^2}\right)\left(\frac{1}{c^2+d^2}\right)\\ \left(\frac{1}{a^2r^2+a^2r^4}\right)^2=\left(\frac{1}{a^2+a^2r^2}\right)\left(\frac{1}{a^2r^4+a^2r^6}\right)\\ \frac{1}{a^4(r^2+r^4)}=\frac{1}{a^2(1+r^2)}\times\frac{1}{a^2(r^4+r^6)}\\ \frac{1}{a^4r^4(1+r^2)^2}=\frac{1}{a^2r^4(1+r^2)(1+r^2)}\\ \frac{1}{a^4r^4(1+r^2)^2}=\frac{1}{a^2r^4(1+r^2)^2}

L.H.S = R.H.S

\Rightarrow\left(\frac{1}{b^2+c^2}\right)^2=\left(\frac{1}{a^2+b^2}\right)\left(\frac{1}{c^2+d^2}\right)\\ 

Hence, proved (\frac{1}{a^2+b^2}),\ (\frac{1}{b^2+c^2}),(\frac{1}{c^2+d^2}) 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.


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