# 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

b

^{2}= ac(b

^{2})^{n}= (ac)^{n}b

^{2}n = a^{n}c^{n}Now, use log on both the sides, we get,

log b

^{2n}= log (a^{n}c^{n})log (b

^{n})^{2}= log a^{n}+ log c^{n}2 log b

^{n}= log a^{n}+ log c^{n}Hence, proved log a

^{n}, log b^{n}, log c^{n}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

b

^{2}= acOn applying log on both sides with base m, we get

log

_{m}b^{2}= log_{m}acUsing property of log

log

_{m}b^{2}= log_{m}a + log_{m}c2log

_{m}b = log_{m}a + log_{m}c2/log

_{b}m = 1/log_{a}m + 1/log_{c}mHence, 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

b

^{2}= ac(k − 6)

^{2}= 4(k + 9)k

^{2}– 12k + 36 = 4k + 36k

^{2}– 16k = 0k = 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

a

_{1}+ a_{2}+ a_{3}= 15Here, 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)a

^{2}+ d^{2}+ 9 + 2ad + 6d + 6a = a^{2}+ a + 2da + 2d + 9a + 9(5 – a)

^{2}– 4a + 4(5 – a) = 025 + a

^{2}– 10a – 4a + 20 – 4a = 0a

^{2}– 18a + 45 = 0a

^{2}– 15a – 3a + 45 = 0a(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

a

_{1}+ a_{2}+ a_{3}= 21Here, 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)a

^{2}+ d^{2}+ 1 + 2ad – 2d – 2a = a^{2}+ a + 2da(7 – a)

^{2}– 3a + 1 – 2(7 – a) = 049 + a

^{2}– 14a – 3a + 1 – 14 + 2a = 0a

^{2}– 15a + 36 = 0a

^{2}– 12a – 3a + 36 = 0a(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)a

^{2}+ d^{2}+ 16 + 8a + 2ad + 8d = a^{2}+ 4a + 2da + 36a + 144 + 8dd

^{2}– 32a – 128(6 – a)

^{2}– 32a – 128 = 036 + a

^{2}– 12a – 32a – 128 = 0a

^{2}– 44a – 92 = 0a

^{2}– 46a + 2a – 92 = 0a(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, ar

^{2}a + ar + ar

^{2}= 56 -(1)Now, subtract 1, 7, 21 from the numbers, we get,

(a – 1), (ar – 7), (ar

^{2}– 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 + ar

^{2}– 21= (ar

^{2}+ a) – 222ar – 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 + 16r

^{2})/r = 5616 + 16r + 16r

^{2}= 56r16r

^{2}– 40r + 16 = 02r

^{2}– 5r + 2 = 02r

^{2}– 4r – r + 2 = 02r(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, ar

^{2}) is (8, 16, 32)

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

### (i) a(b2 + c^{2}) = c(a^{2} + b^{2})

### (ii) a^{2}b^{2}c^{2} [1/a^{3} + 1/b^{3} + 1/c^{3}] = a^{3} + b^{3} + c^{3}

### (iii) (a+b+c)^{2} / (a^{2} + b^{2} + c^{2}) = (a+b+c) / (a-b+c)

### (iv) 1/(a^{2} – b^{2}) + 1/b^{2} = 1/(b^{2} – c^{2})

### (v) (a + 2b + 2c) (a – 2b + 2c) = a^{2} + 4c^{2}

**Solution:**

(i)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= acLet LHS: a(b

^{2}+ c^{2})Now, on substituting b

^{2}= ac, we geta(ac + c

^{2})a

^{2}c + ac^{2}c(a

^{2}+ ac)On Substituting ac = b

^{2}we get,c(a

^{2}+ b^{2}) = RHSLHS = RHS

Hence, proved.

(ii)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= acLet LHS: a

^{2}b^{2}c^{2}[1/a^{3}+ 1/b^{3}+ 1/c^{3}]a

^{2}b^{2}c^{2}/a^{3}+ a^{2}b^{2}c^{2}/b^{3}+ a^{2}b^{2}c^{2}/c^{3}b

^{2}c^{2}/a + a^{2}c^{2}/b + a^{2}b^{2}/c(ac)c

^{2}/a + (b^{2})^{2}/b + a^{2}(ac)/c -(∵ b^{2}= ac)ac

^{3}/a + b^{4}/b + a^{3}c/cc

^{3}+ b^{3}+ a^{3}= RHSLHS = RHS

Hence, proved.

(iii)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= acLet LHS: (a + b + c)

^{2}/ (a^{2}+ b^{2}+ c^{2})(a + b + c)

^{2}/ (a^{2}+ b^{2}+ c^{2}) = (a + b + c)^{2}/(a^{2}– b^{2}+ c^{2}+ 2b^{2})= (a + b + c)

^{2}/ (a^{2}– b^{2}+ c^{2}+ 2ac) -(∵ b^{2}= ac)= (a + b + c)

^{2}/ (a + b + c)(a – b + c) -(∵ (a + b + c)(a – b + c) = a^{2}– b^{2}+ c^{2}+ 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,

b

^{2}= acLet LHS: 1/(a

^{2}– b^{2}) + 1/b^{2}On taking LCM, we get

1/(a

^{2}– b^{2}) + 1/b^{2}= (b^{2}+ a^{2}– b^{2})/(a^{2}– b^{2})b^{2}= a

^{2}/ (a^{2}b^{2}– b^{4})= a

^{2}/ (a^{2}b^{2}– (b^{2})^{2})= a

^{2}/ (a^{2}b^{2}– (ac)^{2}) -(∵ b^{2}= ac)= a

^{2}/ (a^{2}b^{2}– a^{2}c^{2})= a

^{2}/ a^{2}(b^{2}– c^{2})= 1/ (b

^{2}– c^{2})= RHS

LHS = RHS

Hence, proved.

(v)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= acLet LHS: (a + 2b + 2c) (a – 2b + 2c)

Now, on expanding, we get

(a + 2b + 2c) (a – 2b + 2c) = a

^{2}– 2ab + 2ac + 2ab – 4b^{2}+ 4bc + 2ac – 4bc + 4c^{2}= a

^{2}+ 4ac – 4b^{2}+ 4c^{2}= a

^{2}+ 4ac – 4(ac) + 4c^{2}-(∵ b^{2}= ac)= a

^{2}+ 4c^{2}= RHS

LHS = RHS

Hence, proved.

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

### (i) (ab – cd) / (b^{2} – c^{2}) = (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,

b

^{2}= acbc = ad

c

^{2}= bdLet LHS: (ab – cd) / (b

^{2}– c^{2})(ab – cd) / (b

^{2}– c^{2}) = (ab – cd) / (ac – bd)= (ab – cd)b / (ac – bd)b

= (ab

^{2}– bcd) / (ac – bd)b= [a(ac) – c(c

^{2})] / (ac – bd)b= (a

^{2}c – c^{3}) / (ac – bd)b= [c(a

^{2}– c^{2})] / (ac – bd)b= [(a + c) (ac – c

^{2})] / (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,

b

^{2}= acbc = ad

c

^{2}= bdLet 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}= a

^{2}+ b^{2}+ 2ab + 2(c^{2}+ b^{2}+ 2cb) + c^{2}+ d^{2}+ 2cd= a

^{2}+ b^{2}+ c^{2}+ d^{2}+ 2ab + 2(c^{2}+ b^{2}+ 2cb) + 2cd= a

^{2}+ b^{2}+ c^{2}+ d^{2}+ 2(ab + bd + ac + cb +cd) -(∵ c^{2}= bd, b^{2}= ac)(a + b + c)

^{2}+ d^{2}+ 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,

b

^{2}= acbc = ad

c

^{2}= bdLet LHS: (b + c) (b + d)

Now, on expanding, we get

(b + c) (b + d) = b

^{2}+ bd + cb + cd= ac + c

^{2}+ 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) a^{2}, b^{2}, c^{2}

### (ii) a^{3}, b^{3}, c^{3}

### (iii) a^{2} + b^{2}, ab + bc, b^{2} + c^{2}

**Solution:**

(i)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= acOn squaring both the sides we get,

(b

^{2})^{2}= (ac)^{2}(b

^{2})^{2}= a^{2}c^{2}Hence, proved a

^{2}, b^{2}, c^{2}are in G.P.

(ii)Given: a, b, c are in GP.By using the property of geometric mean,

b

^{2}= acOn squaring both the sides, we get

(b

^{2})^{3}= (ac)^{3}(b

^{2})^{3}= a^{3}c^{3}(b

^{3})^{2}= a^{3}c^{3}Hence, proved a

^{3}, b^{3}, c^{3}are in G.P.

(iii)Given: a, b, c are in GP.Using the property of geometric mean,

b

^{2}= aca

^{2}+ b^{2}, ab + bc, b^{2}+ c^{2}or (ab + bc)^{2}= (a^{2}+ b^{2}) (b^{2}+ c^{2})Let LHS: (ab + bc)

^{2}Now, on expanding, we get

(ab + bc)

^{2}= a^{2}b^{2}+ 2ab^{2}c + b^{2}c^{2}= a

^{2}b^{2}+ 2b^{2}(b^{2}) + b^{2}c^{2}-(∵ ac = b^{2)}= a

^{2}b^{2}+ 2b^{4}+ b^{2}c^{2}= a

^{2}b^{2}+ b^{4}+ a^{2}c^{2}+ b^{2}c^{2}-(∵ b^{2}= ac)= b

^{2}(b^{2}+ a^{2}) + c^{2}(a^{2}+ b^{2})= (a

^{2}+ b^{2})(b^{2}+ c^{2})= RHS

LHS = RHS

Hence, a

^{2}+ b^{2}, ab + bc, b^{2}+ c^{2}are in GP.

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

### (i) (a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d^{2}) are in G.P.

### (ii) (a^{2} – b^{2}), (b^{2} – c^{2}), (c^{2} – d^{2}) are in G.P.

### (iii) are in G.P.

### (iv) (a^{2} + b^{2} + c^{2}), (ab + bc + cd), (b^{2} + c^{2} + d^{2}) are in G.P.

**Solution:**

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

^{2}, d = ar^{3}Now,

(b

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

^{2}r^{2}+ a^{2}r^{4})^{2}= (a^{2}+ a^{2}r^{2})(a^{2}r^{4}+ a^{2}r^{6})a

^{4}(r^{2}+ r^{4}) = a^{2}(1 + r^{2})a^{2}r^{4}(1 + r^{2})a

^{4}r^{4}(1 + r^{2})^{2}= a^{4}r^{4}(1 + r^{2})^{2}L.H.S = R.H.S

⇒ (b

^{2}+ c^{2})^{2}= (a^{2}+ b^{2})(c^{2}+ d^{2})Hence, proved (a

^{2}+ b^{2}), (b^{2}+ c^{2}), (c^{2}+ d^{2}) are in G.P.

(ii)Given: a, b, c, d are in G.P.So, a, b = ar, c = ar

^{2}, d = ar^{3}Now,

(b

^{2}– c^{2})^{2}= (a^{2}– b^{2})(c^{2}– d^{2})(a

^{2}r^{2}– a^{2}r^{4}) = (a^{2}– a^{2}r^{2})(a^{2}r^{4}– a^{2}r^{6})a

^{4}(r^{2}– r^{4})^{2}= a^{2}(1 – r^{2}) a^{2}r^{4}(1 – r^{2})a

^{4}r^{4}(1 – r^{2})^{2}= a^{4}r^{4}(1 – r^{2})^{2}L.H.S = R.H.S

⇒ (b

^{2}– c^{2})^{2}= (a^{2}– b^{2})(c^{2}– d^{2})Hence, proved (a

^{2}– b^{2}), (b^{2}– c^{2}), (c^{2}– d^{2}) are in G.P.

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

^{2}, d = ar^{3}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 = ar

^{2}, d = ar^{3}Now,

(ab + bc + cd)

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

^{2}r + a^{2}r^{3}+ a^{2}r^{5})^{2}= (a^{2}+ a^{2}r^{2}+ a^{2}r^{4})(a^{2}r^{2}+ a^{2}r^{4}+ a^{2}r^{6})a

^{4}(r + r^{3}+ r^{5})^{2}= a^{2}(1 + r^{2}+ r^{4}) a^{2}r^{2}( 1 + r^{2}+ r^{4})a

^{4}r^{2}(1 + r^{2}+ r^{4})^{2}= a^{4}r^{2}(1 + r^{2}+ r^{4})^{2}L.H.S = R.H.S

⇒ (ab + bc + cd)

^{2}= (a^{2}+ b^{2}+ c^{2})(b^{2}+ c^{2}+ d^{2})Hence, proved (a

^{2}+ b^{2}+ c^{2}), (ab + bc + cd), (b^{2}+ c^{2}+ d^{2}) are in G.P.

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