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RD Chapter 20- Geometric Progressions Ex-20.5 Interview Questions Answers

Question 1 :

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

Answer 1 :

It is given that a, band c are in G.P.

b2 = ac {usingproperty of geometric mean}

(b2)n = (ac)n

b2n =an cn

Now, apply log on both the sideswe get,

log b2n =log (an cn)

log (bn)2 = log an +log cn

2 log bn =log an + log cn

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.

Answer 2 :

Given:

a, b and c are in GP

b2 =ac {property of geometric mean}

Apply log on bothsides with base m

logm b2 =logm ac

logm b2 =logm a + logm c {using property of log}

2logm b= logm a + logm c

2/logb m= 1/loga m + 1/logc m

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 form three consecutive terms of a G.P.

Answer 3 :

Let a = k + 9; b =k−6; and c = 4;

We know that a, b andc are in GP, then

b2 =ac {using property of geometric mean}

(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 from a G.P. find the numbers.

Answer 4 :

Let the first term ofan A.P. be ‘a’ and its common difference be‘d’.

a1 + a2 +a3 = 15

Where, the threenumber are: a, a + d, and a + 2d

So,

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

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

d = 5 – a … (i)

Now, according to thequestion:

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

they are in GP, thatis:

(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,the A.P is 3, 5, 7

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

 The numbers are3, 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.

Answer 5 :

Let the first term ofan A.P. be ‘a’ and its common difference be‘d’.

a1 + a2 +a3 = 21

Where, the threenumber are: a, a + d, and a + 2d

So,

3a + 3d = 21 or

a + d = 7.

d = 7 – a …. (i)

Now, according to thequestion:

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,the A.P is 3, 7, 11

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

 The numbers are3, 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.

Answer 6 :

Let the first term ofan A.P. be ‘a’ and its common difference be‘d’.

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

Given:

a + b + c = 18

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

d = 6 – a … (i)

Now, according to thequestion:

a + 4, a + d + 4, anda + 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,the A.P is -2, 6, 14

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

 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.

Answer 7 :

Let the three numbersbe a, ar, ar2

According to thequestion

a + ar + ar2 =56 … (1)

Let us subtract 1,7,21we get,

(a – 1), (ar – 7), (ar2 –21)

The above numbers arein AP.

If three numbers arein AP, by the idea of the arithmetic mean, we can write 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 thevalue of a in equation (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 valueof r in equation (2) we get,

a = 16/r

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

= 8 or 32

 The threenumbers 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

Answer 8 :

(i) a(b2 +c2) = c(a2 + b2)

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

Let us consider LHS:a(b2 + c2)

Now, substituting b2 = ac,we get

a(ac + c2)

a2c + ac2

c(a2 +ac)

Substitute ac = b2 weget,

c(a2 +b2) = RHS

 LHS = RHS

Hence proved.

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

Let us consider 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 [by substituting the b2 =ac]

ac3/a + b4/b+ a3c/c

c3 + b3 +a3 = RHS

 LHS = RHS

Hence proved.

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

Let us consider 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) [Since, b2 =ac]

= (a+b+c)2 /(a+b+c)(a-b+c) [Since, (a+b+c)(a-b+c) = a2 – b2 +c2 + 2ac]

= (a+b+c) / (a-b+c)

= RHS

 LHS = RHS

Hence proved.

(iv) 1/(a2 –b2) + 1/b2 = 1/(b2 – c2)

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

Let us consider LHS:1/(a2 – b2) + 1/b2

Let us take LCM

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

= a/(a2b2 – b4)

= a2 /(a2b2 – (b2)2)

= a2 /(a2b2 – (ac)2) [Since, b2 =ac]

= a2 /(a2b2 – a2c2)

= a2 /a2(b2 – c2)

= 1/ (b2 –c2)

= RHS

 LHS = RHS

Hence proved.

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

Let us consider LHS:(a + 2b + 2c) (a – 2b + 2c)

Upon expansion 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 [Since, 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)

Answer 9 :

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

bc = ad

c2 =bd

Let us consider 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) (a + b + c + d)2 =(a + b)2 + 2(b + c)2 + (c + d)2

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

bc = ad

c2 =bd

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

Let us expand

(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) [Since, c2 = bd, b2 = ac]

You can visualize theabove expression by making separate terms for (a + b + c)2 + d2 +2d(a + b + c) = {(a + b + c) + d}2

 RHS = LHS

Hence proved.

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

bc = ad

c2 =bd

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

Upon expansion we get,

(b + c) (b + d) = b2 +bd + cb + cd

= ac + c2 +ad + cd [by using property of geometric mean]

= 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

Answer 10 :

(i) a2, b2,c2

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

on squaring both thesides we get,

(b2)2 =(ac)2

(b2)2 =a2c2

a2, b2,c2 are in G.P.

(ii) a3, b3,c3

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

on squaring both thesides we get,

(b2)3 =(ac)3

(b2)3 =a3c3

(b3)2 =a3c3

a3, b3,c3 are in G.P.

(iii) a2 + b2,ab + bc, b2 + c2

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

a2 + b2,ab + bc, b2 + c2 or (ab + bc)2 =(a2 + b2) (b2 + c2) [byusing the property of GM]

Let us consider LHS:(ab + bc)2

Upon expansion we get,

(ab + bc)2 =a2b2 + 2ab2c + b2c2

= a2b2 +2b2(b2) + b2c2 [Since, ac = b2]

= a2b2 +2b4 + b2c2

= a2b2 +b4 + a2c2 + b2c2 {againusing 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.


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