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

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Question 1 :

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

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.

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.

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.

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.

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.

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.

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

(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)

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

Given that a, b, c arein GP.

By using the propertyof geometric mean,

b2 =ac

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

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

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

(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.

krishan

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