CHAPTER-12 - Geometrical Progression - Amazon AWS

Class XI www.vedantu.com RS Aggarwal Solutions

CHAPTER-12 - Geometrical Progression

Exercise 12A

Solution 1 : Given: GP is 2, 6, 18, 54…. The standard form of GP is, a, ar, ar2, ar3….

The common ratio is r . First term in the given GP, a1= a = 2 Second term in GP, a2 = 6 The common ratio, 2

1

ara

= 6r 32

= =

The , nth term of GP is, an = arn-1

So, the 6th term in the GP, a6 = ar5 = 2 x 35= 486 nth term in the GP, an= arn-1

= 2.3n-1 Hence, the 6th term = 486 and nth term = 2.3n-1 of the given G.P.

Solution 2: Given GP is 2, 2√2, 4, 8√2 ….. The standard form of GP is , a, ar, ar2, ar3….

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The common ratio is r . First term in the given GP, a1 = a = 2 Second term in GP, a2 = 2√2

The common ratio, 2

1

ara

=2 2r 2

2= =

Now, nth term of GP is, an = arn-1

So, the 17th term in the GP, a17 = ar16 = 2 x (√2)16 = 512 nth term in the GP, an = arn-1

= 2(√2) n – 1= (√2) n + 1 Hence, the 17th term = 512 and the nth term = (√2) n + 1 of the given G.P.

Solution 3 : Given GP is 0.4, 0.8, 1.6…. The standard form of GP is, , a, ar, ar2, ar3…. The common ratio is r . First term in the given GP, a1 = a = 0.4 Second term in GP, a2 = 0.8

The common ratio, 2

1

ara

= 0.8r 20.4

= =

Now, nth term of GP is, an = arn – 1 So, the 7th term in the GP,



a7 = ar6 = 0.4 x 26= 25.6 nth term in the GP, an = arn – 1 = (0.4)(2) n – 1= (0.2)2n Hence, the 7th term = 25.6 and the nth term = (0.2)2n. of the given G.P.

Solution 4 : Given GP is 3 1 1 2, , , ...4 2 3 9

− −

The standard form of GP is, , a, ar, ar2, ar3…. The common ratio is r .

The first term in the gives GP, a=a1= 34

−

The second term of the GP a2= 12

The common ratio, 2

1

ara

=

122r 3 3

4

= − = −

Now, nth term of GP is, an = arn-1 So, the 10th term, a10 = ar9

9 910

3 2 128a ar ( )( )4 3 6561

= = − − =

Now, the required nth term, an = arn-1 n 1 n 1 n

n3 2 9 2a ar ( )( ) ( )( )4 3 8 3

− −= = − − = −



Hence, the 10th term 10a = 1286561

and the nth term na n9 2( )( )8 3

= − of the

given G.P. Solution 5 : Given GP is 3, 6, 12, 24…. The standard form of GP is, a, ar, ar2, ar3…. The common ratio is r . First term in the given GP, a1 = a = 3 Second term in GP, a2 = 6

The common ratio, 2

1

ara

= r = 2

Let us consider 3072 as the nth term of the GP. Now, nth term of GP is, n 1

na ar −=

3072 = 3.2n-1 2n = 211 (on comparing)

n = 11 Hence, 3072 is the 11th term of the given G.P. Solution 6 : Given GP is The standard form of GP is, a, ar, ar2, ar3…. The common ratio is r .

The first term in the gives GP, a=a1= 14

−



The second term of the GP a2 = - 12

The common ratio, 2

1

ara

= 4r2

= − = -2

Let us consider -128 as the nth term of the GP. Now, nth term of GP is, n 1

na ar −=

n 11128 ( 2)4

−− = −

(−2)n = 1024 = (-2)10 (on comparing) n = 10 Therefor, -128 is the 10th term of the given G.P. Solution 7 : Given GP is √3, 3, 3√3…. The standard form of GP is, a, ar, ar2, ar3…. The common ratio is r . First term in the given GP, a1 = a = √3 Second term in GP, a2 = 3

Now, the common ratio 2

1

ara

=3r 33

= =

Now, nth term of GP is, an = arn – 1 729 = √3 (√3) n – 1 √3n = √312 (on comparing) n = 12 Therefor, 729 is the 12th term of the given G.P.



Solution 8 : The nth term of a GP is an = arn-1 It’s given in the question that 5th term of the GP is 80 and 8th term of GP is 640. So, a5 = ar4 = 80 → (1) a8 = ar7 = 640 → (2)

73

4( ar 640r 8

802

(1 ar))→ = = =

Common ratio, r = 2, ar4 = 80 16a = 80 a = 5

Our required GP is of the form 2 3a, ar, ar , ar .…

First term of GP, a = 5 Second term of GP, ar = 5 x 2 =10 Third term of GP, ar2 = 5 x 22 = 20 Fourth term of GP, ar3 = 5 x 23 = 40 Fifth term of GP, ar4 = 5 x 24 = 80 And so on... The required GP is 5, 10, 20, 40, 80… Solution 9 : The nth term of a GP is n 1

na ar −=



It’s given in the question that 4th term of the GP is and 7th term of GP is .

Our required GP is of form 2 3a, ar, ar , ar .…

So 34

1a ar28

= = ……(1)

67

1a ar486

= = …..(2)

63

3ar 1r

27a(2)(1) r

→ = = −

Common ratio 1r3

= −

3 1ar18

= −

3a2

= −

The first term of GP, a= 32

−

The second term of GP ar = 3 1 12 3 2

− ×− =

And so on

The required GP is 3 1 1, , ...2 2 6

− −

Solution 10 : Let GP is A, AR, AR2, and AR3…. Now , nth term of GP is an = ARn-1

Now , 5th term a5 = AR4 = a …. (1)



8th term a8 = AR7 = b ……… (2) 11th term a11 =AR10 = c ………. (3) Dividing equation (3) by (2) and (2) by (1), We get

103

7AR cR

bAR= = …………. (4)

73

4AR bR

aAR= = …………. (5)

From 4 and 5, we get c bb a=

2b ac∴ = Hence proved. Solution 11 : It is given that the first term of GP is -3. So, a = -3 It is also given that the square of the second term is equal to its 4th term. ∴ (a2)2 = a4 nth term of GP, n 1

na ar −=

So, a2 = ar; a4 = ar3 (ar) 2 = ar3→ a = r = -3 (Given) Now, the 7th term in the GP, a7 = ar6 a7 = (-3)7 = -2187 Therefor, the 7th term of GP is -2187.



Solution 12 : The given GP is 8, 4, 2, 11024

……(1)

First term in the GP, a1 = a = 8 Second term in the GP, a2 = ar = 4 The common ratio, r = 1/2

The last term in the given GP is 11024

Second last term in the GP = an-1 = arn-2 On reversing the current G.P. the new series forms another new G.P. is in the form , arn-1, arn-2, arn-3….ar3, ar2, ar, a → (2)

Common ratio of this new GP is ' 1rr

= .

So, our new common ratio 'r 2=

a = 11024

So, 6th term of the GP (2), a6 = ar'5

= 51 1.21024 32

=

Therefor, the 6th term from the end of the given GP is 132

.

Solution 13: The given GP is 2 2 2, , ......,16227 9 3

….(1)



First term in the GP, a1 = a = 227

Second term in the GP, a2 = ar = 29

The common ratio, r = 3 The last term in the given GP is an = 162. Second last term in the GP = an-1 = arn-2 On reversing the current G.P. the new series forms another new G.P. is in the form , arn-1, arn-2, arn-3….ar3, ar2, ar, a → (2)

Common ratio of this new GP is ' 1rr

=

So ' 1r3

=

So, 4th term of the GP (2), a4 = ar3

31162 63

× =

Hence, the 4th term from the end of the given GP is 6. Solution 14 : According to the question, a, b and c are the pth, qth and rth term of GP. Let us assume the required GP as A, AR, AR2, AR3… Now, the nth term in the GP, an = ARn-1 pth term, ap = ARp-1 = a → (1)



qth term, aq = ARq-1 = b → (2) rth term, ar = ARr-1 = c → (3)

p 1p q

q 1AR aR

bA(1)(2) R

−−

−→ = = …(i)

q 1q r

r 1AR bR

cA(2)(3) R

−−

−→ = = …..(ii)

r 1r p

p 1AR cR

aA(3)(1) R

−−

−→ = = ….(iii)

Taking logarithm on both sides of equation (i), (ii) and (iii). (p q)log R loga lo

loga log b(p q)lo R

b

g

g−

∴ − =

− = − ….(4)

(q r) log R log b lolog b logc(q r)

lo R

c

g

g−

∴ − =

− = − ……(5)

(r p)log R logc lologc loga(r p)

lo R

a

g

g−

∴ − =

− = − ………(6)

Now multiply log c with(4), log a with (5) , log b with (6) then add all. We get, (p q)logc (q r) loga (r p)log b

loga log b log b logc logc loga( ) logc ( ) loga ( ) log blog R log R log R

− + − + −− − −

= + +



On solving the above equation, we will get, (p – q) log c + (q – r) log a + (r – p) log b = 0 Hence proved. Solution 15 : Given that the third term of the GP, a3 = 4 Let us assume the GP mentioned in the question be,

22

A A, ,A,AR,AR ...RR

With the first term 2AR

and common ratio R

Now, the third term in the assumed GP is A. So, A = 4 (given) Now, Product of the first five terms of GP

2 52

A A A AR AR ARR

× × × × =

So the required product = A5 =45 = 1024 Therefor the product of first five terms of a GP with its third term 4 is 1024. Solution 16 : Let us first consider a finite GP. A, AR, AR2….ARn -1, ARn. Where n is finite.



Product of first and last terms of the GP = A.ARn = A2Rn → (a) Now, nth term of the GP from the beginning = ARn-1 → (1) Now, starting from the end, First term = ARn

Last term = A and Common ratio = 1r

So an nth term from the end of GP An=(ARn)( n 11

R − )= AR

So, the product of nth terms from the beginning and the end of the considered GP, from (1) and (2) = (ARn-1) (AR) = A2Rn → (b) So, from (a) and (b) its proved that the product of the terms equidistant from the beginning and the end is the product of first and last terms of a finite GP.

Solution 17 : Given that a bx b cxa bx b cx+ +

=− −

……(1)

On multiplying (1) and solving. We get, (a + bx)(b – cx) = (b + cx)(a-bx) ab – acx + b2x – bcx2 = ba –b2x + acx – bcx2 2b2x = 2acx b2 = ac → (i) → b2 = ac So, from (i) b, is the geometric mean of a and b.



So, a, b, c are in GP. b cx c dxb cx c dx+ +

=− −

( given)…….(2)

On multiplying (2) and solving, (b + cx)(c – dx) = (c + dx)(b – cx) bc – bdx + c2x – cdx2 = cb – c2x + bdx – dcx2 2c2x = 2bdx c2 = bd → (ii) So, from (ii), c is the geometric mean of b and d. Therefore , b, c, d is in G.P. Hence, a, b, c, d are in GP. Solution 18 : Given that x2 – 3x + p = 0 Now, a and b are roots So, a + b = 3 and ab = p Given that , x2 – 12x + q = 0 Now c and d are roots So, c + d = 12 and cd = q a, b, c, d are in GP. (Given) Similarly A, AR, AR2, AR3 also forms a GP with common ratio R. a + b = 3



A + AR = 3 3 1 RA= + ….(1)

c + d = 12 AR2 + AR3 = 12 AR2 (1 + R) = 12……. (2) Substituting value of (1 + R) in (2). R = 2 Now, A = 1 Now, the GP required is A, AR, AR2, and AR3 1, 2, 4, 8…is the required GP. So, a = 1, b = 2, c = 4, d = 8 From (2) and (4), ab = p and cd = q So, p = 2, and q = 32. q p cd ab 34 17q p cd ab 30 15+ +

= = =− −

So (q+p) : (q-p) = 17:15.

Exercise 12C

Solution 1 A : Given that 1 + 3 + 9 + 27 + …. To 7 terms



Here, a = 1 and r = 3 ÷ 1 = 3 n = 7 terms

7

n3 1S 13 1−

=−

→ n2187 1S

3 1−

=−

→ n2186S

2= → nS 1093=

So the sum of nth term of G.P is 1093.

Solution 1 B : 1 + + 3 + 3 +….. to 10 terms Here,

a = 1 and r = 3 ÷ 1= 3 = 1.732 n = 10 terms 10

n( 3) 1S 1.

3 1−

=−

→ n242.929 1S

0.732−

= → n241.929S0.732

=

nS 330.504=

So the sum of nth term of G.P is 330.504. Solution 1 C : 0.15 + 0.015 + 0.0015 + …. To 6 terms Here, a = 0.15 and r = 0.015 ÷ 0.15 = 0.1 and n = 6 terms

6

n1 0.1S 0.151 0.1−

= ×−

n1 0.000001S 0.15

0.9−

= ×

n0.999999S 0.15

0.9= ×



nS 16.67=

So the sum of nth term of G.P is 16.67.

Solution 1 D : to 9 terms Here,

a = 1 and r = - 1 112 2÷ = − and n = 9 terms

n

nr 1S ar 1−

=−

9

n

11 ( )2S 1 11 ( )2

−−= ×

−−

n

11512S 11

2

+=

+

nS 171∴ =

So the sum of nth term of G.P is171.

Solution 1 E: ……to 8 terms Here,



a = 2 and r = 1 1222

÷ = n = 8 terms

n

nr 1S ar 1−

=−

, when |r|<1

8

n

11 ( )2S 11

2( )2

−= ×

−

n252 5S128

= ×

n255S

82

12=

So the sum of nth term of G.P is 255 2128

.

Solution 1 F: …. to 6 terms Here, a = 2

9 and r = 1 2 3 1.5

3 9 2− ÷ = − =

n = 6 terms n

nr 1S ar 1−

=−

, when |r|<1

6

n2 1.5 1S9 1.5 1

−= ×

−

nS 4.62=

So the sum of nth term of G.P is 4.62.



Solution 2 A : + … to n terms

Here, a = 7

r = 21 7 3÷ =

n terms n

nr 1S ar 1−

=−

, when |r|<1

n

n3 17

13S −

∴ = ×−

n

n3 373

1 1S1 13−

∴ = × ×++−

n

n1)(( 3 3 1S

17

3)+−

= ×−

n

n1)(7( 3 3 1)S

13−

∴+

=−

Solution 2 B : … to n terms

Here, a = 1 And r = - 13 3

1 1÷ = − n terms = ?

n

nr 1S ar 1−

=−

, when |r|<1



n

n

113S 1 113

−∴ = ×

−

n

n

113S 2

3

−=

Solution 2 C : … to n terms Here, a = 1 and r = -a/1 = -a n terms= ?

∴n

n(( a) 1)S 1

a 1−

= ×− −−

[Multiplying both numerator and denominator by -1] n

n1 ( a)S

1 a− −

=+

Solution 2 D : x3 + x5 + x7 + …. To n terms Here, a = x3 and r = 5 3 2x x x÷ = and n terms =?

n

nr 1S ar 1−

=−

, when |r|<1



2 n3

n 2((x ) 1)S x

x 1−

= ×−

2 nn

n1)( 1)S

(x 1)(x )x x x

1( +−

∴ =− +

Solution 2 E : The given expression can be written as = (x2+ xy) + (x4 + x2y2 ) + (x 6 + x3y3 ) + …. To n terms = (x2 + x4 + x6 + … to n terms ) + ( xy + x2y2 + x3y3 + … to n terms )

∴2n n n

2n 2

x 1 x y 1S x xyxy 1x 1

− −= × + ×

−−

2 n 1 n 1

n

nnx (x 1)( 1) x y 1S(x 1)

x(x 1) xy 1

+ +− −= +

− + −+

Solution 3 :(i) This can be written as

4 64 6

2 4 62 6

2

4

2 1 1( ) (x 2) (x 2) ...to _ n _ termsx x

1 1 1(x x x ......n _ terms) ( .....n _ terms) (2 2 2 ...n _ terms)x x x

1x 2x

+ + + + + + +

= + + + + + + + + + +

+ +

2 4 6n 2 4 6

1 1 1S (x x x ......n _ terms) ( .....n _ terms) 2nx x x

= + + + + + + +

.a = 2x , 21x

nn 22

2n 2 2

2

1( ) 11) 1 xS x 1x 1((

x 1x

x ) −−= × + ×

− −+2n



On solving n n

2

n

2 n 11)( 1) ( ) 1xS 2n

(x 1)

x (x x

(x 1)

− + −∴ = +

− +

+

(ii) If we divide and multiply the terms by (x-y)

2 2 2 2 2 2(x y)(x y) (x y)(x xy y ) (x y)(x x y xy y ) ...to _ n _ terms(x y)

− + + − + + + − + + + +−

2 2 3 3 4 4(x y ) (x y ) (x y )....to _ n _ terms(x y)

− + − + −−

2 3 4 2 3 4(x x x ....n _ terms) (y y y ....n _ terms)(x y)

+ + + − + + +−

a = 2x ,y2 r = x, y and n terms =?

n n2 2

n

1) (y) 1x yx 1 y 1S

x y

(x − −× + ×− −=

−

Solution 4:.Given …. To 2n terms We can write it as

3 23 3 4 4( ....to _ n _ terms) ( ....to _ n _ terms)5 55 5

= + + + + +



Here, a = 35

, 45

And r = 23 23 1 an3 4 4 1

5 55 5d

5 5÷ = ÷ =

n terms = ?

n

n

n

2

2

11 ( )3 5S 1

114 5155 )5

15

1(

−= ×

−×

−

−+

n

n

n

2

2

11 ( )3 5S 2

114 545

554

5

−×

−= × +

2n 1 n

n

155S (1 )8

15

−−= + −

Solution 5: (i) We can write this as (2 + 31)+(2+32) + (2 +33)+… to 10 terms = ( 2+2+2+… to 10 terms) + ( 3+32+33+… to 10 terms) = 2×10 + (3+32+33+… to 10 terms) = 20 + (3+32+33+… to 10 terms) Now , a = 3 and r = 3 and n = 10 terms

n

nr 1S ar 1−

=−

, when |r|<1

10

n)

1(3 1S 3

3−

= ×−

n59049 1S 3

2−

= × = 88572



Thus, sum of the given expression is = 20 + (3+32+33+… to 10 terms) = 20 + 88572 = 88592 So the sum of the series is 88592. (ii) The given expression can be written as, (21 + 31-1) + (22 + 32-1) + …to n terms = (2 + 30) + (22+ 31) + …to n terms = (2 + 1) +(22 + 3 ) + …to n terms = (2 + 22 + …to n/2 terms) + (1 + 3 + … to n/2 terms)

Here, a = 2, 1 And r = 2, 3 and n = n2

terms

n

nr 1S ar 1−

=−

, when |r|<1

n n2 2

n1) 1)S 2 1

2 1(2

3 1(3− −

= × + ×− −

nn 212

n3 1S 2 2

2+ −

= − +

(iii) We can rewrite the given expression as (51 + 52 + 53+ …to 8 terms) Here, a = 5 And r = 5 and n = 8 terms



n

nr 1S ar 1−

=−

, when |r|<1

8

n1)S 5 5

5 1( −

= ×−

n(390625 1)S 5

4−

= ×

nS =488280

So the sum of nth term of G.P is 488280. Solution 6 : The expression can be written as 8(1+ 11 + 111+ … to n terms)

= 89

(9 + 99+ 999 + … to n terms)

= 89

((10-1) + (100-1) + (1000-1) + … to n terms )

= 89

( ( 10 + 100 + 1000 + … to n terms) – ( 1+1+1+ … to n terms)

= 89

( ( 10 + 100 + 1000 + … to n terms) – n)

Here, a = 10 r = 10 n = n terms n

nr 1S ar 1−

=−

, when |r|<1

n

n(10 1)S 1010 1

−= ×

−



n 1

n(10 10)S

9

+ −=

∴ The sum of the given expression is

= 89

( ( 10 + 100 + 1000 + … to n terms) – n)

= 89

(n 1

91 )( 0 10+ − -n)

(ii) The given expression can be written as = 3( 1+11+111+ …to n terms)

= 39

( 9+99+999+ … to n terms )

= 39

( (10-1) + (100-1) + (1000-1) + … to n terms )

= 39

( ( 10+100+1000+ …to n terms ) – (1+1+1+ … to n terms) )

= 39

( (10+100+1000+ to n terms) – n )

Here, a = 10 And r = 10 and n = n terms n

nr 1S ar 1−

=−

, when |r|<1

n

n

n 1

n

1)S 1010

(10

(101

10)S9

+

−= ×

−−

=




= 39

( (10+100+1000+ to n terms) – n )

= 39

(n 1

91 )( 0 10+ − -n)

(iii) The given expression can be written as = 7(0.1+0.11+0.111+ … to n terms)

= 79

( 0.9+0.99+0.999+ … to n terms )

= 79

( (1-0.1)+(1-0.01)+(1-0.001)+ … to n terms)

= 79

( (1+1+1+ … to n terms )–(0.1+0.01+0.001+… to n terms ))

= 79

( n – (0.1+0.01+0.001+ … to n terms ) )

Here, a = 0.1 And r = 0.1 and n = n terms n

nr 1S ar 1−

=−

, when |r|<1

n

n(1 0.1 )S 0.11 0.1

= ×−−

n

n( )S 1 00 1

0..1.9

−= ×

n

n1 0S

9.1−

=




= 79

( n – (0.1+0.01+0.001+ … to n terms ) )

= 79

(n-(n1

90.1− ))

Solution 7 : It is given that Sn = ( 2n – 1)…..(i) As we know that

n

nr 1S ar 1−

=−

……………………..(ii)

From (i) and (ii) n

n r 1(2 1) ar 1−

− =−

⇒n n2 1 r 11 a

2 1 r 1− −

× =− −

Now, a = 1 and r = 2 The G.P. series is 1, 2, 4, 8, 16,……to n terms Since, the given progression is a G.P. series with the common ratio being 2. Solution 8 : Given that the ratio of the sum of the first three terms is to first six terms is 125 : 152.

Sum of first 3 terms =3r 1ar 1−−



Sum of first 6 terms =6r 1ar 1−−

∴

3

6

r 1a 125r 1152r 1a

r 1

−− =−−

⇒3

6r 1 125

152r 1−

=−

⇒ 152r3 – 152= 125r6-125 ⇒ 125r6-152r3-125+152 = 0 ⇒ 125r6 – 152r3 + 27 = 0 ⇒ 125r6 – 125r3 – 27r3 + 27 = 0 ⇒ (125r3 - 27) (r3-1)= 0 Either 125r3 -27 = 0 or r3-1 = 0 Either 125r3=27 or r3=1

Either r3 = 27125

or r=1

Either r= 35

or r=1

Since r ≠ 1

∴ r = 35

So the common ratio of the G.P. is 35

.



Solution 9 : Given Series: 3 + 6 + 12 + … + 1536. Here, r = 6 ÷ 3 = 2 Tn = arn-1 ⇒1536 = 3 × 2n-1 ⇒1536 ÷ 3 = 2n ÷ 2 ⇒1536 ÷ 3 × 2 = 2n ⇒1024 = 2n ⇒210 = 2n (on comparing) ∴ n = 10 Now , a = 3 and r = 2 and n = 10 terms

10

n)

1(2 1S 3

2−

= ×−

nS =3x(1024-1)

nS =3069

So, the sum of nth term of G.P is 3069. Solution 10: Given that the series 2 + 6 + 18 + …. And Sn = 728 Here, a = 2 and r = 6 ÷ 2 = 3 Sn = 728

∴ 728 =2 xn3 13 1−−



⇒728 =2 x n3 12− ⇒728 =3n – 1

⇒729= 3n ⇒36=3n ∴n=6 Therefor, 6th terms must be taken to reach the desired answer. Solution 11: Tn = arn-1 ⇒486 = a(3)n-1 ⇒486 = a( 3n ÷ 3) ) ⇒486 × 3 = a(3n) ⇒1458 = a(3n ) ………(i)

∴ 728 =a xn3 13 1−−

⇒728 =a x n3 12−

⇒728 × 2 = a (3n)-a …… [Putting a(3n ) = 1458 from (i)] ⇒1456 = 1458 -a ⇒1456-1458 = -a ⇒-2=-a [Multipying both sides by -1] ⇒a = 2 Therefor the first term of G.P. is 2. Solution12: Tn = arn-1



⇒ 8 11 27 r81

−= ×

⇒ 71 r2187

=

⇒ 7 71( ) r3

=

∴ r = 13

Here, a = 27 And r = 13

and n = 10 terms

n

nr 1S ar 1−

=−

, when |r|<1

10

n

11 ( ) )3S 27

1

(

13

−= ×

−

n

11 )590

(49S 27 2

3

−= × ⇒ n

39524S 27190683

= ×

n39524S

729=

So the sum of nth term of G.P is 39524729

.

Solution: Given that 2nd term = ar2-1 = ar1 and 5th term = ar5-1 = ar4 Dividing the 5th term using the 3rd term



41

ar 161ar

2

=−

⇒ r3 = - 18

⇒ r = -1/2 and a = 1

As we know that n

nr 1S ar 1−

=−

, when |r|<1

8

n

11 ( ) )2

(S 1 11

2

−−= ×

−−

n

11 )256S 1

2

(

3

−= × ⇒ n

170S256

=

So the sum of nth term of G.P is170256

.

Solution 14 : Given that 4th term = ar4-1 = ar3 = 127

7th term = ar7-1= ar6 = 1729

Dividing the 7th term by the 4th term,

6

3

1ar 729

1ar27

= ⇒ r3 = 127

⇒ r = 13

a. 127

= 127

⇒ a= 1



Here, a = 1 and r = 13

and n = n terms

As we know that n

nr 1S ar 1−

=−

, when |r|<1.

n

n

11 ( ) )3S 1 1

(

13

−= ×

− ⇒

n 1

n

1( )3S

3

2−−

=

Solution 15 : Let the terms of the G.P. be a, ar, ar2, ar3, … , arn-2, arn-1

As we know that

n

nr 1S ar 1−

=−

, when |r|<1

n

nr 1S ar 1−

=−

The odd terms of this series are a, ar2, ar4, … , arn-2 Sum of the odd terms

⇒n2( )2

n 21)S a

r(r

1−

= ×−

⇒n

n1)S a

(r 1) r(

( 1)r −

= ×− +



By the problem,

⇒nr 1ar 1−−

= a x n 1)a

(r 1)((

1r

r )−

×− +

⇒1= 5r 1+

⇒ r +1 =5 ⇒∴ r = 4 Thus, the common ratio (r) = 4 Solution 16 :

As we know that n

nr 1S ar 1−

=−

Sum of (n+1)th term to 2nth term = Sum of the first 2nth term – the sum of 1st term to nth term

= 2n 2nr 1 r 1a ar 1 r 1− −

−− −

=2n n(ar a) (ar a)

r 1− − −

− =

n nar (r 1)r 1

−−

The ratio of the sum of first n terms of the G.P. to the sum of the terms from (n + 1)th to (2n)th term

=

n

2n n

r 1ar 1

ar (r 1)r 1

−−−

−

= n1r

Hence Proved.

Exercise 12 D



Solution 1 : Given that : (i) Principal – 15625

(ii) Time – 3 years (iii) Rate – 8% per annum

By using this Formula : trA P(1 )100

= +

3 38 108A 15625(1 ) 15625( )100 100

= + =

A = 19683 The amount after three Years is Rs. 19683 Solution 2 : Given that : (i) Principal – 80000 (ii) Time – 3 days (iii) Rate – 15% per annum Deduction = P × R × T

15 380000100 365

= × × = 98.63

The final amount after deduction = 80000 – 98.63 = 79901.37 The value of the machine after 3 days is Rs. 79901.37 Solution 3 : Given that : (i) Three years back population -10000



(ii) Time – 3 years (iii) Rate – 20% per annum Number of people migrated on the very first year is 20% of 10000

⇒10000 20 2000100

×=

People left after migration in the very first year = 10000 – 2000= 8000 Number of people migrated in the second year is 20% of 8000

⇒ 8000 20 1600100×

=

People left after migration in the second year = 8000 – 1600 = 6400 Number of people migrated in the third year is 20% of 6400

⇒ 6400 20 1280100×

=

People left after migration in the third year = 6400 – 1280 = 5120 The present population is 5120 Solution 4 : Given that : (i) Principal – 5000 (ii) Time – 10 years (iii) Rate – 10% per annum


= +



10

10

10A 5000(1 )100

110A 5000( )100

= +

=

A= 12970 The amount after years will be Rs. 12970 Solution 5 : Given: (i) Principal – 156250 (ii) Time – 5 years (iii) Rate – 20% per annum


= −

5

5

20A 156250(1 )100

80A 156250( )100

= −

=

A =156250 x 0.32 = 51200 The amount after five years will be Rs.51200 Solution 6 : Given: (i) Initially, there were 50 bacteria (ii) Rate – 100% per hour


= +



(i) For 2nd hour

⇒ No of bacteria = 210050(1 )100

+

⇒ No of bacteria = 2200A 50( )100

=

⇒ No of bacteria =50(2)2 ⇒ No of bacteria =200 (ii) For 5th hour

⇒ No of bacteria = 510050(1 )100

+

⇒ No of bacteria = 5200A 50( )100

=

⇒ No of bacteria =50(2)5 ⇒ No of bacteria =1600 (iii) For nth hour

⇒ No of bacteria = n10050(1 )100

+

⇒ No of bacteria = n200A 50( )100

=

⇒ No of bacteria =50(2)n ⇒ No of bacteria =2n 50



Number of bacteria in a 2nd hour will be 200, the number of bacteria in a 5th hour will be 1600 and number of bacteria in an nth hour will be 2n.50. Exercise 12 E Solution 1 : To prove: pth, qth and rth terms of any GP are in GP. Given that : (i) p, q and r are in AP As p, q, r are in A.P. ⇒ q – p = r – q = d = common difference … (i) By using this Formula of GP, n 1

nT ar −=

Then, the pth term will be arp-1

The qth term will be arq-1

The rth term will be arr-1

Considering pth term and qth term th q 1

th p 1q _ termp _ te

ararm r

−

−=

q 1 p 1th

thq _ term rp _ term

− − +=

th

thq pq _ term r

p _ term−=



From eqn. (i) q – p = d th

thdq _ term r

p _ term=

Considering qth term and rth term th r 1

th q 1

r 1th

tq 1

h

r _ termq _ term

r _ term rq _ ter

aa

m

rr

−

−

− +

−=

=

th

thr qr _ term r

q _ term−=

From eqn. (i) r – q = d th

thdr _ term r

q _ term=

We can see that pth, qth and rth terms have common ratio i.e rd

Therefor they are in G.P. Hence Proved. Solution 2 : To prove: log an, log bn, log cn are in AP. Given: a, b, c are in GP As a, b, c are in GP ⇒ b2 = ac Taking power n on both sides ⇒ b2n = (ac)n Taking log both side



⇒ logb2n = log(ac)n ⇒ logb2n = log(ancn) ⇒ 2logbn = log(an) + log(cn) ( log ab = log a + log b) As we know that whenever a,b,c are in AP then 2b = a+c….(i) from above equation (i), we can say that log an, log bn, log cn are in AP. Hence Proved

Solution 3 : To prove:a b c

1 1 1, ,log m log m log m

are in AP.

Given that : a, b, c are in GP As, a, b, c are in GP

⇒ b ca b=

Taking log both side b clog loga b=

⇒ log b – log a = log c – log b ⇒ 2log b = log a + log c Dividing by log m

⇒ log b loga logc2( )log m log m log m

= + ( ma

1 logalog alog m log m

= = )

⇒ m m m2log b log a log c= +



⇒b a c

1 1 12log m log m log m

= +

As we know that whenever a,b,c are in AP then 2b = a+c….(i)

From above equation (i), we can say that a b c

1 1 1, ,log m log m log m

are in

AP. Hence proved Solution 4 : To find: Value of k Given that : k + 12, k – 6 and 3 are in GP As, k + 12, k – 6 and 3 are in GP ⇒ (k – 6)2 = (k + 12) (3) ⇒ k2 – 12k + 36 = 3k + 36 ⇒ k2 – 15k = 0 ⇒ k (k – 15) = 0 ⇒ k = 0 , Or k = 15 The two values of k are 0 or 15 Solution 5 : To find: The three numbers Given that : Three numbers are in A.P. Their sum is 15 Let the numbers be a - d, a, a + d According to the question a + d + a +a – d = 15 ⇒ 3a = 15 ⇒ a = 5 Hence numbers are 5 - d, 5, 5 + d



As per question 1, 4, 19 be added to the above numbers respectively then the numbers become – 5 – d + 1, 5 + 4, 5 + d + 19 ⇒ 6 – d, 9, 24 + d and these above numbers are in GP So, 92 = (6 – d) (24 + d) ⇒ 81 = 144 – 24d +6d – d2 ⇒ 81 = 144 – 18d – d2 ⇒ d2 + 18d – 63 = 0 ⇒ d2 + 21d – 3d – 63 = 0 ⇒ d (d + 21) -3 (d + 21) = 0 ⇒ (d – 3) (d + 21) = 0 ⇒ d = 3 Or d = - 21 When taking d = 3, the numbers are 5 - d, 5, 5 + d = 5 - 3, 5, 5 + 3 = 2, 5, 8 When taking d = -21, the numbers are 5 - d, 5, 5 + d = 5 – (-21), 5, 5 + (-21) = 26, 5, -16 The two sets of triplet are 2, 5, 8 and 26, 5, -16. Solution 6 : Given: Three numbers are in A.P. Their sum is 21 Let the numbers be a - d, a, a + d According to the question a + d + a +a – d = 21



⇒ 3a = 21 ⇒ a = 7 Hence numbers are 7 - d, 7, 7 + d As per question when second number is reduced by 1 and third is increased by 1 then the numbers become – 7 – d , 7 – 1 , 7 + d + 1 ⇒ 7 – d , 6 , 8 + d These above numbers are in GP So, 62 = (7 – d) (8 + d) (When a,b,c are in GP, b2 = ac)

⇒ 36 = 56 + 7d – 8d – d2 ⇒ d2 + d – 20 = 0 ⇒ d2 + 5d – 4d – 20 = 0 ⇒ d (d + 5) – 4 (d + 5) = 0 ⇒ (d – 4) (d + 5) = 0 ⇒ d = 4, Or d = -5 When taking d = 4, the numbers are 7 - d, 7, 7 + d = 7 - 4, 7, 7 + 4 = 3, 7, 11 When taking d = -5, the numbers are 7 - d, 7, 7 + d = 7 – (-5), 7, 7 + (-5) = 12, 7, 2 The two sets of triplet are 3, 7, 11 and 12, 7, 2. Solution 7 : Given that : The three numbers are in G.P. Their sum is 56 Let the three numbers in GP be a, ar, ar2 According to the questions:- a + ar + ar2 = 56



a(1 + r + r2) = 56 … (i) As per question 1, 7, 21 be subtracted from a, ar, ar2 respectively, we get as :- a – 1, ar – 7, ar2 – 21 According to question the above numbers are in AP ⇒ ar – 7 – (a – 1) = ar2 – 21 – (ar – 7) ⇒ ar – 7 – a + 1 = ar2 – 21 – ar + 7 ⇒ ar – a – 6 = ar2 – ar – 14 ⇒ 8 = ar2 – 2ar + a ⇒ 8 = a(r2 – 2r + 1) ⇒ 56 = 7a(r2 – 2r + 1) ( Multiplying by 7) ⇒ a(1 + r + r2) = 7a(r2 – 2r + 1) ⇒ 1 + r + r2 = 7r2 – 14r + 7 ⇒ 6r2 – 15r + 6 = 0 ⇒ 6r2 – 12r – 3r + 6 = 0 ⇒ 6r(r – 2) -3 (r – 2) = 0 ⇒ (6r – 3) (r – 2) = 0

⇒ r =3/6 = 12

Or r = 2

Putting r = 12

in eqn. (i)

a(1 + r + r2) = 56

21 1a(1 ) 562 2

+ + =

4 2 1a( ) 564

+ +=



a = 32 The numbers are a, ar, ar2 ⇒ 32, 16, 8 Putting r = 2 in eqn. (i) a(1 + r + r2) = 56

2a(1 2 2 ) 56+ + =

a = 8 The numbers are a, ar, ar2 ⇒ 8, 16, 32 The two sets of triplet are 32, 16, 8 and 8, 16, 32. Solution 8 : Given: a, b, c are in GP a, b, c are in GP, ⇒ b2 = ac … (i)

a bc= … (ii)

Taking LHS =2 2a ab b

ab bc ca+ ++ +

Substituting the value b2 = ac from eqn. (i)

LHS = a(a b c)b(a b c)

+ ++ +

= ab

Substituting the value b = ac from eqn. (ii)



a aac c

= =

Multiplying and dividing with( a c)+

a ( a b)

c( a b )

+=

+ = (a ac)

ac c++

= a bb c++

=RHS

Hence Proved Solution 9: To prove: (a + b + c)2 = 3(ab + bc + ca). Given that : (a – b), (b – c), (c – a) are in GP By using Formula: When a,b,c are in GP, b2 = ac

As, (a – b), (b – c), (c – a) are in GP ⇒ (b – c)2 = (a – b) (c – a) ⇒ b2 -2cb + c2 = ac – a2 – bc + ab ⇒ a2 + b2 + c2 – bc – ac – ab = 0 Adding 3(ab + bc + ac) both side ⇒ a2 + b2 + c2 – bc – ac – ab + 3(ab + bc + ac) = 3(ab + bc + ac) ⇒ a2 + b2 + c2 + 2bc + 2ac + 2ab = 3(ab + bc + ac) ⇒ (a + b + c)2 = 3(ab + bc + ac) Hence Proved Solution 10 : (i) Given that a(b2 + c2) = c(a2 +b2)



To prove: a(b2 + c2) = c(a2 + b2) Given: a, b, c are in GP By using Formula: When a,b,c are in GP, b2 = ac

Proof: When a,b,c are in GP, b2 = ac Taking LHS = a(b2 + c2) = a(ac + c2) [b2 = ac] = (a2c + ac2) = c(a2 + ac) = c(a2 + b2) = RHS [b2 = ac] Hence Proved (ii) Given that a, b, c are in GP To prove: a(b2 + c2) = c(a2 + b2) Proof: When a,b,c are in GP, b2 = ac

Taking LHS = 2 2 21 1

a b b+

−

=2 2 2

2 2 2b a bb (a b )+ −

−=

2

2 2 2a

b (a b )−=

2

2 2a

ac(a b )−=

2

3 2 2a

(a c a c )−

= 2 21

b c−( b2 = ac)

Hence Proved (iii) Given that a, b, c are in GP To prove: (a + 2b + 2c)(a – 2b + 2c) = a2 + 4c2 By using Formula: When a,b,c are in GP, b2 = ac

Proof: When a,b,c are in GP, b2 = ac



Taking LHS = (a + 2b + 2c)(a – 2b + 2c) ⇒ [(a + 2c) + 2b] [(a + 2c) – 2b] ⇒ [(a + 2c)2 – (2b)2] [(a + b) (a – b) = a2 – b2] ⇒ [(a2 + 4ac + 4c2) – 4b2] ⇒ [(a2 + 4ac + 4c2) – 4b2] [ b2 = ac] ⇒ [(a2 + 4ac + 4c2 – 4ac] ⇒ a2 + 4c2 = RHS Hence Proved (iv)Given that a, b, c are in GP

To prove: 2 2 2 3 3 33 3 3

1 1 1a b c ( ) a b ca b c

+ + = + +

Proof: When a,b,c are in GP, b2 = ac

Taking LHS = 2 2 2a b c 3 3 3 3 3 3

3 3 3b c a c a b( )

a b c+ +

=3 3 3 3 3 3b c a c a b( )

abc+ +

=2 3 2 3 2b bc (ac) ac a b b

abc+ + [b2 = ac]

=3 2 2 3acbc (b ) ac a acb

abc+ + take abc common and divide

We get

= 3 3 3a b c+ + = RHS Hence Proved



Solution 11 : (i) Given that a, b, c, d are in GP To prove: (b + c)(b + d) = (c + a)(c + a) Proof: When a,b,c,d are in GP then b c da d c= =

From the above, we can have the following conclusion ⇒ bc = ad … (i) ⇒ b2 = ac … (ii) ⇒ c2 = bd … (iii) Taking LHS = (b + c)(b + d) = b2 + bd + bc + cd Using eqn. (i) , (ii) and (iii) = ac + c2 + ad + cd = c(a + c) + d(a + c) = (a + c) (c + d) Hence Proved (ii)Given that a, b, c, d are in GP

To prove: 2 2ab cd a c

bb c− +

=−

Proof: When a,b,c,d are in GP then

= b c da d c= =

From the above, ⇒ bc = ad … (i) ⇒ b2 = ac … (ii)



⇒ c2 = bd ⇒ d = 2c

b … (iii)

Taking LHS = 2 2ab cdb c−−

=

2

2 2

cab cb

b c

−

− [From eqn. (iii)]

=

3

2 2

cabb

b c

−

−=

2 3

2 2ab c

b(b c )−−

=3

2aac c

b(ac c )−−

[From eqn. (ii)]

=2 2

2c(a c )b(ac c )

−−

= 2c(a c)(a c)

b(ac c )− +

−

=2

2(ac c )(a c)

b(ac c )− +

−a c

b+

= = RHS

Hence Proved (iii) Given that: a, b, c, d are in GP To prove: (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2

Proof: When a,b,c,d are in GP then From the above, we can have the following conclusion ⇒ bc = ad … (i) ⇒ b2 = ac … (ii) ⇒ c2 = bd … (iii)



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

⇒ (a + b + c + d) (a + b + c + d) ⇒ a2 + ab + ac + ad + ba + b2 + bc + bd + ca + cb + c2 + cd + da + db + dc + d2 On rearranging ⇒ a2 + ab + ba +b2 + ac + ad + bc + bd + ca + cb + c2 + cd + da + db + dc + d2 On rearranging ⇒ (a + b)2 + ac + ad + bc + bd + ca + cb + da + db + c2 + cd + dc + d2

On rearranging ⇒ (a + b)2 + ac + ad + bc + bd + ca + cb + da + db + (c + d)2 On rearranging ⇒ (a + b)2 + ac + ca + ad + bc + cb + da + bd + db + (c + d)2 Using eqn. (i) ⇒ (a + b)2 + ac + ca + bc + bc + bc + bc + bd + db + (c + d)2 Using eqn. (ii) ⇒ (a + b)2 + b2 + b2 + bc + bc + bc + bc + bd + db + (c + d)2 Using eqn. (iii) ⇒ (a + b)2 + 2b2 + 4bc + c2 + c2 + (c + d)2 On rearranging ⇒ (a + b)2 + 2b2 + 4bc + 2c2 + (c + d)2 ⇒ (a + b)2 + 2[b2 + 2bc + c2 ] + (c + d)2 ⇒ (a + b)2 + 2(b + c)2 + (c + d)2 = RHS



Hence proved Solution 12: Given: a, b, c are in GP

To prove: 1 1 1( , , )a b 2b b c+ +

are in AP

By using Formula :

When a,b,c are in GP, b2 = ac

Taking 1a b+

and 1b c+

1a b+

+ 1b c+

⇒ b c a b(a b)(b c)+ + ++ +

c a 2b(a b)(b c)

+ ++

⇒+

2

2c a b

ab ac b bc+ +

+ + +⇒ [ b2 = ac ]

2

2 2c a b

ab b b bc+ +

+ + +⇒

2

2a b c

b(a b c)+ +

⇒+ +

⇒1b⇒

22b

We can see that 1a b+

+ 1b c+

=2 x 12b

Hence, the terms 1 1 1( , , )a b 2b b c+ +

are in AP.

Solution 13: Given: a, b, c are in GP To prove: a2, b2, c2 are in GP



Proof: As a, b, c are in GP ⇒ b2 = ac … (i) Considering b2, c2

=2

2cb

= common ratio = r

⇒ 2c

ac[From eqn. (i)]

⇒ ca

= r

Considering a2, b2

= 2

2ba

= common ratio = r

⇒ 2aca

[From eqn. (i)]

⇒ ca

= r

We can see that in both the cases we have obtained a common ratio. Hence, the terms a2, b2, c2 are in GP. Solution 14: To prove: a3, b3, c3 are in GP Given: a, b, c are in GP Proof: As a, b, c are in GP ⇒ b2 = ac Cubing both sides



= 2 3 3(b ) (ac)=

=3 3

3 3b ca b

= = common ratio = r

From the above equation, we can say that a3, b3, c3 are in GP Hence, the following terms a3, b3, c3 are in GP. Solution 15 : To prove: (a2 + b2), (ab + bc), (b2 + c2) are in G.P. Given: a, b, c are in G.P. Formula used: When a,b,c are in GP, b2 = ac Proof: When a,b,c are in G.P., b2 = ac … (i) Considering (a2 + b2), (ab + bc), (b2 + c2) (ab + bc)2 = (a2b2 + 2ab2c + b2c2) = (a2b2 + ab2c + ab2c + b2c2) = (a2b2 + b4 + a2c2 + b2c2) [From eqn. (i)] = [b2 (a2 + b2)+ c2 (a2 + b2)] (ab + bc)2 = [(b2 + c2) (a2 + b2)] From the above equation we can say that (a2 + b2), (ab + bc), (b2 + c2) are in G.P. Hence, the following terms (a2 + b2), (ab + bc), (b2 + c2) are in G.P. Solution 16 : To prove: (a2 – b2), (b2 – c2), (c2 – d2) are in GP. Given: a, b, c are in G.P.



By using Formula : When a,b,c are in G.P., b2 = ac

Proof: When a,b,c,d are in G.P. then From the above, we can have the following conclusion ⇒ bc = ad … (i) ⇒ b2 = ac … (ii) ⇒ c2 = bd … (iii) Considering (a2 – b2), (b2 – c2), (c2 – d2) (a2 – b2) (c2 – d2) = a2c2 – a2d2 – b2c2 + b2d2 = (ac)2 – (ad)2 – (bc)2 + (bd)2 From eqn. (i) , (ii) and (iii) = (b2)2 – (bc)2 – (bc)2 + (c2)2 = b4 – 2b2c2 + c4 (a2 – b2) (c2 – d2) = (b2 – c2)2 From the above equation we can say that (a2 – b2), (b2 – c2), (c2 – d2) are in G.P. Hence, the following terms (a2 – b2), (b2 – c2), (c2 – d2) are in G.P.

Solution 16 : To prove: 2 2 2 2 2 21 1 1( , , )

a b b c c d+ + + are in GP.

Given: a, b, c, d are in GP Proof: When a,b,c,d are in GP then

= b c da b c= =

From the above, we can have the following conclusion



⇒ bc = ad … (i) ⇒ b2 = ac … (ii) ⇒ c2 = bd … (iii)

Considering 2 2 2 2 2 21 1 1( , , )

a b b c c d+ + +

= 2 2 2 2 2 2 2 2 2 2 2 21 1 1

a b b c a c a d b c b d× =

+ + + + +

= 2 2 2 21

(ac) (ad) (bc) (bd)+ + +

From eqn. (i) , (ii) and (iii)

= 2 2 2 2 2 21

(b ) (bc) (bc) (c )+ + +

= 4 2 2 41

b 2b c c+ +

= 2 2 21

(b c )+

From the above equation, we can say that 2 2 2 2 2 21 1 1( , , )

a b b c c d+ + +

are in GP.

Hence, the following terms 2 2 2 2 2 21 1 1( , , )

a b b c c d+ + +are in G.P.

Solution 18 : To prove: p, q, r are in GP Given: (p2 + q2), (pq + qr), (q2 + r2) are in GP By using Formula : When a,b,c are in GP, b2 = ac



Proof: When (p2 + q2), (pq + qr), (q2 + r2) are in GP, (pq + qr)2 = (p2 + q2) (q2 + r2) p2q2 + 2pq2r + q2r2 = p2q2 + p2r2 + q4 + q2r2 2pq2r = p2r2 + q4 pq2r + pq2r = p2r2 + q4 pq2r - q4 = p2r2 - pq2r

q2(pr – q2) = pr (pr – q2) q2 = pr From the above equation we can say that p, q and r are in G.P. Hence, the following terms p, q and r are in G.P. Solution 19: Given: a, b, c are in AP, and a, b, d are in GP To prove: a, (a – b) and (d – c) are in GP. Proof: As a,b,d are in GP then b2 = ad … (i) As a, b, c are in AP 2b = (a + c) … (ii) Considering a, (a – b) and (d – c) (a – b)2 = a2 – 2ab + b2 = a2 – (2b)a + b2 From eqn. (i) and (ii) = a2 – (a+c)a + ad = a2 – a2 - ac + ad = ad – ac (a – b)2 = a (d – c)



From the above equation we can say that a, (a – b) and (d – c) are in GP. Hence, the following terms a, (a – b) and (d – c) are in G.P. Solution 20 : Given: a, b, c are in AP, and a, x, b and b, y, c are in GP To prove: x2, b2, y2 are in AP. Proof: As, a,b,c are in AP ⇒ 2b = a + c … (i) As, a,x,b are in GP ⇒ x2 = ab … (ii) As, b,y,c are in GP ⇒ y2= bc … (iii) Considering x2, b2, y2 x2 + y2 = ab + bc [From eqn. (ii) and (iii)] = b (a + c) = b(2b) [From eqn. (i)] x2 + y2 = 2b2 From the above equation we can say that x2, b2, y2 are in A.P. Hence, the following terms x2, b2, y2 are in A.P.

Exercise 12F Solution 1 : (i) A.M. = 25 and G.M. = 20



To find: Two positive numbers a and b Given: A.M. = 25 and G.M. = 20 By using Formula:

(i) Arithmetic mean between a and b= a b2+

(ii) Geometric mean between a and b = ab

Arithmetic mean of two numbers a b2+

a b2+ =25

⇒ a + b = 50 ⇒ b = 50 – a … (i)

Geometric mean of two numbers = ab

ab = 20 ab = 400 On Substituting value of b from eqn. (i) a(50 – a) = 400 ⇒ 50a – a2 = 400 On rearranging ⇒ a2 – 50a + 400 = 0 ⇒ a2 – 40a – 10a + 400 = 0 ⇒ a(a – 40) – 10(a – 40) = 0 ⇒ (a – 10) (a – 40) = 0 ⇒ a = 10, 40 Substituting, a = 10 Or a = 40 in eqn. (i) b = 40 Or b = 10 So two numbers are 10 and 40



(ii) Given AM = 10 and GM = 8 To find: Two positive numbers a and b By using Formula:

(i) Arithmetic mean between a and b= a b2+

(ii) Geometric mean between a and b = ab

Arithmetic mean of two numbers= a b2+

a b2+ =10

⇒ a + b = 20 ⇒ a = 20 – b … (i)

Geometric mean of two numbers= ab

ab = 8 ⇒ ab = 64 Substituting value of a from eqn. (i) b(20 – b) = 64 ⇒ 20b – b2 = 64 On rearranging ⇒ b2 – 20b + 64 = 0 ⇒ b2 – 16b – 4b + 64 = 0 ⇒ b(b – 16) – 4(b – 16) = 0 ⇒ (b – 16) (b – 4) = 0 ⇒ b = 16, 4 Substituting, b = 16 Or b = 4 in eqn. (i) a = 4 Or b = 16 Hence, two numbers are 16 and 4. Solution 2 : (i) Given: 5 and 125



To find: Geometric Mean of two numbers.

By using Formula: (i) Geometric mean between a and b = ab


5 25× = 625 = 25 Hence, The geometric mean between 5 and 125 is 25

(ii) Given: 1 and 96

To find: Geometric Mean of two numbers.


Geometric mean of two numbers ab

= 916

1× = 916

= 34

Hence, The geometric mean between 1 and 916

is 34

.

(iii) Given: 0.15 and 0.0015 To find: Geometric Mean of two numbers.



= 0.15 0.0015× = 0.015 Hence, The geometric mean between 0.15 and 0.0015 is 0.015.



(iv) Given: -8 and -2 To find: Geometric Mean of two numbers.



= −8×−2 = 16 =±4 Hence, The geometric mean between -8 and -2 is -4. (v) Given: -6.3 and -2.8 To find: Geometric Mean of two numbers.



= 6.3 2.8− ×− = 4.2± Hence, The geometric mean between -6.3 and -2.8 is -4.2. (vi) Given: a3b and ab3 To find: Geometric Mean of two numbers.



= 3 3a b ab× = 4 4a b = a2b2 Hence, The geometric mean between a3b and ab3 is a2b2. Solution 3 : To find: Two geometric Mean between two numbers.



Given: The numbers are 9 and 243

By using Formula :1

n 1br ( )a

+= where n is the number of geometric

mean Let G1 and G2 be the three geometric mean

Then 1

n 1br ( )a

+=

⇒ 1

2 1243r ( )9

+= ⇒ 13r 27= ⇒ r = 3

G1 = ar = 9×3 = 27 G2 = ar2= 9×32 = 9×9 = 81 Hence, two geometric mean between 9 and 243 are 27 and 81.

Solution 4: Given: The numbers 13

and 432

To find: Three geometric Mean between two numbers. By using Formula :

(i) 1

n 1br ( )a

+= , where n is the number of geometric mean.

Let G1, G2 and G3 be the three geometric mean

Then 1

n 1br ( )a

+=

⇒ r 1

3 1432( )13

+=



⇒ r = 14432 2( )

1× ⇒ r = 6

G1 = ar = 13

×6 = 2

G2 = ar2= 13

×62 = ×36 = 12

G3 = ar3= 13

×63 = 13

×216 = 72

Hence, Three geometric mean between and 432 are 2, 12 and 72. Solution 5 : Given: The numbers 6 and 192 To find: Four geometric Mean between two numbers. By using Formula:

(i) 1

n 1br ( )a

+= where n is the number ofgeometric mean

Let G1, G2, G3 and G4 be the three geometric mean

Then1

n 1br ( )a

+=

14 1192r ( )

6+= ⇒

15r 32= ⇒ r= 2

G1 = ar = 6×2 = 12 G2 = ar2= 6×22 = 24 G3 = ar3= 6×23 = 48 G4 = ar4= 6×24 = 96 Four geometric mean between 6 and 192 are 12, 24, 48 and 96.



Solution 6 : Given: Arithmetic mean is twice of Geometric mean.

To Prove that a:b= (2 3) : (2 3)+ −

By using Formula:

(i) Arithmetic mean between a and b = a b2+

(ii) Geometric mean between a and b = ab AM = 2(GM) a ab

2b2+

=

a + b = 4 ab Squaring both side ⇒ (a + b)2 = 16ab … (i) We know that (a – b)2 = (a + b)2 – 4ab From eqn. (i) ⇒ (a – b)2 = 16ab – 4ab ⇒ (a – b)2 = 12ab … (ii) Dividing eqn. (i) and (ii)

2

2(a b) 16ab

12ab(a b)+

=−

Taking square root both side

⇒ a b 2a b 3+

=−



Applying componendo and dividendo a b a b 2 3a b a b 2 3+ + − +

=+ − + −

a 2 3b 2 3

+=

−

Hence Proved Solution 7 : Given: (i) a, b, c are in AP (ii) x is the GM between a and b (iii) y is the GM between b and c To prove: b2 is the AM between x2 and y2. By using Formula:


(ii) Geometric mean between a and b = ab As a, b, c are in A.P. ⇒ 2b = a + c … (i) As x is the GM between a and b

⇒ x = ab ⇒ x2 = ab … (ii) As y is the GM between b and c

⇒ y = cb ⇒ y2 = bc … (iii)



Arithmetic mean of x2 and y2 = 2 2x y

2+

Substituting the value from (ii) and (iii) 2 2x y ab bc

2 2+ +

= = b(a c)2+

Substituting the value from eqn. (i)

= b(2b)2

=b2

Hence Proved Solution 8 : To prove: Product of n geometric means between a and b is equal to the nth power of the single GM between a and b. By using Formula:

(i) Geometric mean between a and b = ab

(ii) Sum of n terms of A.P =. n(n 1)2+

Let, the n geometric means between and b be G1, G2, G3, … Gn Hence a, G1, G2, G3, … Gn, b are in GP ⇒ G1 = ar, G2 = ar2 and so on … Now, we have n+2 term ⇒ b = arn+2-1 ⇒ b = arn+1

⇒ 1

n 1br ( )a

+= … (i)

The product of n geometric means is G1× G2× G3× … Gn



= ar × ar2 × ar3 × … arn

= an × r(1+2+3… + n)

= an ×n 1(n)( )

2r+

[Sum of n terms of A.P =. n(n 1)2+ ]

Substituting the value of r from eqn. (i)

= an ×1 n 1(n)( )

n 1 2b( )a

++ =

n n2 2a b = n( ab) … (ii)

Single geometric mean between a and b = ab

nth power of single geometric mean between a and b = n( ab)

Hence Proved Solution 9 : Given: (i) AM of roots of quadratic equation is 10

(ii) GM of roots of quadratic equation is 8 To find: The quadratic equation. By using Formula:


(ii) Geometric mean between a and b = ab Let the roots be p and q

Arithmetic mean of roots p and q = p q2+ =10

⇒ p q2+ = 10 ⇒ p + q = 20 = sum of roots … (i)



Geometric mean of roots p and q = pq = 8

⇒ pq = 64 = product of roots … (ii) Quadratic equation = x2 – (sum of roots)x + (product of roots) From equation (i) and (ii) Quadratic equation = x2 – (20)x + (64) = x2 –20x + 64 Hence , the required quadratic is x2 –20x + 64.

Exercise 12G

Solution 1: Given that ∞ Here, a=8 and 4 2 1r

8 2= =

By using Formula: Sum of an infinite Geometric series, S∞a

1 r=

−

sum = 9112

− = 6 2

2 1−

Hence, the Sum of Infinite Geometric Series = 6 22 1−

.

Solution 2 : 6 + 1.2 + 0.24 + ….. ∞



Here, a=6 and 1.2 2r 0.26 10

= = =

By using Formula: Sum of an infinite Geometric series, S∞ a1 r

=−

Sum = 61 0.2−

=152

Hence, the Sum of given Infinite Geometric Series = 152

.

Solution 3 : ∞

Here, a=√2 and

112r

22

−−

= =

Sum of an infinite Geometric series, S∞ a1 r

=−

a 211 r 1

2

= =−− −

= 2 23

Hence, the Sum of Infinite Geometric Series = 2 23

.

Solution 4 : Given 10 – 9 + 8.1 - ……∞ Here, a=10 and 9r

10−

= 0.9= −



Sum of an infinite Geometric series, S∞ a 101 r 1 ( 0.09)

= =− − −

= 10019

Hence, the Sum of Infinite Geometric Series = 10019

.

Solution 5 : Given that

Above geometric series is the sum of two geometric series:

3 52 2 2 .......5 5 5+ + ∞ and 2 4 6

3 3 4 .............5 5 5

+ + + ∞

Sum of geometric series 3 52 2 2 .......5 5 5+ + ∞

Here a = 25

and r = 125


=−

= 512

Sum of geometric series 2 4 63 3 4 .............5 5 5

+ + + ∞

Here a= 235

and r = 125


=−

= 18

Sum of the given infinite series = 512

+ 18

= 1324

Hence, the Sum of Given infinite Geometric Series = 1324

.



Solution 6 : Given that 91/3 × 91/9 × 91/27 × …..∞ = 3 L.H.S=91/3 × 91/9 × 91/27 × …..∞ = 9(1/3)+(1/9)+(1/27)+…∞ The series in the exponent is an infinite geometric series

Whose, a= 13

and r = 13

∴Sum of the series in the exponent S∞ a1 r

=−

= 12

∴L.H.S = 91/2 = 3 = R.H.S Hence, Proved that 91/3 × 91/9 × 91/27 × …..∞ = 3

Solution 7 : (i) Given Let, x=0.3333… ⇒ x=0.3+0.03+0.003+… ⇒ x=3(0.1+0.01+0.001+0.0001+…∞)

⇒ x=3( 1 1 1 1 ............10 100 1000 10000

+ + + + ∞ )

This is an infinite geometric series. Here, a =1/10 and r =1/10

Sum of the series in the exponent S∞ = a1 r−

= 19

X = 3x 19

= 13

=0.3333..

Hence, the Sum of infinite Geometric Series is 0.3333..



(ii) Given Let, x=0.231231231…. ⇒ x=0.231+0.000231+0.000000231+…∞ ⇒ x=231(0.001+0.000001+0.000000001+…∞)

⇒ x=231( 3 6 91 1 1 ............

10 10 10+ + + ∞ )

This is an infinite geometric series.

Here, a = 31

10 and r = 3

110

Sum of the series in the exponent S∞ = a1 r−

= 1999

X =231 x 1999

= 231999

Hence, the Sum of Infinite Geometric Series is 231999

.

(iii) Given Let, x=3.525252552… ⇒ x=3+0.52+0.0052+0.000052+…∞ ⇒ x=3+52(0.01+0.0001+…∞)

⇒ x=3 + 52( 2 4 61 1 1 ............

10 10 10+ + + ∞ )



Here, a = 21

10 and r = 2

110

Sum of the series in the exponent, S∞ = a1 r−

= 199

X=3 + (52 x 199

) = 34999

Hence, the Sum of infinite Geometric Series = 34999

Solution 8 : Given that 0.125125125 …. = Let, x=0.125125125… …(i) 1000x=125.125125125… …(ii) By equation (ii)-(i), ⇒ 1000x-x=125.125125125 - 0.125125125=125

⇒ 999x=125 ⇒ x =125999

Hence, the value of x = 125999

Solution 9 : Given Let, x=0.423423423… …(i) 1000x=423.423423423… …(ii) By equation (ii)-(i), ⇒ 1000x-x=423.423423423-0.423423423=423



⇒ 999x=423 ⇒x = 423999


Solution 10 : Given that Let, x = 2.134134134… …(i) 1000x=2134.134134134… …(ii) By equation (ii)-(i), ⇒ 1000x-x=2134.134134134-2.134134134=2132

⇒ 999x=2132 ⇒x= 2132999


.

Solution11: Given that : a1 r−

= 6, a= 2 and r =?

Sum of the series in the exponent, S∞ = a1 r−

= 2 61 r

=−

⇒ r = 23

Hence, the common ratio r = 23

Solution 12 :

Given : a1 r−

=20 and 2

2a 100

1 r=

−



To find: the series a=20(1-r)…(i)

2

2a 100

1 r=

−=

2(20 (1 r))(1 r)(1 r)

× −− +

⇒ 100(1+r)=400(1-r) ⇒ 100+100r=400-400r ⇒ 100r+400r=400-100

⇒ 500r=300 ⇒ 5r=3 ⇒r = 35

Put this value of r in equation (i) we get 3a 20(1 ) 85

= − =

Hence, The infinite geometric series is: 8,24/5 ,72/25…. Solution 13: The sum of an infinite GP is 57, and the sum of their cubes is 9747 Let the first term Of G.P. be a, and common ratio be r.

By using Formula a1 r−

=57…….(i)

On cubing each term will become, a3, a3r3, ….

Sum = 2

2a 9747

1 r=

−



a=57(1-r) put this in equation 2 we get 3

3[57 (1 r)] 9747

1 r× −

=−

⇒ 19(1-2r + r2)=1+ r + r2 ⇒ 19r2-r2-38r-r + 19 - 1= 0 ⇒ 18r2-39r + 18=0 ⇒ 6r2-13r + 6=0 ⇒ (2r-3)(3r-2)=0 ⇒ r = 2/3, 3/2 But -1< r < 1 ⇒ r = 2/3 Substitute this value of r in equation 1 we get

a= 57(1- 23

)= 19

Thus the first term of G.P. is 19, and the common ratio is 2/3 Hence, the infinite GP series is 19,38/3,76/9

Exercise 12 H Solution 1 : Given that : 5th term of a GP is 2. To find: the product of its first nine terms. ∴ ar4 = 2 We have to find the value of: a × ar1 × ar2 × ar3 × … × ar8 = a9r1 + 2 + 3 + 4 + … + 8 = a9r36 = (ar4)9 = (2)9 = 512



Solution 2 : Given that tp + q = m and tp - q = n tp + q = m = Arp + q - 1 = Arp - 1rq And tp - q = n = Arp - q - 1 = Arp - 1r - q We know that pth term = Arp - 1 ∴ m × n = A2r2p - 2

⇒ Arp - 1 = (mn)1/2 ⇒ pth term = (mn)1/2 Ans: pth term = (mn)1/2 Solution 3 : Let the three consecutive terms of the G.P. be a, ar, ar2. Since the common difference of the A.P. = ar - a And the sixth term of the A.P. = ar2 by using the formula : t = a + (n - 1)d ∴ ar2 = a + 4(ar - a)…(the difference between 2nd and 6th term is 4(ar - a)) ⇒ ar2 = a + 4ar - 4a ⇒ ar2 + 3a - 4ar = 0 ⇒ a(r2 - 4r + 3) = 0 ⇒ a(r - 1)(r - 3) = 0 Here, we have 3 possible options:



(i) a = 0 which is not expected because all the terms of A.P. and G.P. will be 0. (ii) r = 1,which is also not expected because all th terms would be equal to first term. (iii) r = 3,which is the required answer. Solution 4 : Let the roots of the required quadratic equation be a and b. The arithmetic and geometric means of roots are A and G respectively. ⇒ A = (a + b)/2…(i) And G = ab …(ii) We know that the equation whose roots are given is = X2-(a+b)x+ab=0 From (i) and (ii) we get: X2-2A+G2=0 Thus, X2-2A+G2=0 is the required quadratic equation. Solution 5 : It is given that: a1/x = b1/y = c1/z Let a1/x = b1/y = c1/z = k ⇒ a1/x = k ⇒ (a1/x)x = kx…(Taking power of x on both sides.) ⇒ a1/x × x = kx



⇒ a = kx

Similarly b = ky and c = kz It is given that a,b,c are in G.P. ⇒ b2 = ac Substituting values of a,b,c calculated above, we get: ⇒ (ky)2 = kxkz ⇒ k2y = kx + z

Comparing the powers we get, 2y = x + z Which is the required condition for x,y,z to be in A.P. x,y,z, are in A.P Hence Proved Solution 6 : It is given that a,b,c are in A.P. To prove: xb - c. yc - a. za - b = 1….(i) ⇒ 2b = a + c…(ii) ⇒ y2 = xz (x,y,z, are in G.P.) ⇒ x = y2/z Putting the value of x in equation (i),we get

L.H.S =2

b c c a a by( ) y zz

− − −× ×

⇒y2b-2c+c-a za-b-2b+2c ( 2b = a + c ) ⇒ y0.z0…(Using equation (i))



= 1 = R.H.S Hence Proved that . If a, b, c are in AP and x, y, z are in GP then xb - c. yc - a. za - b= 1

Solution 7: Given that

Here, a = 1 and r = 1

131 3

−−

=

As we know that Sum of an infinite Geometric series= a a 3

11 r 413

= =−− −

= R.H.S.

Hence Proved


CHAPTER-12 - Geometrical Progression - Amazon AWS

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