Geometrical Progression

Given: GP is 2, 6, 18, 54….

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

First term in the given GP, a₁ = a = 2

Second term in GP, a₂ = 6

Now, the common ratio,

Now, n^th term of GP is, a_n = ar^{n – 1}

So, the 6^th term in the GP,

a₆ = ar⁵

= 2 x 3⁵

= 486

n^th term in the GP,

a_n = ar^{n – 1}

= 2.3^{n – 1}

Hence, 6^th term = 486 and n^th term = 2.3^{n – 1}

Given GP is 2, 2√2, 4, 8√2 …..

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

First term in the given GP, a₁ = a = 2

Second term in GP, a₂ = 2√2

Now, the common ratio,

Now, n^th term of GP is, a_n = ar^{n – 1}

So, the 17^th term in the GP,

a₁₇ = ar¹⁶

= 2 x (√2)¹⁶

= 512

n^th term in the GP,

a_n = ar^{n – 1}

= 2(√2) ^{n – 1}

= (√2) ^{n + 1}

Hence, 17^th term = 512 and n^th term = (√2) ^{n + 1}

Given GP is 0.4, 0.8, 1.6….

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

First term in the given GP, a₁ = a = 0.4

Second term in GP, a₂ = 0.8

Now, the common ratio,

Now, n^th term of GP is, a_n = ar^{n – 1}

So, the 7^th term in the GP,

a₇ = ar⁶

= 0.4 x 2⁶

= 25.6

n^th term in the GP,

a_n = ar^{n – 1}

= (0.4)(2) ^{n – 1}

= (0.2)2ⁿ

Hence, 7^th term = 25.6 and n^th term = (0.2)2ⁿ

Given GP is.

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

The first term in the given GP,

The second term in GP,

Now, the common ratio,

Now, n^th term of GP is, a_n = ar^{n – 1}

So, the 10^th term, a₁₀ = ar⁹

Now, the required n^th term, a_n = ar^n-1

Hence, the 10^th term, and n^th term,
.

Given GP is 3, 6, 12, 24….

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

First term in the given GP, a₁ = a = 3

Second term in GP, a₂ = 6

Now, the common ratio,

Let us consider 3072 as the n^th term of the GP.

Now, n^th term of GP is, a_n = ar^{n – 1}

3072 = 3.2^{n – 1}

ⁿ

2ⁿ = 2¹¹

n = 11

So, 3072 is the 11^th term in GP.

Given GP is ….

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

The first term in the given GP,

The second term in GP,

Now, the common ratio,

Let us consider -128 as the n^th term of the GP.

Now, n^th term of GP is, a_n = ar^{n – 1}

^{n – 1}

2)ⁿ = 1024 = (-2)¹⁰

n = 10

So, -128 is the 10^th term in GP.

Given GP is √3, 3, 3√3….

The given GP is of the form, a, ar, ar², ar³….

Where r is the common ratio.

First term in the given GP, a₁ = a = √3

Second term in GP, a₂ = 3

Now, the common ratio,

Let us consider 729 as the n^th term of the GP.

Now, n^th term of GP is, a_n = ar^{n – 1}

729 = √3 (√3) ^{n – 1}

√3ⁿ = √3¹²

n = 12

So, 729 is the 12^th term in GP.

The n^th term of a GP is a_n = ar^n-1

It’s given in the question that 5^th term of the GP is 80 and 8^th term of GP is 640.

So, a₅ = ar⁴ = 80 → (1)

a₈ = ar⁷ = 640 → (2)

→

Common ratio, r = 2,

ar⁴ = 80

16a = 80

a = 5

The required GP is of the form a, ar, ar², ar³, ar⁴….

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…

The n^th term of a GP is a_n = ar^n-1

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

So, → (1)

→ (2)

→

Common ratio,

The required GP is of form a, ar, ar², ar³, ar⁴….

The first term of GP,

The second term of GP,

The third term of GP,

The fourth term of GP,

The fifth term of GP,

And so on...

The required GP is …

It is given in the question that 5^th, 8^th and 11^th terms of GP are a, b and c respectively.

Let us assume the GP is A, AR, AR², and AR3….

So, the n^th term of this GP is a_n = AR^n-1

Now, 5^th term, a₅ = AR⁴ = a → (1)

8^th term, a₈ = AR⁷ = b → (2)

11^th term, a₁₁ = AR¹⁰ = c → (3)

Dividing equation (3) by (2) and (2) by (1),

→ (4)

→ (5)

So, both equation (4) and (5) gives the value of R³. So we can equate them.

,

∴ b² = ac,

Hence proved.

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 4^th term.

∴ (a₂)² = a₄

n^th term of GP, a_n = ar^n-1

So, a₂ = ar; a_{4 =} ar³

(ar) ² = ar³→ a = r = -3

Now, the 7^th term in the GP, a₇ = ar⁶

a₇ = (-3)⁷ = -2187

Hence, the 7^th term of GP is -2187.

The given GP is 8, 4, 2….→ (1)

First term in the GP, a₁ = a = 8

Second term in the GP, a₂ = ar = 4

The common ratio,

The last term in the given GP is.

Second last term in the GP = a_n-1 = ar^n-2

Starting from the end, the series forms another GP in the form,

ar^n-1, ar^n-2, ar^n-3….ar³, ar², ar, a → (2)

Common ratio of this GP is .

So, common ratio = 2

So, 6^th term of the GP (2),

a₆ = ar⁵

=

Hence, the 6^th term from the end of the given GP is.

The given GP is .→ (1)

The first term in the GP,

The second term in the GP,

The common ratio,r = 3

The last term in the given GP is a_n = 162.

Second last term in the GP = a_n-1 = ar^n-2

Starting from the end, the series forms another GP in the form,

ar^n-1, ar^n-2, ar^n-3….ar³, ar², ar, a → (2)

Common ratio of this GP is .

So,.

So, 4^th term of the GP (2),

a₄ = ar³

=

Hence, the 4^th term from the end of the given GP is 6.

As per the question, a, b and c are the p^th, q^th and r^th term of GP.

Let us assume the required GP as A, AR, AR², AR³…

Now, the n^th term in the GP, a_n = AR^n-1

p^th term, a_p = AR^p-1 = a → (1)

q^th term, a_q = AR^q-1 = b → (2)

r^th term, a_r = AR^r-1 = c → (3)

→ (i)

→ (ii)

→ (iii)

Taking logarithm on both sides of equation (i), (ii) and (iii).

(p – q) log R = log a – log b,

∴ → (4)

(q – r) log R = log b – log c

∴ → (5)

(r – p) log R = log c – log a

∴ → (6)

Now, multiply equation (4) with log c,

→ (7)

Now, multiply equation (5) with log a,

→ (8)

Now, multiply equation (6) with log b,

→ (9)

Now, add equations (7), (8) and (9).

On solving the above equation, we will get,

(p – q) log c + (q – r) log a + (r – p) log b = 0

Hence proved.

Given that the third term of the GP, a₃ = 4

Let us assume the GP mentioned in the question be,

…

With the first term and common ratio R.

Now, the third term in the assumed GP is A.

So, A = 4 (given data)

Now,

Product of the first five terms of GP= = A⁵

So, the required product = A⁵ = 4⁵ = 1024

∴ The product of first five terms of a GP with its third term 4 is 1024.

We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

Let us first consider a finite GP.

A, AR, AR²….AR^{n -1}, ARⁿ.

Where n is finite.

Product of first and last terms in the given GP = A.ARⁿ

= A²Rⁿ → (a)

Now, n^th term of the GP from the beginning = AR^n-1 → (1)

Now, starting from the end,

First term = ARⁿ

Last term = A

So, an n^th term from the end of GP, = AR → (2)

So, the product of n^th terms from the beginning and end of the considered GP from (1) and (2) = (AR^n-1) (AR)

= A²Rⁿ → (b)

So, from (a) and (b) its proved that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

(Given data in the question) → (1)

Cross multiplying (1) and expanding,

(a + bx)(b – cx) = (b + cx)(a-bx)

ab – acx + b²x – bcx² = ba –b²x + acx – bcx²

2b²x = 2acx

b² = ac → (i)

If three terms are in GP, then the middle term is the Geometric Mean of first term and last term.

→ b² = ac

So, from (i) b, is the geometric mean of a and b.

So, a, b, c are in GP.

(Given data in the question) → (2)

Cross multiplying (2) and expanding,

(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.

So, b, c, d is in GP.

∴a, b, c, d are in GP.

Given data is,

x² – 3x + p = 0 → (1)

a and b are roots of (1)

So, (x + a)(x + b) = 0

x² - (a + b)x + ab = 0

So, a + b = 3 and ab = p → (2)

Given data is,

x² – 12x + q = 0 → (3)

c and d are roots of (1)

So, (x + c)(x + d) = 0

x² - (c + d)x + cd = 0

So, c + d = 12 and cd = q → (4)

a, b, c, d are in GP.(Given data)

Similarly A, AR, AR², AR³ also forms a GP, with common ratio R.

From (2),

a + b = 3

A + AR = 3

→ (5)

From (4),

c + d = 12

AR² + AR³ = 12

AR² (1 + R) = 12 → (6)

Substituting value of (1 + R) in (6).

R = 2

Now, substitute value of R in (5) to get value of A,

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.

So, (q + p): (q – p) = 17: 15.

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

r = (ratio between the n term and n-1 term) 3 ÷ 1 = 3

n = 7 terms

∴

⇒

⇒

⇒

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

r = (ratio between the n term and n-1 term) ÷ 1 = = 1.732

n = 10 terms

∴

⇒

⇒

⇒

⇒

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 0.15

r = (ratio between the n term and n-1 term) 0.015÷ 0.15 = 0.1

n = 6 terms

⇒

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

r = (ratio between the n term and n-1 term)

n = 9 terms

∴

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a =

r = (ratio between the n term and n-1 term)

n = 8 terms

∴

⇒

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a =

r = (ratio between the n term and n-1 term)

n = 6 terms

∴

⇒

∴

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a =

r = (ratio between the n term and n-1 term)

n terms

∴ [ Rationalizing the denominator]

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

r = (ratio between the n term and n-1 term)

n terms

∴

⇒

∴

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

r = (ratio between the n term and n-1 term)

n terms

∴

[ Multiplying both numerator and denominator by -1 ]

⇒

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = x³

r = (ratio between the n term and n-1 term)

n terms

∴

⇒

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 )

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

a= x² first part and xy for the second part

r = (ratio between the n term and n-1 term) x² for the first part and xy for the second part

n terms

∴

⇒

This can also be written as

=

=

=

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

a =

r = (ratio between the n term and n-1 term)

n terms

∴

⇒

⇒

⇒

∴

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

=

=

=

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = x², y²

r = (ratio between the n term and n-1 term) x, y

n terms

∴

⇒

We can split the above expression into 2 parts. We will split 2n terms into 2 parts also which will leave it as n terms and another n terms .

=

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a =

r = (ratio between the n term and n-1 term)

n terms

∴

⇒

⇒

⇒

∴

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)

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 3

r =(ratio between the n term and n-1 term) 3

n = 10 terms

⇒

⇒

⇒

Thus, sum of the given expression is

= 20 + (3+3²+3³+… to 10 terms)

= 20 + 88572

=88592

(ii) The given expression can be written as,

( 2¹ + 3^1-1) + (2² + 3^2-1) + …to n terms

= (2 + 3⁰) + ( 2²+ 3¹) + …to n terms

= (2 + 1) +(2² + 3 ) + …to n terms

= (2 + 2² + …to terms) + ( 1 + 3 + … to terms)

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 2, 1

r = (ratio between the n term and n-1 term)2, 3

terms

⇒

⇒

(iii) We can rewrite the given expression as

( 5¹ + 5² + 5³+ …to 8 terms)

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 5

r =(ratio between the n term and n-1 term) 5

n = 8 terms

⇒

⇒

⇒

The expression can be rewritten as

[ Taking 8 as a common factor ]

8( 1+ 11 + 111+ … to n terms)

[Multiplying and dividing the expression by 9]

= ( 9 + 99+ 999 + … to n terms)

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

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

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

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 10

r = (ratio between the n term and n-1 term) 10

n terms

⇒

⇒

∴ The sum of the given expression is

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

= ()

(ii) The given expression can be rewritten as

[ taking 3 common ]

= 3( 1+11+111+ …to n terms)

[ multiplying and dividing the expression by 9 ]

= ( 9+99+999+ … to n terms )

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

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

= ( (10+100+1000+ to n terms) – n )

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 10

r = (ratio between the n term and n-1 term) 10

n terms

⇒

⇒

∴ The sum of the given expression is

= ( (10+100+1000+ to n terms) – n )

= ( - n )

(iii) We can rewrite the expression as

[ taking 7 as a common factor]

= 7(0.1+0.11+0.111+ … to n terms)

[ multiplying and dividing by 9 ]

= ( 0.9+0.99+0.999+ … to n terms )

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

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

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

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 0.1

r = (ratio between the n term and n-1 term) 0.1

n terms

⇒

[multiplying both numerator and denominator by 10]

⇒

∴ The sum of the given expression is

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

= ( n – ( ) )

In this question, we will try to rewrite the given sum of the progression like the formula for the sum a G.P. series.

It is given that S_n = ( 2ⁿ – 1)

The formula for the sum of a G.P. series is,

By solving the 2 equations together, we can say that

⇒

By corresponding the numbers with the variables, we can conclude

a = 1

r = 2

The G.P. series will therefore look like ⇒ 1,2,4,8,16,……to n terms

∴ The given progression is a G.P. series with the common ration being 2.

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Sum of first 3 terms =

Sum of first 6 terms =

∴

⇒

⇒ 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 = or r=1

Either r= or r=1

Since r ≠ 1 [ if r is 1, all the terms will be equal which destroys the purpose ]

∴ r =

Tn represents the n^th term of a G.P. series.

r = 6 ÷ 3 = 2

T_n = ar^n-1

⇒1536 = 3 × 2^n-1

⇒1536 ÷ 3 = 2ⁿ ÷ 2

⇒1536 ÷ 3 × 2 = 2ⁿ

⇒1024 = 2ⁿ

⇒2¹⁰ = 2ⁿ

∴ n = 10

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 3

r = 2

n = 10 terms

∴

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when r>1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 2

r = (ratio between the n term and n-1 term) 6 ÷ 2 = 3

S_n = 728

∴

⇒

⇒

⇒

⇒

⇒

∴

∴ 6 terms must be taken to reach the desired answer.

‘T_n’ represents the n^th term of a G.P. series.

T_n = ar^n-1

⇒486 = a(3)^n-1

⇒486 = a( 3ⁿ ÷ 3) )

⇒486 × 3 = a(3ⁿ)

⇒1458 = a(3ⁿ ) ………(i)

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

∴

⇒

⇒728 × 2 = a (3ⁿ)-a …… [ Putting a(3ⁿ ) = 1458 fromk (i) ]

⇒1456 = 1458 -a

⇒1456-1458 = -a

⇒-2=-a [ Multipying both sides by -1]

⇒a = 2

‘T_n’ represents the n^th term of a G.P. series.

T_n = ar^n-1

⇒

⇒

⇒

⇒

⇒

∴

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 27

r = (ratio between the n term and n-1 term)

n = 10 terms

∴

⇒

⇒

⇒

∴

2^nd term = ar^2-1 = ar¹

5^th term = ar^5-1 = ar⁴

Dividing the 5^th term using the 3^rd term

r ³ = -

∴ r = -

∴ a = 1

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

n = 8 terms

⇒

⇒

∴

4^th term = ar^4-1 = ar³ =

7^th term = ar^7-1= ar⁶ =

Dividing the 7^th term by the 4^th term,

⇒ ……(i)

∴

ar³ = [putting from eqn (i) ]

a =

∴ a = 1

Sum of a G.P. series is represented by the formula, , when |r|<1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Here,

a = 1

n terms

∴

⇒

⇒

∴

Let the terms of the G.P. be a, ar, ar², ar³, … , ar^n-2, ar^n-1

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Thus, the sum of this G.P. series is

The odd terms of this series are a, ar², ar⁴, … , ar^n-2

{since the number of terms of the G.P. series is even; the 2^nd last term will be an odd term.}

Here,

No. of terms will be as we are splitting up the n terms into 2 equal parts of odd and even terms. { since the no. of terms is even, we have 2 equal groups of odd and even terms }

Sum of the odd terms ⇒

⇒

By the problem,

⇒

⇒ r +1 =5

⇒∴ r = 4

Thus, the common ratio (r) = 4

Sum of a G.P. series is represented by the formula, , when r≠1. ‘S_n’ represents the sum of the G.P. series upto n^th terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

Thus, the sum of the first n terms of the G.P. series is,

Sum of (n+1)^th term to 2n^th term

= Sum of the first 2n^th term – the sum of 1^st term to n^th term

= -

=

=

=

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

=

[Cancelling out the common factors from the numerator and denominator ⇒ a, (r-1), (rⁿ – 1) ]

=

Hence Proved.

To find: The amount after three years

Given: (i) Principal – 15625

(ii) Time – 3 years

(iii) Rate – 8% per annum

Formula used:

A = 19683

Ans) 19683

To find: The amount after three days

Given: (i) Principal – 80000

(ii) Time – 3 days

(iii) Rate – 15% per annum

Deduction = P × R × T

= 98.63

The final amount after deduction = 80000 – 98.63

= 79901.37

The value of the machine after 3 days is Rs. 79901.37

To find: Present population of the village

Given: (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

⇒

People left after migration in the very first year = 10000 – 2000

= 8000

Number of people migrated in the second year is 20% of 8000

⇒

People left after migration in the second year = 8000 – 1600

= 6400

Number of people migrated in the third year is 20% of 6400

⇒

People left after migration in the third year = 6400 – 1280

= 5120

Ans) The present population is 5120

To find: The amount after ten years

Given: (i) Principal – 5000

(ii) Time – 10 years

(iii) Rate – 10% per annum

Formula used:

Ans) The amount after years will be Rs.12970

To find: The amount after five years

Given: (i) Principal – 156250

(ii) Time – 5 years

(iii) Rate – 20% per annum

Formula used:

Ans) The amount after five years will be Rs.51200

To find: The number of bacteria after

(i) 2^nd hour

(ii) 5^th hour

(iii) nth hour

Given: (i) Initially, there were 50 bacteria

(ii) Rate – 100% per hour

The formula used:

(i) For 2^nd hour

⇒

⇒

⇒

⇒

⇒

(ii) For 5^th hour

⇒

⇒

⇒

⇒

⇒

(iii) For n^th hour

⇒

⇒

⇒

⇒

Ans) Number of bacteria in a 2^nd hour will be 200, the number of bacteria in a 5^th hour will be 1600 and number of bacteria in an n^th hour will be

To prove: p^th, q^th and r^th terms of any GP are in GP.

Given: (i) p, q and r are in AP

The formula used: (i) General term of GP,

As p, q, r are in A.P.

⇒ q – p = r – q = d = common difference … (i)

Consider a G.P. with the first term as a and common difference R

Then, the p^th term will be

the q^th term will be

the r^th term will be

Considering p^th term and q^th term

From eqn. (i) q – p = d

Considering q^th term and r^th term

From eqn. (i) r – q = d

We can see that p^th, q^th and r^th terms have common ration i.e

Hence they are in G.P.

Hence Proved

To prove: log aⁿ, log bⁿ, log cⁿ are in AP.

Given: a, b, c are in GP

Formula used: (i) log ab = log a + log b

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)

Whenever a,b,c are in AP then 2b = a+c, considering this and the above equation we can say that log aⁿ, log bⁿ, log cⁿ are in AP.

Hence Proved

To prove: , are in AP.

Given: a, b, c are in GP

Formula used: (i)

As, a, b, c are in GP

⇒

Taking log both side

⇒ log b – log a = log c – log b

⇒ 2log b = log a + log c

Dividing by log m

⇒

⇒

⇒

Whenever any number a,b,c are in AP then 2b = a+c, considering this and the above equation we can say that , are in AP

Hence proved

To find: Value of k

Given: k + 12, k – 6 and 3 are in GP

Formula used: (i) when a,b,c are in GP b² = ac

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

Ans) We have two values of k as 0 or 15

To find: The numbers

Given: Three numbers are in A.P. Their sum is 15

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

Let the numbers be a - d, a, a + d

According to first condition

a + d + a +a – d = 15

⇒ 3a = 15

⇒ a = 5

Hence numbers are 5 - d, 5, 5 + d

When 1, 4, 19 be added to them respectively then the numbers become –

5 – d + 1, 5 + 4, 5 + d + 19

⇒ 6 – d, 9, 24 + d

The above numbers are in GP

Therefore, 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

Taking d = 3, the numbers are

5 - d, 5, 5 + d = 5 - 3, 5, 5 + 3

= 2, 5, 8

Taking d = -21, the numbers are

5 - d, 5, 5 + d = 5 – (-21), 5, 5 + (-21)

= 26, 5, -16

Ans) We have two sets of triplet as 2, 5, 8 and 26, 5, -16.

To find: Three numbers

Given: Three numbers are in A.P. Their sum is 21

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

Let the numbers be a - d, a, a + d

According to first condition

a + d + a +a – d = 21

⇒ 3a = 21

⇒ a = 7

Hence numbers are 7 - d, 7, 7 + d

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

The above numbers are in GP

Therefore, 6² = (7 – d) (8 + d)

⇒ 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

Taking d = 4, the numbers are

7 - d, 7, 7 + d = 7 - 4, 7, 7 + 4

= 3, 7, 11

Taking d = -5, the numbers are

7 - d, 7, 7 + d = 7 – (-5), 7, 7 + (-5)

= 12, 7, 2

Ans) We have two sets of triplet as 3, 7, 11 and 12, 7, 2.

To find: Three numbers

Given: Three numbers are in G.P. Their sum is 56

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

Let the three numbers in GP be a, ar, ar²

According to condition :-

a + ar + ar² = 56

a(1 + r + r²) = 56 … (i)

1, 7, 21 be subtracted from them respectively, we obtain the numbers 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)

Multiplying the above eqn. with 7

⇒ 56 = 7a(r² – 2r + 1)

⇒ 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

⇒ Or r = 2

Putting in eqn. (i)

a(1 + r + r²) = 56

a = 32

The numbers are a, ar, ar²

⇒ 32,

⇒ 32, 16, 8

Putting r = 2 in eqn. (i)

a(1 + r + r²) = 56

a = 8

The numbers are a, ar, ar²

⇒ 8,

⇒ 8, 16, 32

Ans) We have two sets of triplet as 32, 16, 8 and 8, 16, 32.

To prove:

Given: a, b, c are in GP

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

a, b, c are in GP,

⇒ b² = ac … (i)

… (ii)

Taking LHS =

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

LHS =

Substituting the value b = from eqn. (ii)

Multiplying and dividing with

⇒

⇒ = RHS

Hence Proved

To prove: (a + b + c)² = 3(ab + bc + ca).

Given: (a – b), (b – c), (c – a) are in GP

Formula used: 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

(i) a(b² + c²) = c(a² + b²)

To prove: a(b² + c²) = c(a² + b²)

Given: a, b, c are in GP

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

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) [b² = ac]

= RHS

Hence Proved

(ii)

To prove: a(b² + c²) = c(a² + b²)

Given: a, b, c are in GP

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

Proof: When a,b,c are in GP, b² = ac

Taking LHS =

[b² = ac]

Hence Proved

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

To prove: (a + 2b + 2c)(a – 2b + 2c) = a² + 4c²

Given: a, b, c are in GP

Formula used: 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)

To prove:

Given: a, b, c are in GP

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

Proof: When a,b,c are in GP, b² = ac

Taking LHS =

[b² = ac]

= RHS

Hence Proved

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

To prove: (b + c)(b + d) = (c + a)(c + a)

Given: a, b, c, d are in GP

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 = (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)

To prove:

Given: a, b, c, d are in GP

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

⇒ d = … (iii)

Taking LHS =

= [From eqn. (iii)]

=

=

=

= [From eqn. (ii)]

=

=

=

=

= RHS

Hence Proved

(iii) (a + b + c + d)² = (a + b)² + 2(b + c)² + (c + d)²

To prove: (a + b + c + d)² = (a + b)² + 2(b + c)² + (c + d)²

Given: a, b, c, d are in GP

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

To prove: are in AP

Given: a, b, c are in GP

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

When a,b,c are in GP, b² = ac

Taking and

⇒

⇒

⇒ [ b² = ac ]

⇒

⇒

⇒

⇒

We can see that

Hence we can say that are in AP.

To prove: a², b², c² are in GP

Given: a, b, c are in GP

Proof: As a, b, c are in GP

⇒ b² = ac … (i)

Considering b², c²

= common ratio = r

⇒ [From eqn. (i)]

⇒ = r

Considering a², b²

= common ratio = r

⇒ [From eqn. (i)]

⇒ = r

We can see that in both the cases we have obtained a common ratio.

Hence a², b², c² are in GP.

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

= common ratio = r

From the above equation, we can say that a³, b³, c³ are in GP

To prove: (a² + b²), (ab + bc), (b² + c²) are in GP

Given: a, b, c are in GP

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

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

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 GP

To prove: (a² – b²), (b² – c²), (c² – d²) are in GP.

Given: a, b, c are in GP

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

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)

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 GP

To prove: are in GP.

Given: a, b, c, d are in GP

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)

Considering

=

=

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

=

=

=

From the above equation, we can say that are in GP.

To prove: p, q, r are in GP

Given: (p² + q²), (pq + qr), (q² + r²) are in GP

Formula used: 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.

To prove: a, (a – b) and (d – c) are in GP.

Given: a, b, c are in AP, and a, b, d 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.

To prove: x², b², y² are in AP.

Given: a, b, c are in AP, and a, x, b and b, y, c are in GP

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 AP.

(i) AM = 25 and GM = 20

To find: Two positive numbers a and b

Given: AM = 25 and GM = 20

Formula used: (i) Arithmetic mean between

(ii) Geometric mean between

Arithmetic mean of two numbers

⇒ a + b = 50

⇒ b = 50 – a … (i)

Geometric mean of two numbers

Substituting value of b from eqn. (i)

a(50 – a) = 400

⇒ 50a – a² = 400

On rearranging

⇒ a² – 50a + 400 = 0

⇒ a² – 40a – 10a + 400

⇒ 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

Therefore two numbers are 10 and 40

(ii) AM = 10 and GM = 8

To find: Two positive numbers a and b

Given: AM = 10 and GM = 8

Formula used: (i) Arithmetic mean between

(ii) Geometric mean between

Arithmetic mean of two numbers

⇒ a + b = 20

⇒ a = 20 – b … (i)

Geometric mean of two numbers

Substituting value of a from eqn. (i)

b(20 – b) = 64

⇒ 20b – b² = 64

On rearranging

⇒ b² – 20b + 64 = 0

⇒ b² – 16b – 4b + 64

⇒ 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

Therefore two numbers are 16 and 4

(i) 5 and 125

To find: Geometric Mean

Given: The numbers are 5 and 125

Formula used: (i) Geometric mean between

Geometric mean of two numbers

= 25

The geometric mean between 5 and 125 is 25

(ii) 1 and

To find: Geometric Mean

Given: The numbers are 1 and

Formula used: (i) Geometric mean between

Geometric mean of two numbers

The geometric mean between 1 and is .

(iii) 0.15 and 0.0015

To find: Geometric Mean

Given: The numbers are 0.15 and 0.0015

Formula used: (i) Geometric mean between

Geometric mean of two numbers

= 0.015

The geometric mean between 0.15 and 0.0015 is 0.015.

(iv) -8 and -2

To find: Geometric Mean

Given: The numbers are -8 and -2

Formula used: (i) Geometric mean between

Geometric mean of two numbers

=

Mean is a number which has to fall between two numbers.

Therefore we will take -4 as our answer as +4 doesn’t lie between -8 and -2

The geometric mean between -8 and -2 is -4.

(v) -6.3 and -2.8

To find: Geometric Mean

Given: The numbers are -6.3 and -2.8

Formula used: (i) Geometric mean between

Geometric mean of two numbers

=

Mean is a number which has to fall between two numbers.

Therefore we will take -4.2 as our answer as +4.2 doesn’t lie between -6.3 and -2.8

The geometric mean between -6.3 and -2.8 is -4.2.

(vi) a³b and ab³

To find: Geometric Mean

Given: The numbers are a³b and ab³

Formula used: (i) Geometric mean between

Geometric mean of two numbers

= a²b²

The geometric mean between a³b and ab³ is a²b².

To find: Three geometric Mean

Given: The numbers and 432

Formula used: (i) r , where n is the number of

geometric mean

Let G₁, G₂ and G₃ be the three geometric mean

Then r

⇒ r

⇒ r

⇒ r

⇒ r

⇒ r = 6

G₁ = ar = ×6 = 2

G₂ = ar²= ×6² = ×36 = 12

G₃ = ar³= ×6³ = ×216 = 72

Three geometric mean between and 432 are 2, 12 and 72.

To find: Four geometric Mean

Given: The numbers 6 and 192

Formula used: (i) r , where n is the number of

geometric mean

Let G₁, G_2, G₃ and G₄ be the three geometric mean

Then r

⇒ r

⇒ r

⇒ r

⇒ 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.

To prove: Prove that a:b

Given: Arithmetic mean is twice of Geometric mean.

Formula used: (i) Arithmetic mean between

(ii) Geometric mean between

AM = 2(GM)

⇒ a + b = 4

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)

⇒

Taking square root both side

Applying componendo and dividend

Hence Proved

To prove: b² is the AM between x² and y².

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

Formula used: (i) Arithmetic mean between

(ii) Geometric mean between

As a, b, c are in A.P.

⇒ 2b = a + c … (i)

As x is the GM between a and b

⇒ x =

⇒ x² = ab … (ii)

As y is the GM between b and c

⇒ y =

⇒ y² = bc … (iii)

Arithmetic mean of x² and y² is

Substituting the value from (ii) and (iii)

Substituting the value from eqn. (i)

= b²

Hence Proved

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.

Formula used:(i) Geometric mean between

(ii) Sum of n terms of A.P.

Let the n geometric means between and b be G_1, G_2, G_{3, …} G_n

Hence a, G_1, G_2, G_{3, …} G_n, b are in GP

⇒ G₁ = ar, G₂ = ar² and so on …

Now, we have n+2 term

⇒ b = ar^n+2-1

⇒ b = arⁿ⁺¹

⇒ … (i)

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

= ar × ar² × ar³ × … arⁿ

= aⁿ × r^{(1+2+3… + n)}

= aⁿ ×

Substituting the value of r from eqn. (i)

= aⁿ ×

= aⁿ ×

= aⁿ ×

=

=

= … (ii)

Single geometric mean between a and b

n^th power of single geometric mean between a and b

Hence Proved

To find: The quadratic equation.

Given: (i) AM of roots of quadratic equation is 10

(ii) GM of roots of quadratic equation is 8

Formula used: (i) Arithmetic mean between

(ii) Geometric mean between

Let the roots be p and q

Arithmetic mean of roots p and q = 10

⇒ = 10

⇒ p + q = 20 = sum of roots … (i)

Geometric mean of roots p and q = = 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

x² –20x + 64

To find: Two geometric Mean

Given: The numbers are 9 and 243

Formula used: (i) r , where n is the number of

geometric mean

Let G₁ and G₂ be the three geometric mean

Then r

⇒ r

⇒ r

⇒ r

⇒ r = 3

G₁ = ar = 9×3 = 27

G₂ = ar²= 9×3² = 9×9 = 81

Two geometric mean between 9 and 243 are 27 and 81.

It is Infinite Geometric Series.

Here, a=8,

The formula used:Sum of an infinite Geometric series

∴Sum

Sum

It is Infinite Geometric Series.

Here, a=6,

The formula used:Sum of an infinite Geometric series

∴Sum=

Sum

It is Infinite Geometric Series

Here, a=√2

∴Sum

Sum

It is Infinite Geometric Series

Here, a=10

r

∴Sum

Sum

This geometric series is the sum of two geometric series:

Sum of geometric series:
Here, a

Sum of geometric series:
Here, a

∴Sum of the given infinite series=sum of both the series

Sum

L.H.S=9^1/3 × 9^1/9 × 9^1/27 × …..∞

=9^{(1/3)+(1/9)+(1/27)+…∞}

The series in the exponent is an infinite geometric series

Whose, a

∴Sum of the series in the exponent

∴L.H.S=9^1/2

=3=R.H.S

Hence, Proved that 9^1/3 × 9^1/9 × 9^1/27 × …..∞ = 3

(i) Let, x=0.3333…

⇒ x=0.3+0.03+0.003+…

⇒ x=3(0.1+0.01+0.001+0.0001+…∞)

⇒ x=3()

This is an infinite geometric series.

Here, a=1/10 and r=1/10

∴x

₌

(ii) _{Let, x=0.231231231….}

_{⇒x=0.231+0.000231+0.000000231+…∞}

⇒ x=231(0.001+0.000001+0.000000001+…∞)

⇒ x=231(++…∞)

This is an infinite geometric series.

Here, a and r=

⇒

₌

(iii) Let, x=3.525252552…

⇒ x=3+0.52+0.0052+0.000052+…∞

⇒ x=3+52(0.01+0.0001+…∞)

⇒ x=3+52(++…∞)

Here, a and r=

⇒

Let, x=0.125125125… …(i)

Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get:

1000x=125.125125125… …(ii)

Equation (ii)-(i),

⇒ 1000x-x=125.125125125-0.125125125=125

⇒ 999x=125

⇒

Let, x=0.423423423… …(i)

Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get:

1000x=423.423423423… …(ii)

Equation (ii)-(i),

⇒ 1000x-x=423.423423423-0.423423423=423

⇒ 999x=423

⇒

Let, x=2.134134134… …(i)

Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get:

1000x=2134.134134134… …(ii)

Equation (ii)-(i),

⇒ 1000x-x=2134.134134134-2.134134134=2132

⇒ 999x=2132

⇒

Given:, a=2

To find:r=?

∴

⇒

⇒ 3(1-r)=1

⇒ 3-3r=1

⇒ 3r=3-1

⇒ r

Common ratio r

Given: &

(because on squaring both first term a and common ratio r will be squared.)

To find: the series

a=20(1-r)…(i)

⇒ …(from (i))

⇒

⇒ 100(1+r)=400(1-r)

⇒ 100+100r=400-400r

⇒ 100r+400r=400-100

⇒ 500r=300

⇒ 5r=3

⇒

Put this value of r in equation (i) we get

∴The infinite geometric series is:8,

Let the first term Of G.P. be a, and common ratio be r.

∴

On cubing each term will become,

a^3, a³r³, ….

∴This sum

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

⇒

⇒

⇒ 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

Thus the first term of G.P. is 19, and the common ratio is 2/3

∴G.P=19,

19,

Given: 5^th term of a GP is 2.

To find: the product of its first nine terms.

First term is denoted by a, the common ratio is denote by r.

∴ ar⁴ = 2

We have to find the value of: a × ar¹ × ar² × ar³ × … × ar⁸

= a⁹r^{1 + 2 + 3 + 4 + … + 8}

= a⁹r³⁶

= (ar⁴)⁹

= (2)⁹

= 512

Ans:512.

Let,

t_{p + q} = m = Ar^{p + q - 1} = Ar^{p - 1}r^q

and

t_{p - q} = n = Ar^{p - q - 1} = Ar^{p - 1}r ^{- q}

We know that p^th term = Ar^{p - 1}

∴ m × n = A²r^{2p - 2}

⇒ Ar^{p - 1} = (mn)^1/2

⇒ p^th term = (mn)^1/2

Ans: p^th term = (mn)^1/2

We have been given that 2^nd, 3^rd and 6^th terms of an AP are the three consecutive terms of a GP.

Let the three consecutive terms of the G.P. be a,ar,ar².

Where a is the first consecutive term and r is the common ratio.

2^nd, 3^rd terms of the A.P. are a and ar respectively as per the question.

∴ The common difference of the A.P. = ar - a

And the sixth term of the A.P. = ar²

Since the second term is a and the sixth term is ar²(In A.P.)

We use the formula:t = a + (n - 1)d

∴ ar² = a + 4(ar - a)…(the difference between 2^nd and 6^th 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:

1)a = 0 which is not expected because all the terms of A.P. and G.P. will be 0.

2)r = 1,which is also not expected because all th terms would be equal to first term.

3)r = 3,which is the required answer.

Ans:common ratio = 3

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 = …(ii)

We know that the equation whose roots are given is =

From (i) and (ii) we get:

Thus, is the required quadratic equation.

Ans: is the required quadratic equation.

It is given that:

a^1/x = b^1/y = c^1/z

Let a^1/x = b^1/y = c^1/z = k

⇒ a^1/x = k

⇒ (a^1/x)^x = k^x…(Taking power of x on both sides.)

⇒ a^{1/x × x} = k^x

⇒ a = k^x

Similarly b = k^y

And c = k^z

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

⇒ b² = ac

Substituting values of a,b,c calculated above,we get:

⇒ (k^y)² = k^xk^z

⇒ k^2y = k^{x + z}

Comparing the powers we get,

2y = x + z

Which is the required condition for x,y,z to be in A.P.

Hence, proved that x,y,z, are in A.P.

To prove: x^{b - c}. y^{c - a}. z^{a - b} = 1….(i)

It is given that a,b,c are in A.P.

⇒ 2b = a + c…(ii)

And x,y,z, are in G.P.

⇒ y² = xz

⇒ x = y²/z

Substitute this value of x in equation (i),we get

L.H.S =

⇒

⇒

⇒

⇒

⇒ 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 x^{b - c}. y^{c - a}. z^{a - b} = 1

It is Infinite Geometric Series.

Here,a = 1,

Formula used:Sum of an infinite Geometric series

∴ Sum = = R.H.S.

Hence, Proved that

_{Let,x = 0.123123123….}

_{⇒x = 0.123 + 0.000123 + 0.000000123 + …∞}

⇒ x = 123(0.001 + 0.000001 + 0.000000001 + …∞)

⇒ x = 123( + …∞)

This is an infinite geometric series.

Here,a and r =

⇒

Ans :₌

Let ,x = 0.6666…

⇒ x = 0.6 + 0.06 + 0.006 + …

⇒ x = 6(0.1 + 0.01 + 0.001 + 0.0001 + …∞)

⇒ x = 6()

This is an infinite geometric series.

Here,a = 1/10 and r = 1/10

∴ x

Ans: ₌

Let,x = 0.68686868…

⇒ x = 0.68 + 0.0068 + 0.000068 + …∞

⇒ x = 68(0.01 + 0.0001 + …∞)

⇒ x = 68( + …∞)

Here,a and r =

⇒

Ans: ₌

Given: second term of a GP is 24 and its fifth term is 81.

To find: sum of first five terms of the G.P.

ar = 24 & ar⁴ = 81

dividing these two terms we get:

⇒

⇒

Taking cube root on both the sides we get,

⇒

_{Substituting this value of r in ar = 24 we get}

_{a = 24/(3/2) = (24 × 2)/3 = 16}

∴ Sum of first Five terms of a G.P. = a(rⁿ - 1)/(r - 1)

=

=

Ans:242

The first three terms of a G.P. are:a,ar,ar²

The first six terms of a G.P. are:a,ar,ar²,ar³,ar⁴,ar⁵

It is given that the ratio of the sum of first three terms is to that of first six terms of a GP is 125 : 152.

⇒ a + ar + ar² = 125x & a + ar + ar² + ar³ + ar⁴ + ar⁵ = 152x

⇒ a + ar + ar² + r³( a + ar + ar²) = 152x

⇒ 125x + r³(125x) = 152x

⇒ r³(125x) = 152x - 125x = 27x

⇒

⇒ r = 3/5

Ans:common ratio =

Let the first three terms of G.P. be

It is given that

⇒

⇒ a = 1

And

⇒

⇒ …(a = 1)

⇒

⇒ 10(1 + r²) = 29r

⇒ 10r² - 29r + 10 = 0

⇒ 10r² - 25r - 4r + 10 = 0

⇒ 5r(2r - 5) - 2(2r - 5) = 0

⇒ (2r - 5)(5r - 2) = 0

⇒ r =

Therefore the first three terms are:

i)if r = then

ii)if r = then

Ans:Common ratio r = and the first three terms are:

i)if r = then

ii)if r = then