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(c) \(\frac{1}{100}.\)

Let the six numbers in G.P. be \(\frac{a}{r^5}\), \(\frac{a}{r^3}\), \(\frac{a}{r}\), ar, ar³, ar⁵

given, \(\frac{a}{r^5}\) x \(\frac{a}{r^3}\) x \(\frac{a}{r}\) x ar x ar³ x ar⁵ = 1000

⇒ a⁶ = 1000 ⇒ a = \(\sqrt{10}\)

Given, T₄ = ar = 1 ⇒ \(\sqrt{10}\)r = 1 ⇒ = \(\frac{1}{\sqrt{10}}\)

∴ Last term of G.P. = ar⁵ = \(\sqrt{10}\) x \(\frac{1}{\sqrt{(10)}^5}\) = \(\frac{1}{100}.\)

\(0.2\overline{34}\) = 0.234 34 34 ....... 

= 0.2 + 0.034 + 0.00034 + 0.0000034 + ...... + ∞

= \(\frac{2}{10}\) + \(\frac{34}{1000}\) + \(\frac{34}{100000}\) + \(\frac{34}{10000000}\) + ..... + ∞

= \(\frac{2}{10}\) + \(\frac{34}{10^3}\)\(\bigg[1+\frac{1}{10^2}+\frac{1}{10^4}+.......\infty\bigg]\)

= \(\frac{2}{10}\) + \(\frac{34}{10^3}\) x \(\bigg(\frac{1}{1-\frac{1}{10^2}}\bigg)\)          \(\big(\because\,S_{\infty}=\frac{a}{1-r}.\text{Here}\,a = 1,\,r=\frac{1}{10^2}\big)\)

= \(\frac{2}{10}\) + \(\frac{34}{1000}\) x \(\frac{100}{99}\) = \(\frac{198+34}{990}\) = \(\frac{232}{990}\) = \(\frac{116}{495}.\)

Let the root of the quadratic equation be a and b.

According to the given condition,

⇒ AM = a+b/2 = A

⇒ a + b = 2A …..(1)

⇒ GM = √ab = G

= ab = G²…(2)

The quadratic equation is given by,

x²– x (Sum of roots) + (Product of roots) = 0

x² – x (2A) + (G²) = 0

x² – 2Ax + G² = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x² – 2Ax + G² = 0.

Given that a and b are roots of x² – 3x + p = 0 

⇒ a + b = 3 and ab = p ...(i) 

It is given that c and d are roots of x² – 12x + q = 0 

⇒ c + d = 12 and cd = q...(ii) 

Also given that a, b, c, d are in G.P. 

Let a, b, c, d be the first four terms of a G.P.

⇒ a = a, b = ar c = ar²d = ar³

Now, 

∴a + b = 3 

⇒ a + ar = 3 

⇒ a(1 + r) = 3…(iii) 

c + d = 12 

⇒ ar² + ar³ = 12 

⇒ ar²(1 + r) = 12.....(iv) 

From (iii) and (iv) we get 

3.r² = 12 

⇒ r² = 4 

⇒ r = ±2 

Substituting the value of r in (iii) we get a = 1 

⇒ b = ar = 2

∴ c = ar² = 2² = 4 

d = ar³ = 2³ = 8 

⇒ ab = p = 2 and cd = 4×8 = 32 

⇒ q + p = 32 + 2 = 34 and q−p = 32−2 = 30

⇒ q + p:q−p = 34:30 = 17:15 

Hence, proved.

(a) \(\frac{-10}{31}.\)

Let the first term of the A.P. be a and common difference d.

Given d = 2 ⇒ T₁₁ of A.P. = a + 10d = a + 20. 

Let the first term of the G.P. be b and common ratio r. 

Given r = 2. Now, the middle term of A.P = middle term of G.P 

⇒ T₆ of A.P = a + 5d = T₆ of G.P = br⁵ 

⇒ a + 5d = br⁵ 

⇒ a + 10 = 32b              (∵ r = 2)                   ...(i) 

Also the last term of A.P. is the first term of G.P.

∴ b = T₁₁ of A.P = a + 20                    ...(ii)

∴ From (i) and (ii) 

a + 10 = 32.(a + 20) 

⇒ 31a = – 630 ⇒ a = \(\frac{-630}{31}\)

∴ Middle term of the entire sequence of 21 terms = 11th term 

= a + 10d

= \(\frac{-630}{31}\) + 20 = \(\frac{-630+620}{31}\) = \(\frac{-10}{31}.\)

(c) \(\frac{7}{81}(179+10^{-20})\)

Let S₂₀ = 0.7 + 0.77 + 0.777 + ..... upto 20 terms 

= 7(0.1 + 0.11 + 0.111 + ..... upto 20 terms) 

= \(\frac{7}{9}\) (0.9 + 0.99 + 0.999 + ..... upto 20 terms) 

= \(\frac{7}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) + upto 20 terms) 

= \(\frac{7}{9}\) [(1 + 1 + 1 + ..... upto 20 terms) – (0.1 + 0.01 + 0.001 + ..... upto 20 terms)]

= \(\frac{7}{9}\)\(\bigg[20-\frac{0.1\{1-(0.1)^2\}}{(0-0.1)}\bigg]\)           \(\bigg(\because\,S_n=\frac{a(1-r^n)}{1-r}, \text{when}\,r<1\bigg)\)

= \(\frac{7}{9}\)\(\bigg[20-\frac{1}{9}\bigg(1-\big({\frac{1}{10}\big)^{20}}\bigg)\bigg]\) = \(\frac{7}{9}\)\(\bigg[20-\frac{1}{9}+\frac{10}{9}^{-20}\bigg]\)

= \(\frac{7}{9}\)\(\bigg[\frac{179+10^{-20}}{9}\bigg]\) = \(\frac{7}{81}(179+10^{-20})\).

Given:

a_n = 2/3ⁿ

Let us consider n = 1, 2, 3, 4, … since n is a natural number.

So,

a₁ = 2/3

a₂ = 2/3² = 2/9

a₃ = 2/3³ = 2/27

a₄ = 2/3⁴ = 2/81

In GP,

a₃/a₂ = (2/27) / (2/9)

= 2/27 × 9/2

= 1/3

a₂/a₁ = (2/9) / (2/3)

= 2/9 × 3/2

= 1/3

∴ Common ratio of consecutive term is 1/3. Hence n ∈ N is a G.P.

Sum of a G.P. series is represented by the formula, S_n = a\(\frac{r^n -1}{r-1}\), when r≠1. 

‘S_n’ represents the sum of the G.P. series up to 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) -a \(\div\) 1 = -a

n terms

\(\therefore\) S_n = 1 \(\times\) \(\frac{(-a)^n -1}{-a-1}\)

[Multiplying both numerator and denominator by -1]

\(\Rightarrow\)  S_n = \(\frac{1-(-a)^n}{1+a}\)

 \(\displaystyle\sum_{n =2}^{10}\) 4ⁿ

Now this term is in GP. 

16, 64, 256…to 10 terms 

∴ Common Ratio = r = \(\frac{64}{16} = 4\)

∴ Sum of GP for n terms = \(\frac{a(r^n -1)}{r-1} \) ........(1)

⇒ a = 16, r = 4, n = 10 

∴ Substituting the above values in (1) we get

⇒ \(\cfrac{16(4^{10} - 1)}{4-1}\)

⇒ 5592400

(a) \(\frac{10}{9}\) (10¹⁰⁰ – 1) – 100 

Let S₁₀₀ = 9 + 99 + 999 + ...... upto 100 terms 

= (10 – 1) + (100 – 1) + (1000 – 1) + ..... + upto 100 terms 

= (10 + 10² + 10³ + .... upto 100 terms) – (1 + 1 + 1 + ..... upto 100 terms)

= \(\frac{10(10^{100}-1)}{10-1}-100\)                  \(\bigg(\because\,S_n=\frac{a(r^n-1)}{r-1}\,\text{when}\,r>1\bigg)\)

= \(\frac{10}{9}\) (10¹⁰⁰ – 1) – 100.

Let the three numbers be a/r, a, ar

So, according to the question

a/r + a + ar = 13/12 … equation (1)

a/r × a × ar = -1 … equation (2)
From equation (2) we get,

a³ = -1

a = -1

From equation (1) we get,

(a + ar + ar²)/r = 13/12

12a + 12ar + 12ar² = 13r … equation (3)

Substituting a = – 1 in equation (3) we get

12( – 1) + 12( – 1)r + 12( – 1)r² = 13r

12r² + 25r + 12 = 0

12r² + 16r + 9r + 12 = 0… equation (4)

4r (3r + 4) + 3(3r + 4) = 0

r = -3/4 or r = -4/3 

Now the equation will be

-1/(-3/4), -1, -1×-3/4 or -1/(-4/3), -1, -1×-4/3

4/3, -1, 3/4 or 3/4, -1, 4/3

∴ The three numbers are 4/3, -1, 3/4 or 3/4, -1, 4/3

Let the three numbers be a/r, a, ar

So, according to the question

a/r × a × ar = 125 … equation (1)

From equation (1) we get,

a³ = 125

a = 5

a/r × a + a × ar + ar × a/r = 87 1/2

a/r × a + a × ar + ar × a/r = 195/2

a²/r + a²r + a² = 195/2

a² (1/r + r + 1) = 195/2

Substituting a = 5 in above equation we get,

5² [(1 + r² + r)/r] = 195/2

1 + r² + r = (195r/2 × 25)

2(1 + r² + r) = 39r/5

10 + 10r² + 10r = 39r

10r² – 29r + 10 = 0

10r² – 25r – 4r + 10 = 0

5r(2r - 5) – 2(2r  -5) = 0

r = 5/2, 2/5

So G.P is 10, 5, 5/2 or 5/2, 5, 10

∴ The three numbers are 10, 5, 5/2 or 5/2, 5, 10

T_n = ar^n-1

a = \(\frac{1}{a^3x^3}\), r = \(\cfrac{ax}{\frac{1}{a^3x^3}}\) = \({a^4}{x^4}\)

 \(\therefore\) T₁₂ = \(\frac{1}{a^3x^3}(a^4 x^4)^{12-1}\)

= \(\frac{1}{a^3x^3}(a^4x^4)^{11}\)

= a⁴¹x⁴¹

∴ The 12 term is a⁴¹ x⁴¹.

T_n = ar^n-1

a = \(\frac{3}{100}\), r =\(\cfrac{\frac{6}{100}}{\frac{3}{10}}\) = \(\frac{2}{10}\)

\(\therefore\) T₈ = \(\frac{3}{100} \times (\frac{2}{10})^{8-1}\)

= \(\frac{3}{100}\times \frac{2^7}{10^7}\)

= \(\frac{3.2^7}{10^9}\)

= \(\therefore \) The 10 term is \(\frac{3.2^7}{10^9}\)

Let the three numbers be a/r, a, ar

So, according to the question

a/r + a + ar = 39/10 … equation (1)

a/r × a × ar = 1 … equation (2)
From equation (2) we get,

a³ = 1

a = 1

From equation (1) we get,

(a + ar + ar²)/r = 39/10

10a + 10ar + 10ar² = 39r … equation (3)

Substituting a = 1 in 3 we get

10(1) + 10(1)r + 10(1)r² = 39r

10r² – 29r + 10 = 0

10r² – 25r – 4r + 10 = 0… equation (4)

5r(2r – 5) – 2(2r – 5) = 0

r = 2/5 or 5/2

so now the equation will be,

1/(2/5), 1, 1 × 2/5 or 1/(5/2), 1, 1 × 5/2

5/2, 1, 2/5 or 2/5, 1, 5/2

∴ The three numbers are 2/5, 1, 5/2

T_n = ar^n-1

a = \(\frac{-3}{4}\), r = \(\cfrac{\frac{1}{2}}{\frac{-3}{4}}\) = \(\frac{-2}{3}\)

\(\therefore\) T₉ = \(\frac{-3}{4}(-\frac{2}{3})^{10-1}\)

= \(\frac{-3}{4} \times \frac{-2^9}{3^9}\)

= \(\frac{2^7}{3^8}\)

\(\therefore\) The 10 Term is \(\frac{2^7}{3^8}\)

T_n = ar^n-1 

a = 4, r = \(\frac{16}{4}\) = 4

∴ T₉ = 4.(4^9-1) 

= 4.4⁸ 

= 4⁹ 

∴ The 9 term is 4⁹

a, b, c are in AP 

So, 2b = a + c …(1) 

b, c, d are in GP 

So, b² = ad …(2) 

Multiply first equation with a and subtract it from 2nd. 

b² – 2ab = ad – ac – a² 

a² + b² – 2ab = a(d – c) 

⇒ (a – b)² = a(d – c) 

As a, (a – b), (d – c) satisfy the geometric mean relationship 

Hence a, (a – b),(d – c) are in G.P.

a, b, c, d are in G.P. 

Therefore, 

bc = ad … (1) 

b² = ac … (2) 

c² = bd … (3) 

To prove: (a² + b²), (b² + c²), (c² + d²) are in G.P, 

we need to prove that: 

(a² + b²) (c² + d²) = (b² + c²)² {deduced using GM relation} 

∴ RHS = (b² + c²)² = b⁴ + c⁴ + 2b²c² 

= a²c² + b²d² + a²d² + b²c² {using equation 2 and 3} 

= c²(a² + b²) + d²(a² + b²) 

= (a² + b²) (c² + d²) = LHS 

∴ (a² + b²), (b² + c²), (c² + d²) are in G.P 

Hence proved.

Put n = 1,2,3,4… 

a₁, a₂, a₃, a₄

_{\(\Rightarrow\)} a₁ = \(\frac{2}{3}\), a₂ = \(\frac{2}{9}\), a₃ = \(\frac{2}{27}\), a₄ = \(\frac{2}{81}\), .....

If a_1, a₂, a₃......

\(\Rightarrow\) (a₂)² = a₁,a₃

\(\Rightarrow\) \((\frac{2}{9})^2\) = \(\frac{2}{3}\) = \(\frac{2}{27}\)

\(\Rightarrow\) \(\frac{4}{81}\) 

So, It is GP.

Common ratio = r= \(\cfrac{\frac{1}{3}}{\frac{1}{2}}\) = \(\frac{2}{3}\)

Common Ratio = r = \(\frac{(a^2 - b^2)}{(a-b)}\) = \(\frac{(a+b)(a-b)}{(a-b)}\) = a+ b

∴ Sum of GP for n terms = \(\frac{a(r^n-1)}{r-1}\) ...... (1)

⇒ a = (a² - b²), r = (a + b), n = n 

∴ Substituting the above values in (1) we get

⇒ \(\frac{(a^2-b^2)((a+b)^n-1)}{a+b-1}\) 

Let first term of GP be a and common ratio be r 

As n^th term of GP is given as – T_n = ar^{n – 1} 

∴ T₄ = ar^{4 – 1} = ar3 Similarly T₁₀ = ar⁹ 

And T₁₆ = ar¹⁵ 

∴ x = ar³, y = ar⁹ & z = ar¹⁵ 

Clearly we observed that x, y, z have a common ratio. 

∴ x,y,z are in GP with common ratio r⁶.

Hence proved.

Common Ratio = r = \(\frac{2}{4} = \frac{1}{2}\)

∴ Sum of GP for n terms = \(\frac{a(r^n-1)}{r-1}\) ....... (1)

⇒ a = 4, r = \(\frac{1}{2}\), n = 10

∴ Substituting the above values in (1) we get,

⇒ \(\cfrac{\big(4(\frac{1}{2})^8 - 1\big)}{\frac{1}{2} - 1}\)

⇒ \(\frac{-4 \times 255 \times 2}{-1 \times 256}\)

⇒ \(\frac{255}{32}\)

Common Ratio = r = \(\frac{\frac{-1}{2}}{1}\) = \(\frac{-1}{2}\)

∴ Sum of GP till infinity = \(\frac{a}{1-r}\) ........ (1)

⇒ a = 1, r = \(\frac{-1}{2}\)

∴ Substituting the above values in (1), we get,

⇒ \(\frac{1}{1-(\frac{-1}{2})}\)

⇒ \(\frac{1}{\frac{3}{2}}\)

⇒ \(\frac{2}{3}\)

Common Ratio = r = \(\frac{3}{1}\) = 3

\(\therefore\) Sum of GP for n terms = \(\frac{a(r^n -1)}{r-1}\)..... (1)

⇒ a = 1, r = 3, n = 8

∴ Substituting the above values in (1) we get

⇒ \(\frac{1(3^8 -1)}{3-1}\)

⇒ \(\frac{3^8 -1}{2}\)

⇒ 3280

a, b, c are in G.P 

Therefore 

b² = ac … (1) 

We have to prove a² + b², ab + bc, b² + c² are in GP or we need to prove: 

(ab + bc)² = (a² + b²).(b² + c²) {using GM} Take LHS and proceed: 

⇒ LHS = (ab + bc)² = a²b² + 2ab²c + b²c² 

∵ b² = ac 

⇒ LHS = a²b² + 2b²(b²) + b²c² 

⇒ LHS = a²b² + 2b⁴ + b²c² 

⇒ LHS = a²b² + b⁴ + a²c² + b²c² {again using b² = ac } 

⇒ LHS = b²(b² + a²) + c²(a² + b²) 

⇒ LHS = (a² + b²)(b² + c²) = RHS 

Hence a² + b², ab + bc, b² + c² are in GP.

As a, b, c are in G.P. 

Therefore b² = ac … (1) 

We have to prove a³, b³, c³ are in GP or we need to prove: (b³)² = (a³c³) {using idea of GM} 

On cubing equation 1 we get, 

⇒ b⁶ = a³c³ 

⇒ (b³)² = (a³c³) 

Hence a³,b³,c³ are in GP.

Given: A.M = 25, G.M = 20.

G.M = √ab

A.M = (a + b)/2

So,

√ab = 20 ……. (1)

(a + b)/2 = 25……. (2)

a + b = 50

a = 50 – b

Putting the value of ‘a’ in equation (1), we get,

√[(50 - b)b] = 20
50b – b² = 400

b² – 50b + 400 = 0

b² – 40b – 10b + 400 = 0

b(b – 40) – 10(b – 40) = 0

b = 40 or b = 10

If b = 40 then a = 10

If b = 10 then a = 40

∴ The numbers are 10 and 40.

(i) Let a = 4, b = -2, c = 1. 

In GP, b²= ac 

⇒ (-2)² = 4.1 

⇒ 4 = 4 

Common ratio = r = \(\frac{-2}{4}\) = \(\frac{-1}{2}\)

(ii) Let a = \(\frac{-2}{3}\) , b = -6, c = -54. 

In GP, b²= ac 

⇒ (-6)² = \(\frac{-2}{3}\) x -54 

⇒ 36 = 36 

Common ratio = r = \(\cfrac{-6}{\frac{-2}{3}}\) = 9

(iii) Let a = a, b = \(\frac{3a^2}{4}, \) C = \(\frac{9a^2}{16}\)

In GP, b² = ac

\(\Rightarrow\) \((\frac{3a^2}{4})^2\) = \(\frac{9a^2}{16} \times a\)

\(\Rightarrow\) \(\frac{9a^4}{4} = \frac{9a^4}{16}\)

Common ratio = r = \(\frac{\frac{3a^2}{4}}{a}\) = \(\frac{3a}{4}\)

(iv) Let a = \(\frac{1}{2}\), b = \(\frac{1}{3}\), c = \(\frac{2}{9}\)

In GP, b² = ac

\(\Rightarrow\) \((\frac{1}{3})^2\) = \(\frac{1}{2} \times \frac{2}{9}\)

\(\Rightarrow\) \(\frac{1}{9} = \frac{1}{9}\)

Common ratio = r = \(\cfrac{\frac{1}{3}}{\frac{1}{2}}\) = \(\frac{2}{3}\)

a, b, c, d are in G.P. Therefore, 

bc = ad … (1) 

b² = ac … (2) 

c² = bd … (3) 

To prove: (a² – b²), (b² – c²), (c² – d²) are in G.P, we need to prove that: 

(a² – b²) (c² – d²) = (b² – c²)² {deduced using GM relation} 

∴ RHS = (b² – c²)² 

= b⁴ + c⁴ – 2b²c² 

= a²c² + b²d² – a²d² – b²c² {using equation 2 and 3} 

= c²(a² – b²) – d²(a² – b²) 

= (a² – b²) (c² – d²) = LHS 

∴ (a² – b²), (b² – c²), (c² – d²) are in G.P

Hence proved.

(i) 4, -2, 1, -1/2, ….

Let a = 4, b = -2, c = 1

In GP,

b² = ac

(-2)² = 4(1)

4 = 4

So, the Common ratio = r = -2/4 = -1/2

(ii) -2/3, -6, -54, ….

Let a = -2/3, b = -6, c = -54

In GP,

b² = ac

(-6)² = -2/3 × (-54)

36 = 36

So, the Common ratio = r = -6/(-2/3) = -6 × 3/-2 = 9

(iii) a, 3a²/4, 9a³/16, ….

Let a = a, b = 3a²/4, c = 9a³/16

In GP,

b² = ac

(3a²/4)² = 9a³/16 × a

9a⁴/4 = 9a⁴/16

So, the Common ratio = r = (3a²/4)/a = 3a²/4a = 3a/4

(iv) 1/2, 1/3, 2/9, 4/27, …

Let a = 1/2, b = 1/3, c = 2/9

In GP,

b² = ac

(1/3)² = 1/2 × (2/9)

1/9 = 1/9

So, the Common ratio = r = (1/3)/(1/2) = (1/3) × 2 = 2/3

a, b, c, d are in G.P. 

Therefore,

bc = ad … (1) 

b² = ac … (2) 

c² = bd … (3) 

LHS = b² + bd + bc + cd

 ⇒ LHS = ac + bd + bc + cd {on substituting value of b² } …(1) 

RHS = c² + cd + ac + ad 

⇒ RHS = bd + cd + ac + bc {putting value of c²} …(2) 

From equation 1 and 2 we can say that 

LHS = RHS 

Hence proved

We know that GM between a and b is √ab

Let a = 2 and b = 1/4

GM = √(2 × 1/4)

= √(1/2)

= 1/√2

∴ value of a is 1/√2

a, b, c, d are in G.P. Therefore, 

bc = ad … (1) 

b² = ac … (2) 

c² = bd … (3)

To prove: \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\) are in G.P, we need to prove that:

\(\frac{1}{(c^2 + d^2)(a^2 + b^2)}\) = \(\frac{1}{(b^2 + c^2)^2}\) {deduced using GM relation}

Or, (b² + c²)² = (a² + b²)(c² + d²) 

Take LHS and proceed to prove 

LHS = (b² + c²)² 

= b⁴ + c⁴ + 2b²c² 

= a²c² + b²d² + a²d² + b²c² {using equation 2 and 3} 

= c²(a ² + b²) + d²(a² + b²) 

= (a² + b²) (c² + d²) = RHS

\(\therefore\)  \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\)  are in GP

Hence proved

Given as (a – b), (b – c), (c – a) are in G.P

\(\therefore\) \(\frac{b-c}{a-b}\) = \(\frac{c-a}{b-c}\) = common ratio

⇒ (b – c)² = (a – b)(c – a) 

As we have to prove :(a + b + c)² = 3(ab + bc + ca) so we proceed as follows: 

⇒ b² + c² – 2bc = ac – a² – bc + ab 

⇒ a² + b² + c² = ac + ab + bc 

Add 2(ac + ab + bc) to both sides: 

⇒ a² + b² + c² + 2(ac + ab + bc) = ac + ab + bc + 2(ac + ab + bc)

⇒ (a + b + c)² = 3(ab + bc + ca) 

Hence Proved.

(i) a², b², c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

on squaring both the sides we get,

(b²)² = (ac)²

(b²)² = a²c²

∴ a², b², c² are in G.P.

(ii) a³, b³, c³

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

on squaring both the sides we get,

(b²)³ = (ac)³

(b²)³ = a³c³

(b³)² = a³c³

∴ a³, b³, c³ are in G.P.

(iii) a² + b², ab + bc, b² + c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

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

Let us consider LHS: (ab + bc)²

Upon expansion we get,

(ab + bc)² = a²b² + 2ab²c + b²c²

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

= a²b² + 2b⁴ + b²c²

= a²b² + b⁴ + a²c² + b²c² {again using b² = ac }

= b²(b² + a²) + c²(a² + b²)

= (a² + b²)(b² + c²)

= RHS

∴ LHS = RHS

Hence a² + b², ab + bc, b² + c² are in GP.

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

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

Given:

a + b + c = 18

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

d = 6 – a … (i)

Now, according to the question:

a + 4, a + d + 4, and a + 2d + 36

they are now in GP, that is:

(a + d + 4)/(a + 4) = (a + 2d + 36)/(a + d + 4)
(a + d + 4)² = (a + 2d + 36)(a + 4)

a² + d² + 16 + 8a + 2ad + 8d = a² + 4a + 2da + 36a + 144 + 8d

d² – 32a – 128

(6 – a)² – 32a – 128 = 0

36 + a² – 12a – 32a – 128 = 0

a² – 44a – 92 = 0

a² – 46a + 2a – 92 = 0

a(a – 46) + 2(a – 46) = 0

a = – 2 or a = 46

d = 6 –a

d = 6 – (– 2) or d = 6 – 46

d = 8 or – 40

Then,

For a = -2 and d = 8, the A.P is -2, 6, 14

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

∴ The numbers are – 2, 6, 14 or 46, 6, – 34

(i) 2 and 8

GM between a and b is √ab

Let a = 2 and b = 8

GM = √2 × 8

= √16

= 4

(ii) a³b and ab³

GM between a and b is √ab

Let a = a³b and b = ab³

GM = √(a³b × ab³)

= √a⁴b⁴

= a²b²

(iii) –8 and –2

GM between a and b is √ab

Let a = –2 and b = –8

GM = √(–2 × –8)

= √–16

= -4

Now, as a,b,c are in GP. 

Using the idea of geometric mean we can write

∴ b² = ac …(1) 

Put in the LHS of the given equation to be proved 

LHS = a(ac + c²⁾ {putting b² = ac} 

⇒ LHS = a²c + ac² 

⇒ LHS = c(a² + ac) 

Again put ac = b² 

⇒ LHS = c(a² + b²) = RHS 

∴ L.H.S = R.H.S 

Hence proved

As, 

a, b, c, d are in G.P, let r be the common ratio. 

Therefore, 

b = ar … (1) 

c = ar² … (2) 

d = ar³ … (3) 

If we show that: (ab + bc + cd)² = (a² + b² + c²) (b² + c² + d²) 

we can say that: 

(a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P 

As, (ab + bc + cd)² = (a²r + a²r³ + a²r⁵)² 

⇒ (ab + bc + cd)² = a⁴r²(1 + r² + r⁴)² …(4) 

As, 

(a² + b² + c²)( b² + c² + d²) = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶) 

⇒ (a² + b² + c²)( b² + c² + d ²) = a⁴r²(1 + r² + r⁴)(1 + r² + r⁴) 

⇒ (a² + b² + c²)( b² + c ² + d²) = a⁴r²(1 + r² + r⁴)² …(5) 

From equation 4 and 5, we have: 

(ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²) 

Hence, 

We can say that (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.

a, b, c, d are in G.P. 

Therefore, 

bc = ad … (1) 

b² = ac … (2) 

c² = bd … (3) 

If somehow we use RHS and Make it equal to LHS, our job will be done. 

we can manipulate the RHS of the given equation as 

Note: Here we are manipulating RHS because working with a simpler algebraic equation is easier and this time RHS is looking simpler. 

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

⇒ RHS = a² + b² + 2ab + 2(c² + b² + 2cb) + c² + d² + 2cd 

⇒ RHS = a² + b² + c² + d² + 2ab + 2(c² + b² + 2cb) + 2cd

Put c² = bd and b² = ac, we get

⇒ RHS = a² + b² + c² + d² + 2(ab + ad + ac + cb + cd) 

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

⇒ RHS = (a + b + c + d)² = LHS 

Hence Proved.

Let the three numbers be a, ar, ar² 

According to the question

a + ar + ar² = 56 … (1)

Let us subtract 1,7,21 we get,

(a – 1), (ar – 7), (ar² – 21)
The above numbers are in AP.

If three numbers are in AP, by the idea of the arithmetic mean, we can write 2b = a + c

2 (ar – 7) = a – 1 + ar² – 21

= (ar² + a) – 22

2ar – 14 = (56 – ar) – 22
2ar – 14 = 34 – ar

3ar = 48

ar = 48/3

ar = 16

a = 16/r …. (2)

Now, substitute the value of a in equation (1) we get,

(16 + 16r + 16r²)/r = 56

16 + 16r + 16r² = 56r

16r² – 40r + 16 = 0

2r² – 5r + 2 = 0

2r² – 4r – r + 2 = 0

2r(r – 2) – 1(r – 2) = 0

(r – 2) (2r – 1) = 0

r = 2 or 1/2

Substitute the value of r in equation (2) we get,

a = 16/r

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

= 8 or 32

∴ The three numbers are (a, ar, ar²) is (8, 16, 32)

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

bc = ad

c² = bd

Let us consider LHS: (ab – cd) / (b² – c²)

(ab – cd) / (b² – c²) = (ab – cd) / (ac – bd)

= (ab – cd)b / (ac – bd)b

= (ab² – bcd) / (ac – bd)b

= [a(ac) – c(c²)] / (ac – bd)b

= (a²c – c³) / (ac – bd)b

= [c(a² – c²)] / (ac – bd)b

= [(a+c) (ac – c²)] / (ac – bd)b

= [(a+c) (ac – bd)] / (ac – bd)b

= (a+c) / b

= RHS

∴ LHS = RHS

Hence proved.

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

bc = ad

c² = bd

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

Let us expand

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

= a² + b² + 2ab + 2(c² + b² + 2cb) + c² + d² + 2cd

= a² + b² + c² + d² + 2ab + 2(c² + b² + 2cb) + 2cd

= a² + b² + c² + d² + 2(ab + bd + ac + cb +cd) [Since, c² = bd, b² = ac]

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

∴ RHS = LHS

Hence proved.

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

bc = ad

c² = bd

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

Upon expansion we get,

(b + c) (b + d) = b² + bd + cb + cd

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

= c (a + c) + d (a + c)

= (a + c) (c + d)

= RHS

∴ LHS = RHS

Hence proved.

A = 1 + r^a + r^2a + r^3a + ..... ∞

⇒ A = \(\frac{a}{1-r^a}\)      \(\big(\because\,S_\infty = \frac{a}{1-r}.\text{Here}\,a=1, \text{common ratio}=r^a\big)\)

⇒ 1 - r^a = \(\frac{1}{A}\) ⇒ r^a = 1 - \(\frac{1}{A}\)= \(\frac{A-1}{A}\) ⇒ a log r = log \(\big(\frac{A-1}{A}\big)\)

⇒ a = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\,r}\)                        .....(i)

Now,  B = 1 + r^b + r^2b + r^3b + .... ∞

 ⇒ B = \(\frac{a}{1-r^b}\)  ⇒ 1 - r^b = \(\frac{1}{B}\) ⇒ r^b = 1 - \(\frac{1}{B}\)= \(\frac{B-1}{B}\) ⇒ log r^b = log \(\big(\frac{B-1}{B}\big)\)

⇒ b log r = log \(\big(\frac{B-1}{B}\big)\) ⇒ b = \(\frac{\text{log}\big(\frac{B-1}{B}\big)}{\text{log}\,r}\)                .....(ii)

∴ \(\frac{a}{b}\) = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\,r}\) x \(\frac{\text{log}\,r}{\text{log}\big(\frac{B-1}{B}\big)}\)                  (From (i) and (ii))

⇒ \(\frac{a}{b}\) = \(\frac{\text{log}\big(\frac{A-1}{A}\big)}{\text{log}\big(\frac{B-1}{B}\big)}\) = \(\text{log}_{\big(\frac{B-1}{B}\big)}\)\({\big(\frac{A-1}{A}\big)}\).              \(\bigg[\because\frac{\text{log}\,a}{\text{log}\,b}\) = log_b a\(\bigg].\)

(i) a(b² + c²) = c(a² + b²)

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: a(b² + c²)

Now, substituting b² = ac, we get

a(ac + c²)

a²c + ac²

c(a² + ac)

Substitute ac = b² we get,

c(a² + b²) = RHS

∴ LHS = RHS

Hence proved.

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: a²b²c² [1/a³ + 1/b³ + 1/c³]

a²b²c²/a³ + a²b²c²/b³ + a²b²c²/c³

b²c²/a + a²c²/b + a²b²/c

(ac)c²/a + (b²)²/b + a²(ac)/c [by substituting the b² = ac]

ac³/a + b⁴/b + a³c/c

c³ + b³ + a³ = RHS

∴ LHS = RHS

Hence proved.

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: (a + b + c)² / (a² + b² + c²)

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

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

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

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

= RHS

∴ LHS = RHS

Hence proved.

(iv) 1/(a² – b²) + 1/b² = 1/(b² – c²)

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

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

Let us take LCM

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

= a² / (a²b² – b⁴)

= a² / (a²b² – (b²)²)

= a² / (a²b² – (ac)²) [Since, b² = ac]

= a² / (a²b² – a²c²)

= a² / a²(b² – c²)

= 1/ (b² – c²)

= RHS

∴ LHS = RHS

Hence proved.

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

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

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

Upon expansion we get,

(a + 2b + 2c) (a – 2b + 2c) = a² – 2ab + 2ac + 2ab – 4b² + 4bc + 2ac – 4bc + 4c²

= a² + 4ac – 4b² + 4c²

= a² + 4ac – 4(ac) + 4c² [Since, b² = ac]

= a² + 4c²

= RHS

∴ LHS = RHS

Hence proved.

(d) \(\frac{xy}{x+y-1}\)

Since a and b are proper fractions, | a | < 1, | b | < 1 

∴ \(x\) = 1 + a + a² + ..... ∞ = \(\frac{1}{1-a}\)           \(\big(\because\,S_\infty=\frac{a}{1-r}\big)\)

and y = 1 + b + b² + ..... ∞ = \(\frac{1}{1-b}\) 

Also, 1 + ab + a²b² + .....∞ = \(\frac{1}{1-ab}\)        ....(i)  

Now  \(x\) = \(\frac{1}{1-a}\) ⇒ \(x\) – \(x\)a = 1

⇒ \(x\)a = \(x\) – 1 ⇒ a = \(\frac{x-1}{x}\)                 .....(ii)

y = \(\frac{1}{1-b}\)  ⇒ y – yb = 1

⇒ yb = y – 1 ⇒ b = \(\frac{y-1}{y}\)                  .....(iii)

∴ Putting the values of a & b from (ii) and (iii) in (i), we get 

Reqd. sum = \(\frac{1}{1-ab}\) = \(\frac{1}{1-\big(\frac{x-1}{x}\big)\big(\frac{y-1}{y}\big)}\)

= \(\frac{xy}{xy-(xy-x-y+1)}\) = \(\frac{xy}{x+y-1}\).

As,

a, b, c are in G.P, let r be the common ratio. 

Therefore, 

b = ar … (1) 

c = ar² … (2) 

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

As, LHS = (a + 2b + 2c) (a – 2b + 2c) 

⇒ LHS = (a + 2ar + 2ar²)(a – 2ar + 2ar²) 

⇒ LHS = a²(1 + 2r + 2r²)(1 – 2r + 2r²) 

⇒ LHS = a² (1 + 4r² + 4r⁴ – 4r²) 

⇒ LHS = a²(1 + 4r⁴) 

And RHS = a² + 4a 2r⁴ = a²(1 + 4r⁴) 

Clearly, LHS = RHS 

Hence proved

Given:

Sum of G.P = 3069/512

Where, a = 3, r = (3/2)/3 = 1/2, n = ?

By using the formula,

Sum of GP for n terms = a(rⁿ – 1 )/(r – 1)

3069/512 = 3 ((1/2)ⁿ – 1)/ (1/2 – 1)

3069/512 × 3 × 2 = 1 – (1/2)ⁿ

3069/3072 – 1 = – (1/2)ⁿ

(3069 – 3072)/3072 = – (1/2)ⁿ

-3/3072 = – (1/2)ⁿ

1/1024 = (1/2)ⁿ

(1/2)¹⁰ = (1/2)ⁿ

10 = n

∴ 10 terms are required to make 3069/512

a, b, c are in G.P. ⇒ b² = ac               ...(i) 

4a, 5b, 4c are in A.P. ⇒ 2 x 5b = 4a + 4c 

⇒ 10b = 4a + 4c ⇒ 5b = 2a + 2c               ...(ii) 

Also, given a + b + c = 70                   ...(iii) 

⇒ 2a + 2b + 2c = 140 ⇒ 5b + 2b = 140                 (From (ii)) 

⇒ 7b = 140 ⇒ b = 20. 

Now, from (i), 400 = ac.            (∵ b = 20) 

Also, from (ii), a + 20 + c = 70 ⇒ a + c = 50 

∴ (a – c)² = (a + c)² – 4ac 

= 2500 – 1600 = 900 

⇒ a – c = ± 30             

∴ a + c = 50  and a - c = ± 30 ⇒ a = 40, c = 10 or a = 10, c = 40. 

∴ The least value out of a, b and c is 10.

1 + (1 + x) + (1 + x + x²) + ....... n terms 

⇒ Required sum = \(\frac{1}{(1-x)}\) [(1 – x) + (1 – x) (1 + x) + (1 – x) (1 + x + x²) + ..... n terms]

= \(\frac{1}{(1-x)}\) [(1 – x) + (1 – x²) + (1 – x³) + ..... n terms]

= \(\frac{1}{(1-x)}\) [(1 + 1 + 1 + .... n terms) – (x + x² + x³ + ..... n terms]

= \(\frac{1}{(1-x)}\)\(\bigg[n-\frac{x(1-x^n)}{(1-x)}\bigg]\)       \(\bigg(\because{S_n}=\frac{a(1-r^n)}{(1-r)},\text{Here}\,a= x, r=x\bigg)\)

\(\frac{n(1-x)-x(1-x^n)}{(1-x)^2}\).