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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

Given : \(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\)

To Prove : a, b, c, and d are in G.P

Proof :

Applying component and dividend to the given expression, we get,

\(\frac{a + bx}{a - bx} = \frac{b +cx}{b-cx} = \frac{c+dx}{c- dx}\)

\(\frac{a} {bx} = \frac{b}{cx} = \frac{c}{dx}\)

\(\frac{a}{b} = \frac{b}{a} = \frac{c}{d}\)

Clearly, a, b, c and d are in G.P.

Hence proved.

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

\(\frac{1}{R} \) = Common Ratio

So, an n^th term from the end of GP, A_n = (ARⁿ)\(\big(\frac{1}{R^{n-1}} \big)\) = AR → (2)

So, the product of nth 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.

a, b, c, d are in G.P ⇒ b = ar, c = ar², d = ar³, where r = common ratio 

∴ (ab + bc + cd)² = (a.ar + ar . ar² + ar² . ar³)² = [a²r (1 + r² + r⁴)]²   ...(i) 

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

= a² (1 + r² + r⁴) . a²r² (1 + r² + r⁴) = a⁴r² (1 + r² + r⁴)² 

= [a²r (1 + r² + r⁴)]² = (ab + bc + cd)²                                (From (i))

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

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

Then, Product = 1000

⇒ \(\frac{a}{r^5}\) x \(\frac{a}{r^3}\) x \(\frac{a}{r}\) x ar x ar³ x ar⁵

⇒ a6 = 1000 ⇒ a = \(\sqrt{10}\)

Given, Fourth term = t₄ = ar = 1

⇒ \(\sqrt{10}\) x r = 1 ⇒ \(\frac{1}{\sqrt{10}}\)

∴ Last term = ar⁵ = \(\sqrt{10}\) x\(\bigg(\)\(\frac{1}{\sqrt{10}}\)\(\bigg)\)= \(\frac{1}{100}.\)

S_n = 7 + 77 + 777 + ..... to n terms 

= 7 [1 + 11 + 111 + ..... to n terms]

= \(\frac{7}{9}\) [9 + 99 + 999 + ..... to n terms]

= \(\frac{7}{9}\) [(10 – 1) + (100 – 1) + (1000 – 1) + ..... to n terms]

= \(\frac{7}{9}\) [{10 + 10² + 10³ + ..... to n terms} – {1 + 1 + 1 + ..... to n terms}]

= \(\frac{7}{9}\) \(\bigg[\frac{10(10^n-1)}{10-1}-n\bigg]\) = \(\frac{7(10^{n+1})-10}{81}-\frac{7}{9}n.\)

0.2 + 0.22 + 0.222 + ..... to n terms 

= 2[0.1 + 0.11 + 0.111 + ..... to n terms] 

= \(\frac{2}{9}\) [0.9 + 0.99 + 0.999 + ..... to n terms] 

= \(\frac{2}{9}\) [(1 – 0.1) + (1 – 0.01) + (1 + 0.001) + ..... to n terms]

= \(\frac{2}{9}\) [(1 + 1 + 1 + ..... to n terms) – (0.1 + 0.01 + 0.001 + ..... to n terms)]

= \(\frac{2}{9}\)\(\bigg[n-\frac{0.1(1-(0.1)^n}{(1-0.1)}\bigg]\) = \(\frac{2}{9}\)\(\bigg[n-\frac{\frac{1}{10}\big(1-\frac{1}{10^n}\big)}{1-\frac{1}{10}}\bigg]\) = \(\frac{2}{9}\)\(\bigg[n-\frac{1}{9}\bigg(1-\frac{1}{10^n}\bigg)\bigg]\).

(c) a = 1, r = \(\frac{3}{4}\)

Sum of an infinite G.P = \(\frac{a}{1-r},\) where first term = a, common ratio = r and | r | < 1

Given, \(\frac{a}{1-r}\) = 4 and ar = \(\frac{3}{4}\)

⇒ a = 4 – 4r and a = \(\frac{3}{4r}\)

⇒ \(\frac{3}{4r}\) = 4 - 4r ⇒ 3 = 16r - 16r²

⇒ 16r² – 16r + 3 = 0 

⇒ (4r – 3) (4r – 1) = 0 

⇒ 4r = 3 or 4r = 1 ⇒ r = \(\frac{3}{4}\) or \(\frac{1}{4}\)

Now when r = \(\frac{3}{4}\), a = \(\frac{3}{4\times\frac{3}{4}}\) = 1

r = \(\frac{1}{4}\), a = \(\frac{3}{4\times\frac{1}{4}}\) = 3

∴ (a, r) = \(\big(1,\frac{3}{4}\big)\) or \(\big(3,\frac{1}{4}\big)\).

We know when three terms say a,b,c are in GP

We can write

b² = a.c

∴ According to the given data

We can write

(a^x/2)² = log_xa . log_bx

a^x = log_xa . log_bx

⇒ a^x = \(\frac{log_ba}{log_bx}\) x log_bx

⇒ a^x = log_ba

Multiplying by log_a to both sides we get

⇒ log_a (a^x) = log_a (log_ba)

⇒ x log_aa = log_a (log_ba)

⇒ x = log_a (log_ba)

Let the first term be a and the common ratio be R.

∴ According to the question,

a_p = x.

a_q = t

a_r = z.

We know that a_n = aR^n-1

∴ a_p = aR^p-1= x

a_q = aR^q-1= y

a_r = aR^r-1= z

⇒ x^q-r = (aR^p-1)^q-r

⇒ y^r-p = (aR^q-1)^r-p

⇒ z^p-q = (aR^r-1)^p-q

Multiplying the above three equations we get

x^q-r.y^r-p.z^p-q = (a^q-r.R^pq-pr-q+r). (a^r-p.R^rq-pq-r+p). (a^p-q.R^pr-qr-p+q)

= (a^q-r+r-p+p-q.R^{pq-pr-q+r+rq-pq-r+p+pr-qr-p+q})

= (a⁰.R⁰)

= 1

Let the first term be a and the common ratio be R. 

∴ According to the question, 

a_p = x. 

a_q = t 

a_r = z. 

We know that a_n = aR^n-1 

∴ a_p = aR^p-1= x 

a_q = aR^q-1= y 

a_r = aR^r-1= z 

⇒ x^q-r = (aR^p-1)^q-r 

⇒ y ^r-p = (aR^q-1)^r-p 

⇒ z^p-q = (aR^r-1)^p-q 

Multiplying the above three equations we get 

x ^q-r .y ^r-p.z ^p-q 

= (a^q-r .R^pq-pr-q+r). (a^r-p.R^rq-pq-r+p). (a^p-q.R^pr-qr-p+q) 

=(a ^q-r+r-p+p-q.R ^{pq-pr-q+r+rq-pq-r+p+pr-qr-p+q}) = (a⁰.R⁰) 

= 1.

Let two roots be a and b 

∴ The arithmetic mean is given by \(\frac{a+b}{2}\)

⇒ A = \(\frac{a+b}{2}\)

⇒ Geometric mean is given by √a.b 

⇒ G = √a.b

Quadratic equation can be written as 

⇒ x² - (a + b)x + ab = 0 

Where a and b are roots of given equation

Substituting AM and GM 

⇒ x²-2Ax+G² = 0.

⇒ Let the first term be a and the common ratio be r.

∴ According to the question,

a_p+q = m.

a_p-q = n.

a_n = ar^n-1

a_p+q = a.r^p+q-1

a_p-q = a.r^p-q-1

∴ a.r^p+q-1 = m.

a.r^p-q-1 = n.

Multiplying above two equations we get

a²r^{(p+q-1+(p-q-1)} = a²r^(2p-2)

a²r^(2p-2) = m.n

(ar)^2(p-1) = m.n

∴ ar^p-1 = √m.n

⇒ P^th term is given by a.r^p-1

∴ ar^p-1 = √m.n

⇒ Let the first term be a and the common ratio be r. 

∴ According to the question, 

a_p+q = m. 

a_p-q = n. 

a_n = ar^n-1 

a_p+q = a.r^p+q-1 

a_p-q = a.r^p-q-1 

∴ a.r^p+q-1 = m. 

a.r^p-q-1 = n. 

Multiplying above two equations we get 

a²r^{(p+q-1+(p-q-1)} = a²r^(2p-2) 

a²r^(2p-2) = m.n 

(ar)^2(p-1) = m.n 

∴ ar^p-1 =√m.n 

⇒ P^th term is given by a.r^p-1 

∴ ar^p-1 =√m.n

Let us suppose a and b are two numbers. 

Let us say G is the Geometric mean of a and b. 

∴ a, G and b must be in Geometric Progression or GP. 

This means, common ratio = G/a = b/G 

Or, G₂ = ab 

Or, Gn = n(ab) ............ (1)

Now, let us say G₁ , G₂ , G₃ ,.......G_n are n geomteric means between a and b. 

Which means that 

a , G₁ , G₂ , G₃ ...... G_n , b form a G.P. 

Note that the above GP has n+2 terms and the first term is a and last term is b, 

which is also the (n+2)^th term 

Hence, b = ar^n+2-1 

where a is the first term. 

So, 

b = arⁿ⁺¹

⇒ r = \(\bigg(\frac{b}{a}\bigg)^\frac{1}{n+1}\)

Now the product of GP becomes Product = G₁G₂G₃......G_n 

= (ar)(ar²)(ar³)..(arⁿ) 

= aⁿ.r ^(1+2+3…+n)

= aⁿ. r

= aⁿ. r^{\(\frac{n(n+1)}{2}\)}

Putting the value of r from equation 2 , we get

= aⁿ. \(\bigg(\big(\frac{b}{a}\big)^\frac{1}{n+1}\bigg)^\frac{n(n+1)}{2}\)

= (a.b)^{\(\frac{n}{2}\)}

(i) GM = √ab 

Let a = 2 and b =8 

GM = √2×8 

= √16 

= 4. 

(ii) GM = √xy 

Let x = a³b and y = ab 3 

GM = √a³b × ab³ 

= √a⁴ b⁴ = a²b². 

(iii) GM = √ab

Let a = –2 and b = –8 

GM = √–2×–8 

= √–16 

[we know that √–1 = i(i to a)] 

= 4i.

Let, the first term of G.P. is a and common ratio is r.

We know that common ratio of infinite G.P. is belongs to

[0, 1)

G.P. ⇒ a, ar, ar², ……

Sum of infinite terms of G.P. = a/(1-r) = 4

⇒ a = 4(1 – r)

Cubic terms of a G.P. ⇒ a³, a³r³, a³r⁶, ……

Sum of cubes of terms = \(\frac{a^3}{1-r^3}\) = 192

⇒ a³ = 192(1 – r³)

⇒ 4³(1 – r)³ = 92(1 – r³)

⇒ (1 – r)³ = 3(1 – r)(1 + r + r²)

Case I : 1 – r = 0

⇒ r = 1 (not possible)

Case II : (1 – r)² = 3(1 + r + r²)

⇒ 2r² + 5r + 2 = 0

⇒ (2r + 1)(r + 2) = 0

⇒ r = -2 (not possible) and r = -1/2

So, common ratio of original G.P. is -1/2

Correct answer is D -1/2

Let, the first term of G.P. is a and common ratio is r.

We know that common ratio of infinite G.P. is belongs to [0, 1) 

G.P. ⇒ a, ar, ar², …… Sum of infinite terms of 

G.P. = \(\frac{a}{1-r} = 4\) 

⇒ a = 4(1 – r) 

Cubic terms of a G.P. 

⇒ a³, a³r³, a³r⁶, …… 

Sum of cubes of terms = \(\frac{a^3}{1-r^3}\) = 192 

⇒ a³ = 192(1 – r³) 

⇒ 4³(1 – r)³ = 92(1 – r³) 

⇒ (1 – r)³ = 3(1 – r)(1 + r + r²) 

Case I : 1 – r = 0 

⇒ r = 1 (not possible) 

Case II : (1 – r)² = 3(1 + r + r²) 

⇒ 2r² + 5r + 2 = 0 

⇒ (2r + 1)(r + 2) = 0 

⇒ r = -2 (not possible) and r = -1/2 

So, common ratio of original G.P. is -1/2

We know when three terms say a,b,c are in GP 

We can write 

b² = a.c 

∴ According to the given data 

We can write (a^x/2)² = log_xa . log_bx

a^x = log_xa . log_bx 

⇒ \(a^x = \frac{logba}{logbx} \times log_bx\)

⇒ a^x = log_ba 

Multiplying by log_a to both sides we get 

⇒ log_a(a^x) = log_a (log_ba) 

⇒ x log_aa = log_a (log_ba) 

⇒ x = log_a (log_ba)

(i) GM = √ab

Let a = 2 and b = 8

GM = √2×8

= √16

= 4.

(ii) GM = √xy

Let x = a³b and y = ab³

GM = √a³b × ab³

= √a⁴b⁴

= a²b².

(iii) GM = √ab

Let a = –2 and b = –8

GM = √–2×–8

= √–16

[we know that √–1 = i(iota)]

= 4i.

Let, x = 0.4233333333….. ….equation 1 

As 3 is the repeating term, so in all such problems multiply both sides of the equation with a number such that complete repetitive part of number comes after the decimal.

∴ multiplying equation 1 with 100 in both sides, we have – 

100x = 42.3333333333… …equation 2 

Subtracting equation 1 from equation 2,we get

100x – x = 42.3333333… - 0.423333333… 

⇒ 99x = 41.91 {as letter terms gives zero only 42.33-0.42 gives result} 

∴ x = 41.91/99 

⇒ x = 4191/9900

Note: We can also solve these problems using geometric progression, but the above method is much simpler.

x = 3.522222222 ….. 

x = 3.5+0.02 + 0.002 + 0.0002 + …∞ 

⇒ x = 3.5+2(0.01 + 0.001 + 0.0001 + …∞ ) 

⇒ x = 3.5 + 2 \(\bigg(\frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} + ...... \infty\bigg)\)

⇒ x = 3.5 + 2S 

Where S = \(\frac{1}{100} + \frac{1}{1000} + \frac{1}{10000} + ...... \infty\)

We observe that the above progression possess a common ratio. So it is a geometric progression. 

Common ratio = 1/10 and first term (a) = 1/100 

Sum of infinite GP = \(\frac{a}{1-k}\), where a is the first term and k is the common ratio. 

Note: We can only use the above formula if |k|<1 

∴ we can use the formula for the sum of infinite GP. 

⇒ S = \(\cfrac{\frac{1}{100}}{1-(\frac{1}{10})}\)  = \(\frac{1}{90}\) = \(\frac{1}{90}\)

∴ x = 3.5 + 2(1/90) 

⇒ x = (35/10) + 1/45 = (315+2)/90 = 317/90

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

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 – d 2 

⇒ 81 = 144 – 18d – d 2 

⇒ d 2 + 18d – 63 = 0 

⇒ d 2 + 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 

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

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

Given: a, b, c are in GP 

Formula used: 

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

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

a₁ + a₂ + a₃ = 15

Where, the three number are: a, a + d, and a + 2d

So,

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

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

d = 5 – a … (i)

Now, according to the question:

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

they are in GP, that is:

(a + d + 3)/(a + 1) = (a + 2d + 9)/(a + d + 3)
(a + d + 3)² = (a + 2d + 9) (a + 1)

a² + d² + 9 + 2ad + 6d + 6a = a² + a + 2da + 2d + 9a + 9

(5 – a)² – 4a + 4(5 – a) = 0

25 + a² – 10a – 4a + 20 – 4a = 0

a² – 18a + 45 = 0

a² – 15a – 3a + 45 = 0

a(a – 15) – 3(a – 15) = 0

a = 3 or a = 15

d = 5 – a

d = 5 – 3 or d = 5 – 15

d = 2 or – 10

Then,

For a = 3 and d = 2, the A.P is 3, 5, 7

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

∴ The numbers are 3, 5, 7 or 15, 5, – 5

Let a = k + 9; b = k − 6; and c = 4;

We know that a, b and c are in GP, then

b² = ac {using property of geometric mean}

(k − 6)² = 4(k + 9)

k² – 12k + 36 = 4k + 36

k² – 16k = 0

k = 0 or k = 16

Given:

a, b and c are in GP

b² = ac {property of geometric mean}

Apply log on both sides with base m

log_m b² = log_m ac

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

2log_m b = log_m a + log_m c

2/log_b m = 1/log_a m + 1/log_c m

∴ 1/log_a m , 1/log_b m, 1/log_c m are in A.P.

If a,b,c are in AP it follows that 

a + c = 2b……..(1) 

and a,x,b and b,y,c are in individual GPs which follows 

x² = ab …….(2) 

y² = bc ……..(3) 

Adding eqn 2 and 3 we get, x² + y² = ab + bc

= b(a + c) = b.2b ( from eqn 1) 

= 2b² 

So we get x² + y² = 2b² which shows that they are in AP.

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) 

Hence a, (a – b), (d – c) are in G.P.

Let the A.P. be A, A + D, A + 2 D, ... and G.P be x, xR, xR², ... then 

a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D 

⇒ a – b = (p – q)D 

Also, b – c = (q – r)D 

And, c – a = (r – p)D 

Also a = p^th term of GP 

∴ a = xR^{p – 1} 

Similarly, b = xR^{q – 1} & c = xR^{r – 1} 

Hence, 

(a^{b – c} ).(b^{c – a} ).(c^{a – b}) = [(xR^{p – 1}) ^{(q – r)D}].[(xR ^{q – 1}) ^{(r – p)D}].[(xR ^{r – 1) (p – q)D}] 

= x ^{(q – r + r – p + p – q)D}. R ^{[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]D} 

⇒ (a^{b – c} ).(b^{c – a} ).(c ^{a – b}) = x⁰. R⁰ 

⇒ (a ^{b – c} ).(b ^{c – a} ).(c ^{a – b}) = 1 …proved

Let a be the first term of GP with r being the common ratio. 

∴ b = ar …(1) 

c = ar² …(2) 

Given, (a + b + c) = xb 

⇒ (a + ar + ar²) = x(ar) 

⇒ a(1 + r + r²) = ar 

⇒ (1 + r + r²) = xr 

⇒ r² + (1 – x)r + 1 = 0 

As r is a real number ⇒ Both solutions are real. 

So discriminant of the given quadratic equation 

D ≥ 0 As, D ≥ 0 

⇒ (1 – x)² – 4(1)(1) ≥ 0 

⇒ x² – 2x – 3 ≥ 0 

⇒ (x – 1)(x – 3) ≥ 0 

∴ x < – 1 or x > 3 …proved

Using the properties of exponents: 

The above term can be written as

Let S = \(9\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\) ....... (1)

We observe that above progression(in power of 9) possess a common ratio. So it is a geometric progression.

Let m = \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ..... \infty\)

Common ratio = r = \(\cfrac{\frac{1}{9}}{\frac{1}{3}} =\frac{1}{3} \)

Sum of infinite GP = \(\frac{a}{1-r}\),where a is the first term and r is the common ratio.

Note: We can only use the above formula if |r|<1

Clearly, a = \(\frac{1}{3}\) and r = \(\frac{1}{3}\)

m = \(\cfrac{\frac{1}{3}}{1-\frac{1}{3}}\) = \(\frac{1}{2}\)

From equation 1 we have, 

S = 9^m = 9^1/2 = 3 = RHS 

Hence Proved

Given:

Sum of G.P of 3 terms is 125

By using the formula,

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

125 = a (rⁿ – 1)/(r - 1)

125 = a (r³ – 1)/ (r - 1) … equation (1)

Now,

Sum of G.P of 6 terms is 152

By using the formula,

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

152 = a (rⁿ – 1)/(r - 1)

152 = a (r⁶ – 1)/ (r - 1) … equation (2)

Let us divide equation (i) by (ii) we get,

125/152 = [a (r³ – 1)/ (r - 1)] / [a (r⁶ – 1)/ (r - 1)]

125/152 = (r³ – 1)/(r⁶ – 1)

125/152 = (r³ – 1)/[(r³ – 1) (r³ + 1)]

125/152 = 1/(r³ + 1)

125(r³ + 1) = 152

125r³ + 125 = 152

125r³ = 152 – 125

125r³ = 27

r³ = 27/125

r³ = 3³/5³

r = 3/5

∴ The common ratio is 3/5

Let us consider the LHS

(9^1/3 . 9^1/9 . 9^1/27 ….∞)

This can be written as

9^{1/3 + 1/9 + 1/27 + …∞}

So let us consider m = 1/3 + 1/9 + 1/27 + … ∞

Where, a = 1/3, r = (1/9) / (1/3) = 1/3

By using the formula,

S_∞ = a/(1 – r)

= (1/3) / (1 – (1/3))

= (1/3) / ((3 - 1)/3)

= (1/3) / (2/3)

= 1/2

So, 9^m = 9^1/2 = 3 = RHS

Hence proved.

We observe that the above progression possess a common ratio. So it is a geometric progression. Common ratio = r = \(\frac{-9}{10}\)

Sum of infinite GP = \(\frac{a}{1-r}\), Where a is the first term and r is the common ratio.

Note : we can only use the above formula if |r|<1

Clearly, a = 10 and r = \(\frac{-9}{10}\)

⇒ sum = \(\frac{10}{1-(\frac{9}{10})}\)= \(\frac{100}{9}\)

We observe that the above progression possess a common ratio. So it is a geometric progression. 

Common ratio = r = \(\frac{4\sqrt{2}}{8} = \frac{1}{\sqrt{2}}\)

Sum of infinite GP = \(\frac{a}{1-r}\) ,where a is the first term and r is the common ratio. 

Note: We can only use the above formula if |r|<1 

Clearly, a = 8 and r = \(\frac{1}{\sqrt{2}}\)

⇒ sum = \(\cfrac{8}{1-\frac{1}{\sqrt{2}}}\) = \(\cfrac{8\sqrt{2}}{\sqrt{2}-1}\)

Let the two numbers be a and b. 

GM = √ab 

According to the given condition, 

a + b = 6√ab …(1) 

(a + b)² = 36ab 

Also, 

(a – b)² = (a + b)² – 4ab 

= 36ab – 4ab 

= 32ab 

⇒ a – b = √32ab 

= 4√2ab …..(2) 

Adding (1) and (2), we obtain 

2a = (6 + 4√2)√ab 

a = (3 + 2√2)√ab 

substituting the value of a in (1), we obtain, 

b =(3 – 2√2)√ab

\(\therefore\) \(\cfrac{3 + 2\sqrt{2}}{3-2\sqrt{2}}\)

Thus, the required ratio is (3+2√2) : 3–2√2.

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

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

⇒ a = √2,  r = \(\frac{1}{2}\), n = 8

∴ Substituting the above values in (1) we get

⇒ \(\cfrac{\sqrt{2}\bigg(\frac{1}{2})^8 -1\bigg)}{\frac{1}{2} - 1}\)

⇒ \(\cfrac{-\sqrt{2} \times 255 \times 2}{-1 \times 256}\)

⇒ \(\cfrac{255 \sqrt{2}}{128}\)

T_n = ar^n-1

a = √2, r = \(\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}\) = \(\frac{1}{2}\), T_n = \(\frac{1}{512\sqrt2}n\) = ?

\(\therefore\)\(\frac{1}{512\sqrt2}\) = \(\sqrt{2} \times (\frac{1}{2})^{n-1}\)

\(\Rightarrow\) \(\frac{1}{1024}\) = \((\frac{1}{2})^n\) \(\times 2\)

\(\Rightarrow\) \(\frac{1}{2048}\) = \((\frac{1}{2})^n\)

\(\Rightarrow\) 2ⁿ = 2048

\(\Rightarrow\) n = 11

(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 a and b = \(\frac{a + b }{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

Arithmetic mean of two numbers = \(\frac{a+b}{2}\)

\(\frac{a+b}{2}\) = 25

⇒ a + b = 50 

⇒ b = 50 – a … (i)

Geometric mean of two numbers = \(\sqrt{ab}\)

\(\Rightarrow\) \(\sqrt{ab}\) = 20

\(\Rightarrow\) ab = 400

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 a and b = \(\frac{a + b}{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

Arithmetic mean of two numbers = \(\frac{a+ b }{2}\)

\(\frac{a+ b }{2}\) = 10

⇒ a + b = 20 

⇒ a = 20 – b … (i)

Geometric mean of two numbers = \(\sqrt{ab}\)

⇒ \(\sqrt{ab}\) = 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 

⇒ 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

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 a and b = \(\frac{a+b}{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

As a, b, c are in A.P. 

⇒ 2b = a + c … (i) 

As x is the GM between a and b

⇒ x = \((\sqrt{ab})\)

⇒ x² = ab … (ii) 

As y is the GM between b and c

y = \((\sqrt{bc})\)

⇒ y² = bc … (iii)

Arithmetic mean of x² and y² is \((\frac{x^2 + y^2}{2})\)

Substituting the value from (ii) and (iii)

\((\frac{x^2 + y^2}{2})\) = \((\frac{ab + bc}{2})\)

Substituting the value from eqn. (i)

 =\(\frac{b(a+c)}{2}\)

= b²

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 a and b  = \(\frac{a + b}{2}\)

(ii) Geometric mean between a and b = \(\sqrt{ab}\)

Let the roots be p and q 

Arithmetic mean of roots p and q = \(\frac{p+q}{2} = 10\)

\(\frac{p+q}{2} = 10\)

⇒ p + q = 20 = sum of roots … (i) 

Geometric mean of roots p and q = \(\sqrt{Pq}\) = 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 

x² –20x + 64

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

According to the given condition,

⇒ AM = a+b/2 = 8

⇒ a + b = 16 …..(1)

⇒ GM = √ab = 5

= ab = 25 …(2)

The quadratic equation is given by,

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

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x² – 16x + 25 = 0.

Let us suppose a and b are two numbers.

Let us say G is a number that is the Geometric mean of a and b 

Therefore a, G and b must be in Geometric Progression or GP. 

This means, common ratio = G/a = b/G 

Or, G² = ab 

Or, G = ?(ab)... (1) 

Now, let us say G₁, G₂, G₃, .......G_n are n geometric means between a and b. 

Which means that 

a, G₁, G₂, G₃ ...... G_n, b form a G.P. 

Note that the above GP has n+2 terms and the first term is a and the last term is b, which is also the (n+2)^th term 

Hence, b = ar^n+2–1 

where a is the first term.

So, 

b = arⁿ⁺¹ 

r = (b/a)^1/n+1 ....(2) 

Now the product of GP becomes 

Product = G₁G₂G₃......G_n 

= (ar)(ar²)(ar³).....(arⁿ) 

= aⁿ r(1+2+3+4+.............+n) 

= aⁿ r^n(1+n)/2 

Putting the value of r from equation 2 , we get 

= aⁿ (b/a)ⁿ(1+n )/2(n+1) 

= (ab)^n/2 

= (?ab)ⁿ 

Now, putting the value from equation 1, we get, 

Product = Gⁿ 

Or, G₁G₂G₃......G_n = Gⁿ

Let the GP be a,ar,ar²

Sum of infinite terms is given by

⇒ s_∞ = a/1-r

⇒ 8 = a/1-r

⇒ Also according to the question ar = 2

∴ r = 2/a

∴ Substituting in above equation

⇒ 8 = \(\cfrac{a}{1-\frac{2}{a}}\)

⇒ 8(a - 2) = a²

⇒ a² - 8a+16 = 0

⇒ (a - 4)² = 0

⇒ a = 4.

⇒ AM = a+b/2

⇒ GM = √ab

Given AM = 10, GM = 8.

⇒ a+b/2 = 10

⇒ a + b = 20

⇒ a = 20 – b

⇒ \(\sqrt{(20-b)b}\) = 8

⇒ 20b – b² = 64

⇒ b² – 20b + 64 = 0

⇒ b² – 16b – 4b + 64 = 0

⇒ b(b – 16) – 4(b – 16) = 0

⇒ b = 4 or b = 16

⇒ If b = 4 then a = 16

⇒ If b = 16 then a = 4.

Hence, the numbers are 4 and 16.

Let us suppose a and b are two numbers.

Let us say G is a number that is the Geometric mean of a and b

Therefore a, G and b must be in Geometric Progression or GP.

This means, common ratio = G/a = b/G

Or, G² = ab

Or, G = 1/2(ab)... (1)

Now, let us say G₁, G₂, G₃, .......G_n are n geometric means between a and b.

Which means that

a, G₁, G₂, G₃ ...... G_n, b form a G.P.

Note that the above GP has n+2 terms and the first term is a and the last term is b, which is also the (n+2)^th term

Hence, b = ar^n+2–1

where a is the first term.

So,

b = arⁿ⁺¹

r = (b/a)^1/n+1 ....(2)

Now the product of GP becomes

Product = G₁G₂G₃......G_n

= (ar)(ar²)(ar³).....(arⁿ)

= aⁿ r^{(1+2+3+4+.............+n)}

= aⁿ r^n(1+n)/2

Putting the value of r from equation 2 , we get

= aⁿ (b/a)ⁿ(1+n)^/2(n+1)

= (ab)^n/2

= (1/2ab)ⁿ

Now, putting the value from equation 1, we get,

Product = Gⁿ

Or, G₁G₂G₃......G_n = Gⁿ

Given : Fifth term of GP is 2

⇒ Let the first term be a and the common ratio be r.

∴ According to the question,

T₅ = 2

We know,

a_n = ar^n-1

a₅ = a.r^5-1

2 = ar⁴

GP = a,ar,ar²,…,ar⁸

Product required = a×ar×ar²×…×ar⁸

= a⁹.r³⁶

= (ar⁴)⁹

= (2)⁹

= 512

Let the two numbers be a and b.

GM = √ab

According to the given condition,

⇒ a+b/2 = 2√ab

⇒ a + b = 4√ab …(1)

(a + b)² = 16ab

Also,

(a – b)² = (a + b)² – 4ab

= 16ab – 4ab

= 12ab

⇒ a – b = 2√3ab…(2)

Adding (1) and (2), we obtain

2a = (4 + 2√3 )√ab

a = (2 + √3)√ab

substituting the value of a in (1), we obtain,

b =(2 – √3)√ab

\(\therefore\frac{a}{b}=\frac{2+\sqrt3}{2-\sqrt3}\)

∴ Thus, the required ratio is (2+√3) : (2–√3).