Chapter 9: Sequences and Series Exercise 9.3

Consider a to be the first term and r to be the common ratio of the G.P.

Then,

a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

From the question, we have

a₃ = a₁ + 9

ar² = a + 9 … (i)

a₂ = a₄ + 18

ar = ar³ + 18 … (ii)

So, from (1) and (2), we get

a(r² – 1) = 9 … (iii)

ar (1– r²) = 18 … (iv)

Now, dividing (4) by (3), we get

-r = 2

r = -2

On substituting the value of r in (i), we get

4a = a + 9

3a = 9

∴ a = 3

Therefore, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)², and 3(–2)³

i.e., 3¸–6, 12, and –24.

Let’s take A to be the first term and R to be the common ratio of the G.P.

Then according to the question, we have

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

Then,

a^q–r b^r–p c^p–q

= A^q–r × R^{(p–1) (q–r)} × A^r–p × R^{(q–1) (r–p)} × A^p–q × R^{(r –1)(p–q)}

= Aq ^{– r + r – p + p – q} × R ^{(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)}

= A⁰ × R⁰

= 1

Hence proved.

Given, the first term of the G.P is a, and the last term is b.

Thus,

The G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.

Then,

b = ar^n–1  … (1)

P = Product of n terms

= (a) (ar) (ar²) … (ar^n–1)

= (a × a ×…a) (r × r² × …r^n–1)

= aⁿ r ^{1 + 2 +…(n–1)}  … (2)

Here, 1, 2, …(n – 1) is an A.P.

And, the product of n terms P is given by,

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

Since there are n terms from (n +1)^th to (2n)^th term,

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

a ^n +1 = ar ^{n + 1 – 1} = arⁿ

Thus, the required ratio =

Thus, the ratio of the sum of the first n terms of a G.P. to the sum of terms from (n + 1)^th to (2n)^th term is ^.

Given, a, b, c, d are in G.P.

So, we have

bc = ad  … (1)

b² = ac  … (2)

c² = bd  … (3)

Taking the R.H.S., we have

R.H.S.

= (ab + bc + cd)²

= (ab + ad + cd)²  [Using (1)]

= [ab + d (a + c)]²

= a²b² + 2abd (a + c) + d² (a + c)²

= a²b² +2a²bd + 2acbd + d²(a² + 2ac + c²)

= a²b² + 2a²c² + 2b²c² + d²a² + 2d²b² + d²c²  [Using (1) and (2)]

= a²b² + a²c² + a²c² + b²c² + b²c² + d²a² + d²b² + d²b² + d²c²

= a²b² + a²c² + a²d² + b² × b² + b²c² + b²d² + c²b² + c² × c² + c²d²

= a²(b² + c² + d²) + b² (b² + c² + d²) + c² (b²+ c² + d²)

= (a² + b² + c²) (b² + c² + d²)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

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

Let’s assume G₁ and G₂ to be two numbers between 3 and 81 such that the series 3, G₁, G₂, 81 forms a G.P.

And let a be the first term and r be the common ratio of the G.P.

Now, we have the 1^st term as 3 and the 4^th term as 81.

81 = (3) (r)³

r³ = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G₁ = ar = (3) (3) = 9

G₂ = ar² = (3) (3)² = 27

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P. are 9 and 27.

We know that,

The G. M. of a and b is given by √ab.

Then, from the question, we have

By squaring both sides, we get

Performing cross multiplication after expanding, we get,

Consider the two numbers to be a and b.

Then, G.M. = √ab.

From the question, we have

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b.

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here, we have a = 30 and r = 2

So, a₃ = ar² = (30) (2)² = 120

Thus, the number of bacteria at the end of 2^nd hour will be 120.

And, a₅ = ar⁴ = (30) (2)⁴ = 480

The number of bacteria at the end of 4^th hour will be 480.

a_n +1 = arⁿ = (30) 2ⁿ

Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.