Chapter 9: Sequences And Series Miscellaneous Exercise

Let’s consider the terms in the G.P. to be T₁, T₂, T₃, T₄, … T_2n.

The number of terms = 2n

Then, from the question we have

T₁ + T₂ + T₃ + …+ T_2n = 5 [T₁ + T₃ + … +T_2n–1]

T₁ + T₂ + T₃ + … + T_2n – 5 [T₁ + T₃ + … + T_2n–1] = 0

T₂ + T₄ + … + T_2n = 4 [T₁ + T₃ + … + T_2n–1] …… (1)

Now, let the terms in G.P. be a, ar, ar², ar³, …

Then (1) becomes,

[Using sum of terms in G.P.]

ar = 4a

r = 4

Thus, the common ratio of the G.P. is 4.

Let’s consider the terms in A.P. to be a, a + d, a + 2d, a + 3d, … a + (n – 2) d, a + (n – 1)d.

From the question, we have

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

Then according to the given condition,

4a + 6d = 56

4(11) + 6d = 56 [Since a = 11 (given)]

6d = 12

d = 2

Hence, 4a + (4n –10) d = 112

4(11) + (4n – 10)2 = 112

(4n – 10)2 = 68

4n – 10 = 34

4n = 44

n = 11

Therefore, the number of terms of the A.P. is 11.

Given,

On cross-multiplying, we have

Also, given

On cross-multiplying, we have

From (1) and (2), we get

b/a = c/b = d/c

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

Let the terms in G.P. be a, ar, ar², ar³, … ar^{n – 1}…

From the question, we have

Hence, P² Rⁿ = Sⁿ

Let’s assume t and d to be the first term and the common difference of the A.P., respectively.

Then the n^th term of the A.P. is given by, a_n = t + (n – 1) d

Thus,

a_p = t + (p – 1) d = a  … (1)

a_q = t + (q – 1) d = b  … (2)

a_r = t + (r – 1) d = c  … (3)

On subtracting equation (2) from (1), we get

(p – 1 – q + 1) d = a – b

(p – q) d = a – b

On subtracting equation (3) from (2), we get

(q – 1 – r + 1) d = b – c

(q – r) d = b – c

Equating both the values of d obtained in (4) and (5), we get

Therefore, the given result is proved.

Given, a, b, c, and d are in G.P.

So, we have

∴b² = ac … (i)

c² = bd … (ii)

ad = bc … (iii)

Required to prove (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P. i.e.,

(bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)

Taking L.H.S.

(bⁿ + cⁿ)² = b²ⁿ + 2bⁿcⁿ + c²ⁿ

= (b²)ⁿ+ 2bⁿcⁿ + (c²) ⁿ

= (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ  [Using (i) and (ii)]

= aⁿ cⁿ + bⁿcⁿ+ bⁿ cⁿ + bⁿ dⁿ

= aⁿ cⁿ + bⁿcⁿ+ aⁿ dⁿ + bⁿ dⁿ  [Using (iii)]

= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ)

= (aⁿ + bⁿ) (cⁿ + dⁿ)

= R.H.S.

Therefore, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P

– Hence proved.

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

So, we have a + b = 3 and ab = p  … (i)

Also, c and d are the roots of x² – 12x + q = 0

So, c + d = 12 and cd = q  … (ii)

And given a, b, c, d are in G.P.

Let’s take a = x, b = xr, c = xr², d = xr³

From (i) and (ii), we get

x + xr = 3

x (1 + r) = 3

And,

xr² + xr³ =12

xr² (1 + r) = 12

On dividing, we get

When r = 2, x = 3/(1 + 2) = 3/3 = 1

When r = -2, x = 3/(1 – 2) = 3/-1 = -3

Case I:

When r = 2 and x =1,

ab = x²r = 2

cd = x²r⁵ = 32

Case II:

When r = –2, x = –3,

ab = x²r = –18

cd = x²r⁵ = – 288

Therefore, in both the cases, we get (q + p): (q – p) = 17:15\

Let the two numbers be a and b.

A.M = (a + b)/ 2 and G.M. = √ab

From the question, we have

Given a, b, c are in A.P.

Hence, b – a = c – b  … (1)

And, given that b, c, d are in G.P.

So, c² = bd  … (2)

Also, 1/c, 1/d, 1/e are in A.P.

So,

Now, required to prove that a, c, e are in G.P. i.e., c² = ae

From (1), we have

2b = a + c

b = (a + c)/ 2

And from (2), we have

d = c²/ b

On substituting these values in (3), we get

Therefore, a, c, and e are in G.P.