Chapter 9: Sequences and Series Exercise 9.3

Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a³ = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

(1 + r + r²)/r = 39/10

10 + 10r + 10r² = 39r

10r² – 29r + 10 = 0

10r² – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

Therefore, the three terms of the G.P. are 5/2, 1 and 2/5.

Given G.P. is 3, 3², 3³, …

Let’s consider that n terms of this G.P. be required to obtain the sum 120.

We know that,

Here, a = 3 and r = 3

Equating the exponents, we get n = 4

Therefore, four terms of the given G.P. are required to obtain the sum 120.

Let’s assume the G.P. to be a, ar, ar², ar³, …

Then according to the question, we have

a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128

a (1 + r + r²) = 16 … (1) and,

ar³(1 + r + r²) = 128 … (2)

Dividing equation (2) by (1), we get

r³ = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

Given,

a = 729 and a₇ = 64

Let r be the common ratio of the G.P.

Then we know that, a_n = a r^n–1

a₇ = ar^7–1 = (729)r⁶

⇒ 64 = 729 r⁶

r⁶ = 64/729

r⁶ = (2/3)⁶

r = 2/3

And we know that

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

Given, S₂ = -4

Then, from the question, we have

And,

a₅ = 4 x a₃

ar⁴ = 4ar²

r² = 4

r = ± 2

Using the value of r in (1), we have

Therefore, the required G.P is

-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

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

According to the given condition,

a₄ = a r³ = x … (1)

a₁₀ = a r⁹ = y … (2)

a₁₆ = a r¹⁵ = z … (3)

On dividing (2) by (1), we get

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, it can be changed to G.P. by writing as

S_n = 8 + 88 + 888 + 8888 + …………….. to n terms

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/2²]

Now, it’s seen that

4, 2, 1, ½, 1/2² is a G.P.

With the first term, a = 4

Common ratio, r =1/2

We know,

Therefore, the required sum = 64(31/4) = (16)(31) = 496

To be proved: The sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

Now, we have

Therefore, the above sequence forms a G.P., and the common ratio is rR.