Sequences and Series – Exercise 9.3[1-10] – Class XI

1. Find the 20^th and n^th terms of the G.P. ⁵/₂, ⁵/₄, ⁵/₈, …

Solution:

The given G.P. is  ⁵/₂, ⁵/₄, ⁵/₈, …

Here a = first term = ⁵/₂

r = common ratio = (⁴/₅)/(⁵/₂) = ¹/₂

a₂₀ = ar^20-1 = ⁵/₂(¹/₂)^n-1 = ⁵/₍₂₎₍₂₎^n-1 = ⁵/₂ⁿ

a_n = ar^n-1 = ⁵/₂(¹/₂)^n-1 = ⁵/₍₂₎₍₂₎^n-1 = ⁵/₂ⁿ

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2. Find the 12^th term of G.P. whose 8th term is 192 and the common ratio is 2.

Solution:

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a₈ = ar^8–1 = ar⁷

⇒ ar⁷ = 192 a(2)⁷ = 192 a(2)⁷ = (2)⁶(3)

⇒a = [(2)⁶x3]/(2)⁷ =  ³/₂

a₁₂ = ar^12-1 = ³/₂(2)¹¹ = (3)(2) = 3072

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3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Solution:

Let a be the first term and r be the common ratio of the G.P. According to the given condition,

a₅ = ar^5–1 = ar⁴ = p —————–(1)

a₈ = a r^8–1 = a r⁷ = q —————-(2)

a₁₁ = a r^11–1 = a r¹⁰ = s ————– (3)

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

[ar⁷]/[ar⁴] = ^q/_p

r³ = ^q/_p —————(4)

Dividing equation (3) by (2), we obtain

[ar¹⁰]/[ar⁷] = ^s/_q

r³ = ^s/_q —————-(5)

Equating the values of r³ obtained in (4) and (5), we obtain

^q/_p = ^s/_q

q² = ps

Thus the given result is proved.

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4. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.

Solution:

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

∴ a = –3

It is known that, an = ar^n–1

∴ a₄ = ar³ = (–3) r³

a₂ = a r¹ = (–3) r

According to the given condition,

(–3) r³ = [(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3 a₇ = a r^7–1 = a

r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Thus, the seventh term of the G.P. is –2187.

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5. Which term of the following sequences:

(a)2, 2√2, 4, …is 128?

(b) √3, 3, 3√3,…is 729?

(c)¹/₃, ¹/₉, ¹/₂₇,…¹/₁₉₆₈₃ ?

Solution:

(a) The given sequence 2, 2√2, 4, …is 128?

Here a = 2 and r = (2√2)/2 = √2

Let nth term of the given sequence be 128.

a_n = ar^n-1

(2)( √2)^n-1 = 128

(2)(2)^(n-1)/2 = 2⁷

(2)^[(n-1)/2]+1 = 2⁷

^n-1/₂ + 1 = 7

^n-1/₂ = 6

n – 1 = 12

n = 13

Thus the 13^th term of the given sequence is 128.

 

(b) the given sequence is √3, 3, 3 √3

a = √3 and r = ³/_√3 = √3

Let the nth term of the given sequence be 729

a_n = ar^n-1

ar^n-1 = 729

(√3)( √3)^n-1 = 729

(3)^1/2( 3)^(n-1)/2 =3⁶

^1+n-1/₂ = 6

n = 12

Thus the 12^th term of the given sequence 729.

 

(c) The given sequence is ¹/₃, ¹/₉, ¹/₂₇,…

Here, a = ¹/₃ and r = ¹/₉ + ¹/₃ = ¹/₃

Let the nth term of the given sequence be ¹/₁₉₆₈₃

a_n = ar^n-1

(¹/₃).( ¹/₃)^n-1 = ¹/₁₉₆₈₃

(¹/₃)ⁿ = (¹/₃)⁹

n = 9

Thus, the 9th term of the given sequence is ¹/₁₉₆₈₃

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6. For what values of x, the numbers ²/₇, x, –⁷/₂ are in G.P.?

Solution:

The given numbers are –²/₇, x, –⁷/₂

Common ratio =

^x/_(-2/7)=^-7x/₂

Also, common ratio = (^-7/₂)/_x = –⁷/_2x

^-7/_2x = –⁷/_2x

x² = ^-2×7/_-2×7 = 1

x = √1

x = ±1

Thus for x = ±1 the given numbers will be in G.P.

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7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015

Solution:

The given G.P. is 0.15, 0.015, 0.0015

Here a = 0.15 and r = ^0.015/_0.15 = 0.1

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8. Find the sum to n terms in the geometric progression √7 , √21, 3√7, …

Solution:

The given G.P. is √7 , √21, 3√7, …

Here, a = √7 and r =^√21/₇ = √3

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9. Find the sum to n terms in the geometric progression 1, -a, a², -a³…(if a≠-1)

Solution:

The given G.P. is 1, -a, a², -a³…

Here first term = a₁ = 1

Common ratio = r = – a

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10. Find the sum to n terms in the geometric progression x³, x⁵, x⁷…(if x ≠±1)

Solution:

The given G.P. is x³, x⁵, x⁷…

Here a = x³ and r = x²

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Sequences and Series – Exercise 9.3[11-20] – Class XI

Sequences and Series – Exercise 9.3[21-32] – Class XI

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