11th Class Mathematics Exercise 6.5

1.Find the 6th term of the G.P.: -6, -3, -3/2, ...

Given

G.P. -6, -3, -32, ldots

First term and ratio a₁ = -6, quad r = 12

Formula

General term formula a_n = a₁ r^n-1

Working

Finding 6th term a₆ = (-6)left(12right)⁵ = (-6) · 132

= -632 = -316

-316

2.Find the 8th term of the sequence 3, 3², 3³, ...

Given

G.P. 3, 3², 3³, ldots

First term and ratio a₁ = 3, quad r = 3

Formula

General term formula a_n = a₁ r^n-1

Working

Finding 8th term a₈ = 3 · 3⁷ = 3⁸

3⁸

3.The nᵗʰ terms of the sequences 1, 2, 4, 8, ... and 256, 128, 64, ... are equal. Find the value of n.

Given

First sequence 1, 2, 4, 8, ldots; quad a₁ = 1, r = 2; quad a_n = 2^n-1

Second sequence 256, 128, 64, ldots; quad a₁ = 256, r = 12; quad a_n = 256 · left(12right)^n-1 = 2^9-n

Equating terms 2^n-1 = 2^9-n

n - 1 = 9 - n ⇒ 2n = 10

Result

Value of n n = 5

4.(i) Find the first five terms of the sequence with a₁ = 243, r = 1/3. (ii) Find the first five terms of the sequence with a₁ = 579, r = -1/2.

Given

(i) Parameters a₁ = 243, quad r = 13

Formula

General term a_n = a₁ r^n-1

a₁ = 243

a₂ = 243 · 13 = 81

a₃ = 243 · left(13right)² = 27

a₄ = 243 · left(13right)³ = 9

a₅ = 243 · left(13right)⁴ = 3

Result

First five terms 243, 81, 27, 9, 3

Given

(ii) Parameters a₁ = 579, quad r = -12

Formula

General term a_n = a₁ r^n-1

a₁ = 579

a₂ = 579 · left(-12right) = -5792

a₃ = 579 · left(-12right)² = 5794

a₄ = 579 · left(-12right)³ = -5798

a₅ = 579 · left(-12right)⁴ = 57916

Result

First five terms 579, -5792, 5794, -5798, 57916

5.Find the 12th term of the sequence 1+i, 2i, -2+2i, ...

Given

First term a₁ = 1 + i

Common ratio r = 2i1+i = 2i(1-i)(1+i)(1-i) = 2i - 2i²1 - i² = 2i + 22 = 1 + i

Formula

General term a_n = a₁ r^n-1

Working

12th term a₁₂ = (1+i)(1+i)¹¹ = (1+i)¹²

Computing (1+i)⁴ (1+i)² = 2i; quad (1+i)⁴ = (2i)² = -4

Computing (1+i)¹² (1+i)⁸ = [(-4)]² = 16; quad (1+i)¹² = (1+i)⁸ · (1+i)⁴ = 16 · (-4) = -64

a₁₂ = -64

6.If the 4th and 9th term of a G.P. are 54 and 13122 respectively, find the G.P. Also find its general term.

Given

Given terms a₄ = 54, quad a₉ = 13122

General formula a_n = a₁ r^n-1

From given terms a₄ = a₁ r³ = 54 quad (i), quad a₉ = a₁ r⁸ = 13122 quad (ii)

Divide (ii) by (i) a₉a₄ = r⁵ = 1312254 = 243 = 3⁵ ⇒ r = 3

Find first term a₁ r³ = 54 ⇒ a₁ · 27 = 54 ⇒ a₁ = 2

Result

The G.P. 2, 6, 18, 54, ldots

General term a_n = 2 · 3^n-1

8.If the (p+q)ᵗʰ term of a G.P. is m and the (p-q)ᵗʰ term is n, then find the pᵗʰ term.

Given

Given terms a_p+q = m, quad a_p-q = n

Formula

General formula a_n = a₁ r^n-1

Write given conditions a_p+q = a₁ r^p+q-1 = m quad (i), quad a_p-q = a₁ r^p-q-1 = n quad (ii)

Multiply (i) and (ii) a₁² r^{(p+q-1)+(p-q-1)} = mn ⇒ a₁² r^2p-2 = mn

Simplify (a₁ r^p-1)² = mn ⇒ a_p² = mn

Result

pᵗʰ term a_p = sqrt{mn}

9.Find three consecutive numbers in G.P. whose sum is 26 and their product is 216.

Given

Three G.P. terms ar, a, ar

Product condition ar · a · ar = a³ = 216 ⇒ a = 6

Sum condition ar + a + ar = 26 ⇒ 6r + 6 + 6r = 26

Simplify 6r + 6r = 20 ⇒ 6 + 6r² = 20r ⇒ 6r² - 20r + 6 = 0 ⇒ 3r² - 10r + 3 = 0

Factor (3r - 1)(r - 3) = 0 ⇒ r = 13 or r = 3

Result

When r = 3 63 = 2, quad 6, quad 18

When r = 1/3 18, quad 6, quad 2

10.The 3rd term of a G.P. is the square of the 1st term. If the 2nd term is 9 then find the 6th term.

Given

Condition 1 a₃ = a₁²

Condition 2 a₂ = 9

From a₃ = a₁r² a₁ r² = a₁² ⇒ r² = a₁

From a₂ = a₁r a₁ r = 9

Substitute r² = a₁ a₁ sqrt{a₁} = 9 ⇒ a₁^3/2 = 9 ⇒ a₁ = 9^2/3 = sqrt[3]{81}

Common ratio r = 9a₁ = frac{9}{sqrt[3]{81} = sqrt[3]{9}

Compute 6th term a₆ = a₁ r⁵ = sqrt[3]{81} · (sqrt[3]{9})⁵ = sqrt[3]{81} · 9sqrt[3]{81} = 9(sqrt[3]{81})² = 81sqrt[3]{9}

a₆ = 81sqrt[3]{9}

12.If the numbers 1, 4 and 3 are subtracted from three consecutive terms of an A.P., the resulting numbers are in G.P. Find the original numbers if their sum is 21.

Given

A.P. terms a-d, a, a+d

Sum condition (a-d) + a + (a+d) = 21 ⇒ 3a = 21 ⇒ a = 7

After subtraction a-d-1, a-4, a+d-3 are in G.P.

Substitute a = 7 6-d, 3, 4+d are in G.P.

G.P. condition 3² = (6-d)(4+d) ⇒ 9 = 24 + 6d - 4d - d² ⇒ d² - 2d - 15 = 0

Factor (d-5)(d+3) = 0 ⇒ d = 5 or d = -3

Result

When d = 5 a-d = 2, a = 7, a+d = 12

When d = -3 a-d = 10, a = 7, a+d = 4

13.If three consecutive numbers in A.P. are increased by 1, 4, 15 respectively, the resulting numbers are in G.P. Find the original numbers if their sum is 6.

Given

A.P. terms a-d, a, a+d

Sum condition (a-d) + a + (a+d) = 6 ⇒ 3a = 6 ⇒ a = 2

After increment a-d+1, a+4, a+d+15 are in G.P.

Substitute a = 2 3-d, 6, 17+d are in G.P.

G.P. condition 6² = (3-d)(17+d) ⇒ 36 = 51 + 3d - 17d - d² ⇒ d² + 14d - 15 = 0

Factor (d+15)(d-1) = 0 ⇒ d = -15 or d = 1

Result

When d = 1 a-d = 1, a = 2, a+d = 3

When d = -15 a-d = 17, a = 2, a+d = -13

15.If a, 2a+2, 3a+3 are in G.P., then find the fifth term.

G.P. condition (2a+2)² = a(3a+3)

Expand 4a² + 8a + 4 = 3a² + 3a

Simplify a² + 5a + 4 = 0

Factor (a+1)(a+4) = 0 ⇒ a = -1 or a = -4

Check a = -1 G.P. is -1, 0, 0 (invalid, r = 0)

Check a = -4 G.P. is -4, -6, -9 with r = 32

Find first term a₁ = -4

Fifth term a₅ = a₁ r⁴ = (-4)left(32right)⁴ = (-4) · 8116 = -814

a₅ = -814

Unit 6: Sequences and Series — Exercise 6 5