11th Class Mathematics Exercise 6.8

1.Find the 8th term of the arithmetico-geometric sequence, where the arithmetic part is 1, 4, 7,... and the geometric part is 5, 10, 20,...

Given

Arithmetic part (A.P.) 1, 4, 7, ldots

Formula

nth term of A.P. a_n = a₁ + (n-1)d = 1 + (n-1)(3)

a_n = 3n-2

Given

Geometric part (G.P.) 5, 10, 20, ldots

Formula

nth term of G.P. b_n = b₁ r^n-1 = 5 · 2^n-1

nth term of A.G. sequence T_n = a_n b_n = (3n-2)(5 · 2^n-1) = 5(3n-2) · 2^n-1

Working

Put n = 8 T₈ = 5(3 × 8 - 2) · 2^8-1 = 5 × 22 × 2⁷

Result

Required term T₈ = 14080

2.Find the nth term of the arithmetico-geometric sequence, Where the arithmetic part is 3, 7, 11,... and the geometric part is 2, 6, 18,...

Given

Arithmetic part (A.P.) 3, 7, 11, ldots

Formula

nth term of A.P. a_n = a₁ + (n-1)d = 3 + (n-1)(4)

a_n = 4n-1

Given

Geometric part (G.P.) 2, 6, 18, ldots

Formula

nth term of G.P. b_n = b₁ r^n-1 = 2(3)^n-1 = 2 · 3^n-1

nth term of A.G. sequence T_n = a_n b_n = (4n-1)(2 · 3^n-1)

Result

Required nth term T_n = 2(4n-1) · 3^n-1

3.Consider the arithmetico-geometric sequence defined by arithmetic part: a_{n+1} = 2n+5 and geometric part: b_{n-2} = (1/9)(-3)^n. Find the nth term and the sum of first three terms of the arithmetico-geometric sequence.

Given

Arithmetic part a_n+1 = 2n+5

Replace n by n-1 a_n = 2(n-1)+5 = 2n+3

Geometric part b_n-2 = 19(-3)ⁿ

Replace n by n+2 b_n = 19(-3)ⁿ⁺² = 19(-3)² · (-3)ⁿ = 99 × (-3)ⁿ = (-3)ⁿ

Formula

nth term of A.G. sequence T_n = a_n b_n = (2n+3)(-3)ⁿ

Working

Put n = 1, 2, 3 S₃ = (2 × 1+3)(-3)¹ + (2 × 2+3)(-3)² + (2 × 3+3)(-3)³

= -15 + 7 × 9 + 9(-27) = -15+63-243

Result

Sum of first three terms S₃ = -195

4.Sum to n terms the series: (i) 1.2 + 3.4 + 5.8 + 7.16 + ... (ii) 2.3 + 4.3^2 + 6.3^3 + 8.3^4 + ... (iii) 2 + 5/4 + 8/4^2 + 11/4^3 + ... (iv) 1 + 3/5 + 5/5^2 + 7/5^3 + ... (v) 1 + 4/3 + 7/9 + 10/27 + ...

Given

(i) S_n = 1.2 + 3.4 + 5.8 + 7.16 + ldots to n terms

Formula

nth term T_n = [a+(n-1)d][br^n-1] = [1+2(n-1)][2 · 2^n-1] = (2n-1)2ⁿ

Equation (i) S_n = 1.2 + 3.2² + 5.2³ + ldots + (2n-1)2ⁿ

Multiply both sides by r=2, equation (ii) 2S_n = 1.2² + 3.2³ + 5.2⁴ + ldots + (2n-1)2ⁿ⁺¹

Subtract (ii) from (i) -S_n = 2 + 2left[2²+2³+ldots+2ⁿright] - (2n-1)2ⁿ⁺¹ = -6-(4n-6)2ⁿ

Result

Required sum S_n = 6+(4n-6)2ⁿ

Given

(ii) S_n = 2.3 + 4.3² + 6.3³ + 8.3⁴ + ldots to n terms

Formula

nth term T_n = [a+(n-1)d]br^n-1 = [2+(n-1)(2)](3 · 3^n-1) = (2n)(3ⁿ)

Equation (i) S_n = 2.3+4.3²+6.3³+8.3⁴+ldots+(2n)3ⁿ

Multiply both sides by r=3, equation (ii) 3S_n = 2.3²+4.3³+6.3⁴+8.3⁵+ldots+(2n)3ⁿ⁺¹

Subtract (ii) from (i) -2S_n = 6+2left[3²+3³+ldots+3ⁿright]-(2n)3ⁿ⁺¹ = -3+(1-2n)3ⁿ⁺¹

Result

Required sum S_n = 32left[1-(n+1)3ⁿ+n · 3ⁿ⁺¹right]

Given

(iii) S_n = 2+54+84²+114³+ldots to n terms

Formula

nth term T_n = [a+(n-1)d]br^n-1 = [2+(n-1)(3)]left(14right)^n-1 = frac{3n-1}{4^n-1

Multiply both sides by r=1/4 and subtract left(1-14right)S_n = 2+3left[14+14²+ldots+frac{1}{4^n-1right]-3n-14ⁿ

34S_n = 2+left(1-frac{1}{4^n-1right)-3n-14ⁿ = 3-frac{1}{4^n-1-3n-14ⁿ

Result

Required sum S_n = 4-43left(14right)^n-1-43(3n-1)left(14right)ⁿ

Given

(iv) S_n = 1+35+55²+75³+ldots to n terms

Formula

nth term T_n = [a+(n-1)d]br^n-1 = [1+(n-1)(2)]left(15right)^n-1 = frac{2n-1}{5^n-1

Multiply both sides by r=1/5 and subtract left(1-15right)S_n = 1+2left[15+15²+ldots+frac{1}{5^n-1right]-2n-15ⁿ

45S_n = 1+12left(1-frac{1}{5^n-1right)-2n-15ⁿ = 32-frac{1}{2 · 5^n-1-2n-15ⁿ

Result

Required sum S_n = 158-54(2n-1)left(15right)ⁿ-58left(15right)^n-1

Given

(v) S_n = 1+43+73²+103³+ldots to n terms

Formula

nth term T_n = [a+(n-1)d]br^n-1 = [1+(n-1)(3)]left(13right)^n-1 = frac{3n-2}{3^n-1

Multiply both sides by r=1/3 and subtract left(1-13right)S_n = 1+3left[13+13²+ldots+frac{1}{3^n-1right]-3n-23ⁿ

23S_n = 1+32left(1-frac{1}{3^n-1right)-3n-23ⁿ = 52-3²2 · 3ⁿ-3n-23ⁿ

Result

Required sum S_n = 154-94left(13right)^n-1-32(3n-2)left(13right)ⁿ

5.Sum the following infinite series: (i) 1 + 3/2 + 5/4 + 7/8 + ... (ii) 2 + 5/3 + 8/9 + 11/27 + ...

Given

(i) A.P. part a=1, d=3-1=2

G.P. part b=1, r=12

Formula

Sum to infinity of A.G. series S_infty = ab1-r+dbr(1-r)²

Working

S_infty = frac{1 × 1}{1-12+frac{2 × 1 × 12{left(1-12right)²} = 2+4

Result

Required sum S_infty = 6

Given

(ii) A.P. part a=2, d=5-2=3

G.P. part b=1, r=13

Formula

Sum to infinity of A.G. series S_infty = ab1-r+dbr(1-r)²

Working

S_infty = frac{2 × 1}{1-13+frac{3 × 1 × 13{left(1-13right)²} = 3+94

Result

Required sum S_infty = 214

6.Show that 2^(1/2).4^(1/4).8^(1/8).16^(1/16)...∞ = 4

Given

x = 2^{12.4^{14.8^{18.16^{116ldots

Take log of both sides log x = 12log2+14log4+18log8+116log16+ldots = left(12+24+38+416+ldotsright)log2

A.P. part of 1/2+2/4+3/8+4/16+... a=1, d=1; G.P. part: b=12, r=12

Formula

Sum to infinity S_infty = ab1-r+dbr(1-r)² = frac{1 × 12{12+frac{1 × 12 × 12{left(12right)²} = 1+1 = 2

Working

log x = 2log2 = log2²

Result

x = 4

7.Show that √4.⁴√16.⁸√64.¹⁶√256...∞ = 16

Given

x = sqrt{4}.sqrt[4]{16}.sqrt[8]{64}.sqrt[16]{256}ldotsinfty

Take log of both sides log x = 12log2²+14log2⁴+18log2⁶+116log2⁸+ldots = left(22+44+68+816+ldotsright)log2

A.P. part of 2/2+4/4+6/8+8/16+... a=2, d=2; G.P. part: b=12, r=12

Formula

Sum to infinity S_infty = ab1-r+dbr(1-r)² = frac{2 × 12{12+frac{2 × 12 × 12{left(12right)²} = 2+2 = 4

Working

log x = 4log2 = log2⁴

Result

x = 16

8.Sum to n terms the series 2 + 4x + 6x^2 + 8x^3 + ... where x ≠ 1

Given

S_n = 2+4x+6x²+8x³+ldots to n terms

Formula

nth term T_n = [a+(n-1)d] · br^n-1 = [2+(n-1)(2)] · x^n-1 = (2n)x^n-1

Equation (i) S_n = 2+4x+6x²+8x³+ldots+(2n)x^n-1

Multiply both sides by x, equation (ii) xS_n = 2x+4x²+6x³+8x⁴+ldots+(2n)xⁿ

Subtract (ii) from (i) (1-x)S_n = 2+2left[x+x²+ldots+x^n-1right]-(2n)xⁿ = frac{2-2(1+n)xⁿ+(2n)xⁿ⁺¹{1-x}

Result

Required sum S_n = frac{2-2(n+1)xⁿ+2nxⁿ⁺¹{(1-x)²}

9.Find the sum to n terms of the series: (2n+1)/(2n-1) + 3[(2n+1)/(2n-1)]^2 + 5[(2n+1)/(2n-1)]^3 + ...

Given

x = 2n+12n-1

Given series becomes S_n = x+3x²+5x³+ldots to n terms

Formula

nth term T_n = [1+(n-1)2] · xⁿ = (2n-1)xⁿ

Equation (i) minus x times equation, subtract (1-x)S_n = x+2left[x²+x³+ldots+xⁿright]-(2n-1)xⁿ⁺¹

Simplify S_n = 1(1-x)²left[x(1+x)-(2n+1)xⁿ⁺¹+(2n-1)xⁿ⁺²right]

Working

Using x=(2n+1)/(2n-1) 1-x = -22n-1,quad 1+x = 4n2n-1

S_n = left(2n-12right)²left[(2n+1)(4n)(2n-1)²-frac{(2n+1)ⁿ⁺²{(2n-1)ⁿ⁺¹+frac{(2n+1)ⁿ⁺²{(2n-1)ⁿ⁺¹right]

Result

Required sum S_n = n(2n+1)

10.Prove that: 1 + 2(1+1/n) + 3(1+1/n)^2 + ... to n terms = n^2

Given

x = 1+1n

Given series becomes S_n = 1+2x+3x²+ldots to n terms

Formula

nth term T_n = [1+(n-1)1] · x^n-1 = n x^n-1

Subtract xS_n from S_n (1-x)S_n = 1+x+x²+ldots+x^n-1-nxⁿ = 11-xleft[1-(n+1)xⁿ+nxⁿ⁺¹right]

Result

S_n = 1(1-x)²left[1-(n+1)xⁿ+nxⁿ⁺¹right]

Since n+1=nx (from x=1+1/n) S_n = 1(1-x)²left[1-nx · xⁿ+nxⁿ⁺¹right] = 1(1-x)²

Working

Using x=1+1/n, so 1-x=-1/n S_n = frac{1}{left(-1nright)²}

Result

S_n = n² quad Hence proved

11.Sum the series to n terms 2 + 5x + 8x^2 + 11x^3 + ... and deduce the sum to infinity if |x| < 1.

Given

S_n = 2+5x+8x²+11x³+ldots to n terms

Formula

nth term T_n = [2+(n-1)3] · x^n-1 = (3n-1)x^n-1

Equation (i) minus x times equation (ii) (1-x)S_n = 2+3left[x+x²+ldots+x^n-1right]-(3n-1)xⁿ = 2+3left[x(1-x^n-1)1-xright]-(3n-1)xⁿ

Result

Sum to n terms S_n = 21-x+3x(1-x^n-1)(1-x)²-(3n-1)xⁿ1-x

As |x|<1, x^n → 0 when n→∞ S_infty = 21-x+3x(1-x)² = 2(1-x)+3x(1-x)² = 2-2x+3x(1-x)²

Sum to infinity S_infty = x+2(1-x)²

Unit 6: Sequences and Series — Exercise 6 8