Question

if a1,a2,.....an are in continued propotion then show that a1/a2= (a1/a2)^n-1

Answer 1

GIven : a₁,a₂,.....an are in continued propotion

To find : show that a₁/aₙ = (a1/a₂)ⁿ⁻¹  ( correction in Question)

Solution:

a₁  , a₂  , a₃  , a₄ ......................aₙ  are in continued proportion

if a₂ = ka₁    => a₁/a₂  = 1/k

=> aₙ  = k aₙ₋₁

LHS =  a₁/aₙ

aₙ  = k aₙ₋₁

aₙ₋₁ = kaₙ₋₂

=> aₙ = k * kaₙ₋₂

=> aₙ = k²aₙ₋₂

=> aₙ = k³aₙ₋₃

=> aₙ = kⁿ⁻¹aₙ₋₍ₙ₋₁₎ =  kⁿ⁻¹a₁

=> LHS =  a₁/ kⁿ⁻¹a₁

= 1/ kⁿ⁻¹

= (1/k)ⁿ⁻¹

a₁/a₂  = 1/k

= (a1/a₂)ⁿ⁻¹

= RHS

QED

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Answer 2

Step-by-step explanation:

$\: \: \:$

∵ a₁ , a₂ , a₃ , ....... an are n positive real numbers

∴ we have to use AM ≥ GM

Where AM is Arithmetic mean and GM is Geometric mean.

Now, AM of a₁ , a₂, a₃, a₄, ........., an = (a₁ + a₂ + a₃ + ..... + an)/n

GM of a₁ , a₂, a₃, ........, an = (a₁.a₂.a₃.a₄....an)^(1/n)

∴ (a₁ + a₂ + a₃ + ..... + an)/n ≥ (a₁.a₂.a₃.a₄....an)^(1/n)

∵ a₁.a₂.a₃.a₄......an = 1

So, (a₁ + a₂ + a₃ + ..... + an)/n ≥ 1

⇒a₁ + a₂ + a₃+ ...... + an ≥ n

minimum value of a₁ + a₂ + a₃+ ...... + an = n

Now, take , (1 +a₁), (1 + a₂), (1 + a₃) ....... (1 + an)

AM = {(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n

GM = {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

Now, AM ≥ GM

{(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒{(1 + 1 + 1 + 1 ....+ n times) + (a₁ + a₂ + a₃ + .....+ an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒(n + n)/n ≥{ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒2 ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

∴ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) ≤ 2ⁿ

Hence, minimum value of (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) = 2ⁿ