Question

[Maths]

Q1) If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.

Q2) How many terms of the G.P. $\large{3, \dfrac{3}{2}, \dfrac{3}{4},...}$are needed to give the sum $\dfrac{3069}{512}$?
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Answer 1

Step-by-step explanation:

Question 1 :

• If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.

Solution 1 :

Given that :

• 4th, 10th and 16th terms of a G.P. are x, y and z.

To prove :

• x, y and z are in G.P.
• [y/x = z/y]

We know in a G.P :

=> a is the first term.

=> r is the common ratio.

Accordingly :

=> 4rth term = x

=> 10th term = y

=> 16th term = z

That is :

=> a₄ = ar³ = x - (1)

=> a₁₀ = ar⁹ = y - (2)

=> a₁₆ = ar¹⁵ = z - (3)

Now (y/x), (2) by (1) :

=> ar⁹/ar³ = a⁹⁻³ = r⁶

Now (z/y), (3) by (2) :

=> ar¹⁵/ar⁹ = a¹⁵⁻⁹ =  r⁶

∴ y/x = z/y, so x ,y ,z are in G.P

Question 2 :

• How many terms of the G.P 3,(3/2),(3/4).. are needed to give the sum 3069/512?

Solution 2 :

We have :

=> G.P = 3,(3/2),(3/4)

=> First term (a) = 3

=> Common ratio (r) = [(3/2)/3] = [(3/4)/(3/2)]

=> Sum (Sₙ) = 3069/512

We know :

Sum of n terms in a G.P,

=> Sₙ = a(1 - rⁿ)/1 - r

Substituting we get,

=> 3069/512 = 3[1 - (1/2)ⁿ]/1/2 - 1

=> 3069/512 = 6(1 - 1/2ⁿ)

=> 1/2ⁿ = 1 - 3069/512 = 1/1024

=> 2ⁿ = 1024 = 2¹⁰

∴ Value of n is 10, 10 terms muse be added.

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Answer 2

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