Question Details

Let an be nth the term of a decreasing infinite geometric progression. If  a1 + a2 + a3 = 52 and  a1 a2 + a2 a3 + a3 a1 = 624 , then the sum of this geometric progression is

Options

A

57

B

54

C

60

D

63

Show Answer

Correct Answer :

Option B

54

Solution :

The correct option is 54.

Step-by-Step Explanation:

Let the first term of the infinite geometric progression be a1=a and the common ratio be r.

The terms of the geometric progression can be expressed as:

a1=a

a2=ar

a3=ar2

Step 1: Use the first given condition.

We are given that the sum of the first three terms is 52:

a1+a2+a3=52

Substitute the terms in terms of a and r:

a+ar+ar2=52

Factor out a:

a(1+r+r2)=52     --- (Equation 1)

Step 2: Use the second given condition.

We are given that:

a1a2+a2a3+a3a1=624

Substitute a1=a, a2=ar, and a3=ar2:

(a)(ar)+(ar)(ar2)+(ar2)(a)=624

a2r+a2r3+a2r2=624

Factor out a2r:

a2r(1+r2+r)=624

a2r(1+r+r2)=624     --- (Equation 2)

Step 3: Divide Equation 2 by Equation 1.

a2r(1+r+r2)a(1+r+r2)=62452

Simplifying both sides gives:

ar=12

a=12r

Step 4: Find the value of r.

Substitute a=12r into Equation 1:

12r(1+r+r2)=52

Multiply both sides by r and divide by 4:

3(1+r+r2)=13r

3+3r+3r2=13r

3r2-10r+3=0

Factor the quadratic equation:

(3r-1)(r-3)=0

This gives r=13 or r=3.

Since the problem states that the geometric progression is decreasing and infinite, the common ratio must satisfy 0<r<1 for the infinite sum to exist and converge. Therefore:

r=13

Step 5: Calculate the first term a and the infinite sum S.

Now, calculate a:

a=121/3=36

The formula for the sum of an infinite geometric progression is:

S=a1-r

Substitute the values of a=36 and r=13:

S=361-13=3623=36×32=54

Thus, the sum of this infinite geometric progression is 54.

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