The sum of 5 numbers in geometric progression is 2

Question

GMATBusters’ Quant Quiz Question -2  
The sum of 5 numbers in the geometric progression is 24. The sum of their reciprocals is 6. The product of the terms of the geometric progression is 
(a) 36 
(b) 32 
(c) 24 
(d) 18
(e) 16

A  geometric progression , also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

GMATBusters · Answer

Official Solution:

Let the terms in geometric progression are a/r^2, a/r, a, ar & ar^2.

The sum of 5 numbers in the geometric progression is 24.
a/r^2 + a/r + a + ar + ar^2 = 24 ... Eq (1)

The sum of their reciprocals is 6.
1/ar^2 + 1/ar + 1/a + r/a + r^2/a = 6 ... Eq(2)

Multiplying Eq (2) by a^2
ar^2 + ar + a + a/r + a/r^2 = 6a^2 = 24
a^2 = 4; a= +/-2

The product of the terms of the geometric progression is =a^5 = 32 or -32
out of the given answer options. We can select 32

Hence, Answer = B

(PS: in Probalem Solving type Question: even if the answer can be more than one value, we must select the answer from the available answer options.)

Kinshook · Answer

In Problem solving question, we need to select the answer out of the given options.

Given: 
1. The sum of 5 numbers in the geometric progression is 24. 
2. The sum of their reciprocals is 6.

Asked: The product of the terms of the geometric progression is 
(a) 36 
(b) 32 
(c) 24 
(d) 18
(e) 16

Let the 5 numbers in geometric progression be ar^2, ar, a, a/r & a/r^2.

1. The sum of 5 numbers in the geometric progression is 24. 
ar^2 + ar + a + a/r + a/r^2 = 24 (1)

2. The sum of their reciprocals is 6. 
1/ar^2 + 1/ar + 1/a + r/a + r^2/a = 6. (2)

Multiplying (2) by a^2
ar^2 + ar + a + a/r + a/r^2 = 6a^2 = 24
a^2 = 4; a= 2 or a = -2

The product of the terms of the geometric progression is = ar^2*ar*a*a/r*a/r^2 = a^5 = 2^5 = 32

IMO B

Hi  GMATBusters 
If a = - 2; a^5 = -32 
Therefore, -32 is also a solution to the above problem

ashwini2k6jha · Answer

let a,ar,ar2,ar3 and ar4 in gp series
given a+ar+ar2+ar3+ar4=24 -----1
also given , 1/a+1/ar+1/ar2+1/ar3+1/ar4=6
1/a(r4+r3+r2+r+1)/r4=6
putting value from 1
24/a2xr4=6
ar2=2
hence product a5r10=2^5=32Ansswer

Raxit85 · Answer

Sum of 5 terms of geometric progression = A(r^5-1)/(r-1)=24 ---1)

sum of 5 terms of geometric progression = 1/A(r^5-1)/(r-1)= 2=6----2)

Let's divide both the eqations,

Square root of A =24/6

A=2

We need product of 5 terms so taking first term as 2 and the successive terms as 2^r..2^r^2,,,, So, 32.

Ans. B

techwave · Answer

Answer: B

Sum of five numbers in a GP:

A(r to the power5-1)/(r-1)= sum of 5 terms of gp1=24

Sum of Reciprocals:

1/A(r to the power5-1)/(r-1)=sum of 5 terms of gp2=6
dividing both the equations to get
A to the power 2 =24/6 = 4
 Hence A=2
We need the product of 5terms so taking the first term as 2 and the successive terms as 2^r..2^r^2
therefore, the answer is 32

adonouis · Answer

Correct answer is B i.e 32.

let the sum of 5 numbers in a geometric progression = ΣGP 1 = 24, 
let the sum of their reciprocals = ΣGP 2 = 6,

Now, by using the formula for sum of geometric progression we have,
ΣGP 1 = A(r^5 - 1)/(r-1) = 24 equation (i)
ΣGP 2 = 1/A(r^5 - 1)/(r-1)= 6 equation (ii)

dividing equation (i) by equation (ii), we get,

A^2 = 24/6
A^2 = 4
thus, A = 2

So, the GP becomes: 2^r; 2^r^2;.. and so on.

Thus product of the terms are 32.

Archit3110 · Answer

sum of sequence (x)= 24 --(1)
let this be x
and sum reciprocals (1/x)= 6 and let sum of reciprocal ; 1/x ---(2)
so divide 1 by 2
we get ; x^2 = 24/6 ; 4 ; 
so x=2
product of 5 terms ; 2^5 ; 32
IMO B

The sum of 5 numbers in the geometric progression is 24. The sum of their reciprocals is 6. The product of the terms of the geometric progression is
(a) 36
(b) 32
(c) 24
(d) 18
(e) 16

shameekv1989 · Answer

Sum of GP_1 =  A*\frac{(1-r^5)}{(1-r)}  = 24
Sum of GP_1 =  A*\frac{(r^5-1)}{(r-1)}  = 24 ----- i)

Sum of GP_2 =  \frac{1}{A}*\frac{(1-\frac{1}{r^5})}{(1-\frac{1}{r})}  = 6

Sum of GP_2 =  \frac{1}{A}*\frac{\frac{(r^5-1)}{r^5}}{\frac{(r-1)}{r}}  = 6

Sum of GP_2 =  \frac{1}{A}*\frac{(r^5-1)}{(r-1)r^4}  = 6 ----- ii)

Dividing i) by ii)

\frac{Sum of GP_1}{Sum of GP_2} = \frac{A*\frac{(r^5-1)}{(r-1)}}{\frac{1}{A}*\frac{(r^5-1)}{(r-1)r^4}}

A^2*r^4  = 4
 A*r^2  = +-2

Therefore Product of terms =  A^5 r^{10} = +-2^5 = +-32

Answer - B

bM22 · Answer

Answer :  B - 32

Consider the terms of GP as a/r^2, a/r, a, ar, ar^2

Sum of the GP of 5 terms = a(r^5 - 1)/r-1 = 24 --- equation 1
sum of reciprocal of GP : 1/a((1-(r^5))/1-r) = 6 ---- equation 2

Equation 1/equation 2 => a^2 = 4 => a = 2

Now product of the terms GP will be a^5 = 32.

Answer : 32

MARVEL13 · Answer

GIVEN

a + ar + ar^2 +ar^3 +ar^4  = 24

&

1/a + 1/ar + 1/ar^2 + 1/ar^3 + 1/ar^4 = 6

=6

24/a^2r^4=6

a^2r^4 = 4

ar^2=2

Product -->

a^5r^10 = (ar^2)^5 = 2^5 = 32

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The sum of 5 numbers in geometric progression is 2

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