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655-705 (Hard)|   Sequences|      
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Bunuel
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 14 Feb 2016, 05:38
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C
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55% (hard)

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63% (02:02) correct 37% (02:03) wrong based on 370 sessions

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If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

A. 3
B. 1/3
C. 2
D. 9
E. 1/9
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C
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ynaikavde
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 14 Feb 2016, 21:19
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sum of 6 terms in GP: 1 + \(a\) + \(a^2\) + \(a^3\)+ \(a^4\)+ \(a^5\) == 1 + a + \(a^2\) + \(a^3\) * (1 + \(a\) + \(a^2\) )
sum of first 3 terms in GP: 1 + \(a\) + \(a^2\)

x= 1 + \(a\)+ \(a^2\)
given:
(x + \(a^3\)*x)/x = 9

\(a^3\) = 8
a= 2
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chetan2u
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GMAT Focus 1: 735 Q90 V89 DI81
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 14 Feb 2016, 21:00
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Bunuel
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

A. 3
B. 1/3
C. 2
D. 9
E. 1/9
Show more

Hi,

1) formula way


the sum of n terms in a GP = first term *(1-r^n)/(1-r)..
so sum of 6 terms = first term *(1-r^6)/(1-r)..
and first three terms=first term *(1-r^3)/(1-r)..

therefore the ratio = (1-r^6)/(1-r^3)= (1-r^3)(1+r^3)/(1-r^3)= 1+r^3...
but this is given as 9..
so 1+r^3=9..
r^3=8..
or r=2..

substitution


if we have problem in simplifying after (1-r^6)/(1-r^3), substitute the choices and see when you get 9..
A. 3....
B. 1/3
C. 2
D. 9
E. 1/9
we can eliminate the fractions straightway here..B and E out..
the ratio is 9, so we cannot have r as a multiple of 9 or 3 as the numerator (1-r^6) will become [b]6th power of multiple of 3 - 1, which will never be a multile of 9..
only C is left[/b]
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 17 Feb 2016, 14:13
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9=(a1+a2+a3+a4+a5+a6)/(a1+a2+a3)
Factorize the same terms
9=1+(a4+a5+a6)/(a1+a2+a3)
Write every term with respect to r
a1=a1
a2=a1*r^1
a3=a1*r^2
.........
9=1+(a1(r^3+r^4+r^5))/(a1(1+r^1+r^2))
8=(r^3 (1+r^1+r^2))/((1+r^1+r^2))
8=r^3
r=2
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 12 Nov 2018, 17:45
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Bunuel
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

A. 3
B. 1/3
C. 2
D. 9
E. 1/9
Show more

say (1+2+4+8+16+32)/(1+2+4)→
63/7=9
common ratio=2
C
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 21 Nov 2018, 01:42
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chetan2u
Bunuel
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

A. 3
B. 1/3
C. 2
D. 9
E. 1/9

Hi,

1) formula way


the sum of n terms in a GP = first term *(1-r^n)/(1-r)..
so sum of 6 terms = first term *(1-r^6)/(1-r)..
and first three terms=first term *(1-r^3)/(1-r)..

therefore the ratio = (1-r^6)/(1-r^3)= (1-r^3)(1+r^3)/(1-r^3)= 1+r^3...
but this is given as 9..
so 1+r^3=9..
r^3=8..
or r=2..

substitution


if we have problem in simplifying after (1-r^6)/(1-r^3), substitute the choices and see when you get 9..
A. 3....
B. 1/3
C. 2
D. 9
E. 1/9
we can eliminate the fractions straightway here..B and E out..
the ratio is 9, so we cannot have r as a multiple of 9 or 3 as the numerator (1-r^6) will become [b]6th power of multiple of 3 - 1, which will never be a multile of 9..
only C is left[/b]
Show more

Thank you for the explanation. But how do we know if r>1???
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If the ratio of the sum of the first 6 terms of a G.P. to the sum of 14 Jan 2026, 15:09
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Deconstructing the Question
Let the first term be \(a\) and the common ratio be \(r\).
The sum of the first \(n\) terms of a G.P. is given by: \(S_n = \frac{a(r^n - 1)}{r - 1}\).

Given condition:
\(\frac{S_6}{S_3} = 9\)

Step 1: Set up the Ratio
Substitute the formula into the ratio:
\(\frac{\frac{a(r^6 - 1)}{r - 1}}{\frac{a(r^3 - 1)}{r - 1}} = 9\)

Cancel out the common terms \(a\) and \((r - 1)\):
\(\frac{r^6 - 1}{r^3 - 1} = 9\)

Step 2: Factor and Simplify
Notice that the numerator \(r^6 - 1\) is a difference of squares: \((r^3)^2 - 1^2\).
Factor it as: \((r^3 - 1)(r^3 + 1)\).

Substitute back into the equation:
\(\frac{(r^3 - 1)(r^3 + 1)}{r^3 - 1} = 9\)

Cancel \((r^3 - 1)\) from the numerator and denominator:
\(r^3 + 1 = 9\)

Step 3: Solve for r
\(r^3 = 9 - 1\)
\(r^3 = 8\)
\(r = 2\)

Answer: C

Bunuel
If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

A. 3
B. 1/3
C. 2
D. 9
E. 1/9
Show more
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