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# In a sequence with the first term a1, each term after the first is d

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Intern
Joined: 10 Feb 2017
Posts: 36
Location: Viet Nam
GPA: 3.5
WE: General Management (Education)
In a sequence with the first term a1, each term after the first is d  [#permalink]

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15 Aug 2018, 01:12
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Difficulty:

95% (hard)

Question Stats:

40% (02:29) correct 60% (02:43) wrong based on 48 sessions

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In a sequence with the first term a1, each term after the first is d greater than the preceding term, where d is a constant. If a6=11^6 and a8 = 11^8. What is the largest prime divisor of a11?

A. 11
B. 13
C. 17
D. 19
E. 43
CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2711
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: In a sequence with the first term a1, each term after the first is d  [#permalink]

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Updated on: 15 Aug 2018, 02:10
Hungluu92vn wrote:
In a sequence with the first term a1, each term after the first is d greater than the preceding term, where d is a constant. If a6=11^6 and a8 = 11^8. What is the largest prime divisor of a^11?

A. 11
B. 13
C. 17
D. 19
E. 43

$$a_6=11^6$$

$$a_7= a_6 + d = 11^6 +d$$

$$a_8= a_7 + d = 11^6 +2d = 11^8$$

i.e. $$2d = 11^8 - 11^6$$

i.e. $$d = (11^6)*(11^2 - 1)/2 = 11^6*(120)/2 = 60*11^6$$

$$a_11= a_8 + 3d = 11^8 +3d = 11^8 + (3)*(60*11^6) = 11^6 (11^2+180) = 11^6*301 = 11^6*7*43$$

Answer: Option E
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Originally posted by GMATinsight on 15 Aug 2018, 01:38.
Last edited by GMATinsight on 15 Aug 2018, 02:10, edited 1 time in total.
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Re: In a sequence with the first term a1, each term after the first is d  [#permalink]

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15 Aug 2018, 02:04
GMATinsight
In your solution :
2d=118−1162d=118−116

i.e. d=(116)∗(112−1)=116∗(120)=120∗11^6

You have not divided by 2.

Value of d would be 60.11^6.

a11= a8+3d
= 11^8+3(60)(11^6)
= 11^6(11^2+180)
=11^6( 301)
=11^6.43^1.7^1

answer would be E 43...
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Re: In a sequence with the first term a1, each term after the first is d  [#permalink]

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16 Aug 2018, 06:36
Hungluu92vn wrote:
In a sequence with the first term a1, each term after the first is d greater than the preceding term, where d is a constant. If a6=11^6 and a8 = 11^8. What is the largest prime divisor of a^11?

A. 11
B. 13
C. 17
D. 19
E. 43

Given $$a_6 = 11^6$$ & $$a_8 = 11^8$$

$$a_8 = a_6 + 2d = 11^8$$

$$11^6 + 2d = 11^8$$

we get $$d = 11^6 (121-1)/2 = 11^6 * 60$$

$$a_{11} = a_8 + 3d = 11^8 + 180 * 11^6 = 11^6 * (121 + 180) = 11^6 * 301 = 11^6 * 7 * 43$$

largest prime divisor is 43.

Answer E

Thanks,
GyM
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Re: In a sequence with the first term a1, each term after the first is d &nbs [#permalink] 16 Aug 2018, 06:36
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# In a sequence with the first term a1, each term after the first is d

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