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# If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?

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RC Moderator
Joined: 24 Aug 2016
Posts: 565
Concentration: Entrepreneurship, Operations
GMAT 1: 630 Q48 V28
GMAT 2: 540 Q49 V16
If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?  [#permalink]

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13 Aug 2018, 10:47
2
00:00

Difficulty:

55% (hard)

Question Stats:

57% (02:29) correct 43% (02:39) wrong based on 21 sessions

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If $$1+4+7+⋯+x=70$$ ; Then $$x=?$$

A) $$11$$
B) $$18$$
C) $$19$$
D) $$20$$
E) $$22$$
Director
Joined: 31 Jul 2017
Posts: 504
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?  [#permalink]

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13 Aug 2018, 11:00
u1983 wrote:
If $$1+4+7+⋯+x=70$$ ; Then $$x$$ = $$?$$

A) $$11$$
B) $$18$$
C) $$19$$
D) $$20$$
E) $$22$$

$$x = 3n-2$$. Implies, n = 7/8 or x = 19/22.. Substitute this in below equation -
$$70 = n(x+1)/2$$

$$x = 19$$
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Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
Re: If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?  [#permalink]

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13 Aug 2018, 18:49
1
u1983 wrote:
If $$1+4+7+⋯+x=70$$ ; Then $$x=?$$

A) $$11$$
B) $$18$$
C) $$19$$
D) $$20$$
E) $$22$$

The given summation series is an Arithmetic Progression(AP) with first term=1, last term=x and common difference=3
nth term of an AP=first term+(n-1)*common difference=1+(n-1)*3=3n-2, So, last term=x=3n-2
We know, sum of 'n' terms =$$\frac{n}{2}$$(first term + last term)
Or, 70=$$\frac{n}{2}$$(1+ 3n-2)
Or, n(3n-1)=140
Or, 3n^2-n-140=0
Or, n=7, $$-\frac{20}{3}$$

So, x=3n-2=3*7-2=19

Ans. (C)
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PKN

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Re: If 1 + 4 + 7 + ⋯ + x = 70; Then x = ? &nbs [#permalink] 13 Aug 2018, 18:49
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