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

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RC Moderator
Joined: 24 Aug 2016
Posts: 804
Location: Canada
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, 11:47
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Difficulty:

45% (medium)

Question Stats:

63% (02:38) correct 38% (02:39) wrong based on 27 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: 515
Location: Malaysia
Schools: INSEAD Jan '19
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, 12:00
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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Joined: 01 Oct 2017
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WE: Supply Chain Management (Energy and Utilities)
Re: If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?  [#permalink]

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13 Aug 2018, 19:49
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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Re: If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?   [#permalink] 13 Aug 2018, 19:49
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# If 1 + 4 + 7 + ⋯ + x = 70; Then x = ?

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