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If 1+4+7+...+ x= 70, then x= ???

Arithmetic progression:
\(\frac{(a_1+ a_n)*n}{2}= 70\)
—>\(( 2a_1+ (n—1)d)*n = 140\)
d= 3 —>\( (2+ 3n—3 )*n= 140\)
\(3n^{2} —n —140 = 0\)
\((3n+20)(n—7) = 0\)
n= 7 —>\( a_7 = a_1 + 6d = 1+ 6*3 = 19\)

The answer is C

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Bunuel
If \(1+4+7+⋯+x=70\), then \(x=?\)

A. 11
B. 18
C. 19
D. 20
E. 22

1+4+7…x=70
a+(a+d1)+(a+d2)+…x=70
d=3, x=a+d(n-1)=1+3(n-1)
sum of ap: num terms * average = n*(1+x)/2=70
n(1+1+3(n-1))/2=70
n(3n-1)=140, 3n^2-n-140=0
(3n+20)(n-7)=0, n=integer=7
x=1+3(7-1)=1+18=19

Ans (C)
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Bunuel
If \(1+4+7+⋯+x=70\), then \(x=?\)

A. 11
B. 18
C. 19
D. 20
E. 22


Are You Up For the Challenge: 700 Level Questions

From the first 3 terms, we see that each term is 1 more than a multiple of 3. Therefore, x should be 1 more than a multiple of 3 as well. This eliminates choices A, B, and D. So we need to consider only the remaining two choices; let’s look at choice C first.

If x = 19, we have 1 + 4 + 7 + 10 + 13 + 16 + 19 = (1 + 19) + (4 + 16) + (7 + 13) + 10 = 20 + 20 + 20 + 10 = 70. Since the sum is indeed 70, we see that x must be 19.

Answer: C
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