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Leno4ka [110]
3 years ago
7

The sum of the first 3 terms of a geometric series is 171 and the common ratio is 2/3.

Mathematics
2 answers:
Paladinen [302]3 years ago
5 0

Answer:

81

Step-by-step explanation:

let the terms be a,ar,ar²

r=2/3

a+a(2/3)+a(2/3)²=171

multiply by 9

9a+6a+4a=171×9

19 a=171×9

a=(171×9)/(19)

a=9×9=81

jok3333 [9.3K]3 years ago
4 0

S_n=\dfrac{b_1(q^n-1)}{q-1} \\q=\dfrac{2}{3} , b_1=x, n=3, S_n=171

Let's find b₁

171=\dfrac{x((\dfrac{2}{3})^3-1) }{\dfrac{2}{3}-1 } \\\\\\171=x*(-\dfrac{19}{27} )*(-3) \\171=\dfrac{19}{9} x\\x=171*\dfrac{9}{19} \\x=81

Answer: b₁=81

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let me edit your question as:

Which two equations are true?

<u>Eq1:</u>

(2×10−4)+(1.5×10−4)=3.5×10−4(3×10−5)+(2.2×10−5)

<u>Eq2:</u>

6.6×10−10(6.3×10−1)−(2.1×10−1)=3×10−1(5.4×103)−(2.7×103)

<u>Eq3:</u>

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Answer:

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Step-by-step explanation:

let's check each equation, if the values on both sides (left and right side) are equal then the equation is true otherwise false.

Using PEMDAS rule we are simplifying the equations as;

<u>Eq1:</u>

(2*10-4)+(1.5*10-4)=3.5*10-4(3*10-5)+(2.2*10-5)\\(16)+(11)=35-4(25)+(17)\\27=35-100+17\\27=-48\\

<u>Eq2:</u>

<u></u>6.6*10-10(6.3*10-1)-(2.1*10-1)=3*10-1(5.4*103)-(2.7*103)\\66-10(62)-(20)=30-1(556.2)-278.1\\66-620-20=30-556.2-278.1\\-574=-804.1<u></u>

<u>Eq3:</u>

2.7*103(7.5*106)-(2.5*106)=5*100\\221089.5-265=500\\220824.5=500\\

<u>we observed that none of the equation has two same values on both sides thus none of the three equations is true.</u>

<u>Also, no value of Eq1, Eq2 or Eq3 are same thus none of the equation is true</u>

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