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Gelneren [198K]
2 years ago
15

Let $a_1$, $a_2$, . . . , $a_{10}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 = 17$ and $a_2 + a_4 + a_6 + a_8 +

a_{10} = 15$, what is the value of $a_1$?
GIVING BRAINLIEST>
Mathematics
1 answer:
vfiekz [6]2 years ago
7 0

So

a be first term and d be common difference

  • a+a+2d+a+4d+a+6d+a+9d=17
  • 5a+21d=17--(1)

And

  • a+d+a+3d+a+5d+a+7d+a+9d=15
  • 5a+25d=15--(2)

Eq(1)-Eq(2)

  • -4d=-2
  • d=1/2

Put in second one

  • 5a+25d=15
  • a+5d=3
  • a+5/2=15
  • a=15-5/2
  • a=25/2
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Answer:

5.33

Step-by-step explanation:

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The formula for the geometric sum is :

First term*(1-multiplier^n)/(1-multiplier), formally \frac{a_1(1-r^n)}{1-r}

Here multiplier is 1/4

First term is (1/4)^(1-1) = 1

Thus the value of sum is 1*(1-(1/4)^1^0)/(1-1/4)which equals (1- (1/4)^1^0)*4/3

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1/4(1-2/3)^2+1/3 Simplify
MrMuchimi

Simplifying the expression gives 13/ 36

<h3>How to solve the expression</h3>

Given;

1/4(1-2/3)^2+1/3

Represented as;

\frac{1}{4} * ( 1  - \frac{2}3} )^2 + \frac{1}{3}

Take the LCM of the bracket

\frac{1}{4} * ( \frac{1}{3} )^2 + \frac{1}{3}

Find the square

\frac{1}{4} (\frac{1}{9} )+ \frac{1}{3}

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\frac{1}{36} + \frac{1}{3}

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Thus, simplifying the expression gives 13/ 36

Learn more about fractions here:

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Step-by-step explanation:

a).

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