Given That 20=20+18+16 + . . . +X, How Many Terms Are On The Right Side Of The Equation?, Find The Sum Of 2212100201399201398201397 2013 . . . +100+10

Given that 20=20+18+16 + . . . +x, how many terms are on the right side of the equation?

Find the sum of −100–99–98–97 – . . . +100+101+102

Please answer these 2 questions.

 

Answer:

Step-by-step explanation:

Sum of arithmetic sequence:

Sn = n/2 2a₁ + (n-1)(d)

Where:

Sn = sum of the sequence

n = number of terms

a₁ = first term

d = common difference

Given:

20 = 20+18+16+...+x

Sn = 20

a₁ = 20

d = -2  (the sequence decreases by 2 per term)

x = nth term

Find n, the number of terms:

20 = n/2 2(20) + (n-1)(-2)

20 = n/2 {40 + 2 - 2n

20 = n 21 - n)

20 = -n² + 21n

Arrange the quadratic equation in standard form:

n² - 21n + 20 = 0

Factor:

(n - 20) (n - 1) = 0

n - 20 = 20

n = 20

n - 1 = 0

n = 1

Choose n = 20.

ANSWER: There are 20 terms.

Find the 20th term:

a₂₀ = 20 + (20-1)(-2)

a₂₀ = 20 + (19)(-2)

a₂₀ = 20 - 38

a₂₀ = -18

The 20th term x = -18.

Check:

Sn = n/2 (a₁ + an)

20 = 20/2 (20 + (-18))

20 = 10 (2)

20 = 20  (True)

----

Find the sum of  −100–99–98–97 – . . . +100+101+102

Where:

a₁ = -100

d = 1 (the sequence is increasing hence the common difference is positive)

an = 102

n = unknowm

Find n, the number of terms:

an = a₁ + (n-1)(d)

102 = -100 + (n-1)(1)

102 = -100 - 1 + n

102 = -101 + n

n = 102 + 101

n = 203

There are 203 terms.

Find the sum:

S₂₀₃ = 203/2 -100 + 102}

S₂₀₃ = 203/2 (2)

S₂₀₃ = 203

ANSWER: The sum is 203.


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