# Determine the sum of the first 100 numbers. (arithmetic sequence)?

In an arithmetic sequence, the first three numbers are 100, 96 and 92.

Determine the sum of the first 100 numbers.

I only know that the answer is -9800, can someone show how I can get this answer?

Relevance

a = 100

d = -4

100th term = 100-4*99 = -396

Average term = (100-396)/2 = -98

Sum = -98*100 = -9800

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• Sn = (n/2) [ 2a + (n - 1) d ]

S100 = 50 [ 2 + 99 x 1 ]

S 100 = 50 x 101

S 100 = 5050

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• 100 + 96 + 92 + ...... + (100 - 4*99)

= 100/2(100 + 100 - 4*99)

= 50(200 - 396)

= 100(-196)/2

= -19600/2

= -9800

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• S(n) = n/2(2a + (n-1)d) is the general equation

where

n = 100

a = 100

d = -4

Substituting

S(100) = 100/2(2(100) + ( 100-1)(-4))

S(100) = 50(200 + 99(-4)

S(100) = 50(200 - 396)

S(100) = 50(-196)

S(100) = -9800 As required.

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• Sn = (n/2)[2a + (n - 1)d]....where a is the first term, d is the common difference and n is the number of terms

i.e. S₁₀₀ = (100/2)[200 - 4(99)]

=> S₁₀₀ = 50(-196)

so, -9800

:)>

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• A(n) = 100 - 4(n-1)

A(1) = 100

A(100) = 100 - 396 = -296

Σ = number of terms multiplied by the average value of each term:

= 100(A(1) + A(100))/2

= 100(100-296)/2

= 50(-196)

Ans: Σ = -9800

Or:

Σ(104-4n) = 100(104) - 4Σn

= 10400 - 400(101)/2

= 10400 - 20200

= -9800

Either way the answer is the same, -9800

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• S[n] = (n/2) * (t + t[n])

This holds true for all arithmetic sequences

S = (100/2) * (t + t)

S = 50 * (t + t)

The first term is 100, or 100 - 0, or 104 - 4, or 104 - 4 * 1

The 2nd term is 96, or 100 - 4, or 104 - 8, or 104 - 4 * 2

The 3rd term is 92, or 104 - 12, or 104 - 4 * 3

See a pattern?

t = 100

t = 104 - 4 * 100 = 104 - 400 = -296

50 * (100 + (-296)) =>

50 * (-196) =>

50 * 2 * (-98) =>

-98 * 100 =>

-9800

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