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The following arithmetic sequence is given: $20 ; 23 ; 26 ; 29 ; ... ; 101$

1. How many terms are there in this sequence?
2. The even numbers are removed from the sequence. Calculate the sumof terms of the remaining sequence.
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## 1 Answer

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1. 23-20=3=d

a=20

101=a+(n-1)d

101=20+(n-1)3

n= (101-20)/3 +1

= 28

since the last term is at position 28, there are 28 terms.

2. Even numbers= 20,26,32,....,100

It is an arithmetic seq with d=6

a= 20

last term=100=20+(n-1)6

n= (100-20)/6 + 1 = 14.33

Sum= 14.33(2(20)+(20-1)6)/2 = 1 103.41

Sum of whole sequence= 20(2(20)+(28-1)3)/2 = 1 210

Hence sum of terms remaining= 1 210-1 103.41 = 106.59
by Diamond (42,256 points)

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