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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.
in Mathematics by Diamond (74,874 points) | 428 views

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 (43,672 points)

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