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The 1st term of an artihmetic sequence is 8 and the common difference is 3. Find the sum of \(a_5+a_6+a_7+\cdots+a_{99}+a_{100}\).

 Feb 3, 2022
 #1
avatar+514 
+6

The 5th term is 20. The pattern is the n-th term is 3(n-1) + 8, so the 100-th term of the pattern is 3(99) + 8 or 305. Thus, all the numbers added from 20 to 305 would be added together, with a difference of 3 between each number. Since we know there are a total of 96 numbers added together, we can subtract 17 from each number, then divide each number by 3. Then the numbers from 1 - 96 would be added together, which equals 4656.

 

Then we multiply 4656 by 3 to get 13968. Then since we subtracted 17 from 96 numbers, we have to add (17 x 96) to 13968. That gets 15600. Thus, the answer is 15600.

 

Correct me if I got it wrong :D

 Feb 3, 2022
edited by proyaop  Feb 3, 2022
 #2
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+3

Use the formula for the sum of an arithmetic sequence:

 

S==96/2 * [2* 20  +  (3*95)], solve for S

 

S ==48 * [40  +  285]

 

S ==48 * 325 ==15,600

 Feb 3, 2022
 #3
avatar+514 
+5

I think applying the formula is the smartest way to do it, I did it a little unorthodoxly, thank you Guest! 

 

smiley

proyaop  Feb 3, 2022
 #4
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a5 = a1 + 4 d = 20

a100 = a1 + 99d = 305      Now use the sum formula for an arith sequence to sum the 96 terms a5 - a100

96/2 ( 20 + 305) = 15600

 Feb 3, 2022
 #5
avatar+63 
+4

All of you got it right!!! Thanks!

 Feb 3, 2022

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