Asked by ying
In an arithmetic sequence , the 9th term is twice the 3rd term and 15th term is 80. Find the common difference and the sum of the terms from from 9th to 15th is inclusive
Answers
Answered by
Reiny
Just change your English to math, using the definitions you learned.
"the 9th term is twice the 3rd term" ---> a + 8d = 2(a + 2d)
a + 8d = 2a + 4d
a = 4d
"15th term is 80" ----> a + 14d = 80
4d + 14d = 80
take it from there.
Use your sum of terms formula for the last part of your question.
Take the sum of 15 terms and subtract the sum of 8 terms from that.
"the 9th term is twice the 3rd term" ---> a + 8d = 2(a + 2d)
a + 8d = 2a + 4d
a = 4d
"15th term is 80" ----> a + 14d = 80
4d + 14d = 80
take it from there.
Use your sum of terms formula for the last part of your question.
Take the sum of 15 terms and subtract the sum of 8 terms from that.
Answered by
Damon
a + r (n-1)
a + 8 r = 2(a+2r) = 2 a +4 r
a + 14 r = 80
a + 14 r = 80
a - 4 r = 0
------------------- suubtract
18 r = 80
r = 4.44
a = 4 r = 17.78
a + 8 r = 2(a+2r) = 2 a +4 r
a + 14 r = 80
a + 14 r = 80
a - 4 r = 0
------------------- suubtract
18 r = 80
r = 4.44
a = 4 r = 17.78
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