Asked by Kenelwe
find the twenty-first term of an A.S of which the 6th term is 3 and the 14th is 19
Answers
Answered by
Damon
a a+d a+2d a+3d .... a+(n-1)d
sixth term = a+5d = 3
fourteenth term = a+13d = 19
subtract to eliminate a
-8d = -16
d = 2
then a + 10 = 3
a = -7
twenty first term = -7 + 20(2)
= -7 + 40 = 33
sixth term = a+5d = 3
fourteenth term = a+13d = 19
subtract to eliminate a
-8d = -16
d = 2
then a + 10 = 3
a = -7
twenty first term = -7 + 20(2)
= -7 + 40 = 33
Answered by
Steve
19+(21-19)(19-3)/(14-6) = 23
Answered by
Steve
Should have said
19+(21-14)(19-3)/(14-6) = 33
19+(21-14)(19-3)/(14-6) = 33
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