find where 0 is a term of the ap 40, 37, 34 ............
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Answered by
33
a= 40
d= a2 - a1 = 37 - 40= -3
an= a+(n-1)d
a0= 40+(0-1)-3
= 40+(-1)-3
= 40+3
=43
d= a2 - a1 = 37 - 40= -3
an= a+(n-1)d
a0= 40+(0-1)-3
= 40+(-1)-3
= 40+3
=43
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Answered by
21
AP . 40, 37 ,34
a = 40
d =( -3)
an = 0
an = a +( n-1) d
0 = 40 + ( n - 1 ) (-3)
0 = 43 - 3n + 3
43 - 3n = 0
3n = 43
n = 43 / 3
but n must be positive and natural no.
therefor, 0 is not a term of this AP.
a = 40
d =( -3)
an = 0
an = a +( n-1) d
0 = 40 + ( n - 1 ) (-3)
0 = 43 - 3n + 3
43 - 3n = 0
3n = 43
n = 43 / 3
but n must be positive and natural no.
therefor, 0 is not a term of this AP.
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