if 3rd term of AP is 1 and 6th term is -11 find the sum of its 32 term
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x3=1
first term=f
common difference=d
f+2d=x3
f+2d=1...(1)
x6=f+5d
-11=f+5d
f+5d= -11 ...(2)
(1)-(2)
f+2d=1-
f+5d= -11
---------------
12=-3d
d= 12/-3= -4
f+2d= -1
f-8=-1
f=7
sn=[tex] \frac{n}{2}[2f+(n-1)d] \\ \\ \frac{32}{2}[-14+ 31*-4) \\ \\ 16(-14-124) \\ \\ 16(-138) =−2208[/tex]
first term=f
common difference=d
f+2d=x3
f+2d=1...(1)
x6=f+5d
-11=f+5d
f+5d= -11 ...(2)
(1)-(2)
f+2d=1-
f+5d= -11
---------------
12=-3d
d= 12/-3= -4
f+2d= -1
f-8=-1
f=7
sn=[tex] \frac{n}{2}[2f+(n-1)d] \\ \\ \frac{32}{2}[-14+ 31*-4) \\ \\ 16(-14-124) \\ \\ 16(-138) =−2208[/tex]
denny1999:
THE ANSWER IS WRONG THE ANSWER SHOULD BE -1856
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