What is Sn for the arithmetic series with d = –4, an = 27, and n = 9?
Hi,
an = a1 + d(n - 1)
27 = a1 - 4(9 - 1)
27 = a1 -4(8)
27 = a1 - 32
a1 = 59
Sn = n/2(a1 + an)
Sn = 9/2(59 + 27)
Sn = 9/2(86)
Sn = 9(43)
Sn = 387 <==ANSWER
I hope that helps!! :-)
an = a1 + (n-1) d
so 27 = a1 + (9-1) (-4)
or a1 = 27 +32 =59
so Sn = n/2 { a1 + an}
= 9/2[ 59 + 27]
= 9/2 *86
Sn =387 ans
1/2(a+l) = Sn
an=27
a=3
9th term is the last term and that is a+8d
1/2(3-29)
1/2*-26 = -13
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Verified answer
Hi,
an = a1 + d(n - 1)
27 = a1 - 4(9 - 1)
27 = a1 -4(8)
27 = a1 - 32
a1 = 59
Sn = n/2(a1 + an)
Sn = 9/2(59 + 27)
Sn = 9/2(86)
Sn = 9(43)
Sn = 387 <==ANSWER
I hope that helps!! :-)
an = a1 + (n-1) d
so 27 = a1 + (9-1) (-4)
or a1 = 27 +32 =59
so Sn = n/2 { a1 + an}
= 9/2[ 59 + 27]
= 9/2 *86
Sn =387 ans
1/2(a+l) = Sn
an=27
a=3
9th term is the last term and that is a+8d
1/2(3-29)
1/2*-26 = -13