To find the sum of an arithmetic series, we use the formula give below.
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
a = first term, d = common difference, n = number of terms and l = last term.
Question 1 :
Find the sum of the series
34 + 32 + 30 + .................. + 10
Solution :
Sn = (n/2) [a + l]
a = 34, d = 32 - 34 = -2, l = 10 = a n
an = a + (n - 1) d
10 = 34 + (n - 1) (-2)
10 - 34 = (n - 1) (-2)
-24 = (n - 1) (-2)
-24/(-2) = (n - 1)
12 = n - 1
n = 12 + 1
n = 13
Hence the number of terms in the sequence is 13.
S13 = (13/2) [34 + 10]
S13 = (13/2) [ 44]
= 13(22)
= 286
Question 2 :
Find the sum of the series
- 5 + (-8) + (-11) + .............+ (-230)
Solution :
S n = (n/2) [a + l]
a = -5 d = -8 - (-5) l = -230 = an
= -8 + 5
= -3
an = a + (n - 1) d
-230 = -5 + (n - 1) (-3)
-230 + 5 = (n - 1) (-3)
-225 = (n - 1) (-3)
225/3 = (n - 1)
75 = n - 1
n = 75 + 1
n = 76
Hence the number of terms in the sequence is 76.
S76 = (76/2) [-8 + (-5)]
S76 = 38 [ -8 - 5]
= 38 (-13)
= -494
Question 3 :
In an AP
Given a = 5, d = 3, a n= 50 find n and Sn
Solution :
an = a + (n - 1) d
50 = 5 + (n - 1) (3)
50 - 5 = (n - 1) (3)
45 = (n - 1) (3)
45/3 = n - 1
n - 1 = 15
n = 15 + 1
n = 16
S n = (n/2) [a + l]
S16 = (16/2) [5 + 50]
= 8 (55)
= 440
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