Evaluate Each Arithmetic Series Described
EVALUATING Arithmetics SERIES
A serial whose terms are in arithmetic sequence is called arithmetic series.
To find sum of terms in arithmetics sequence, we can utilize one of the formulas given below.
Sn = (northward/two) [a + fifty]
Sn = (n/2) [2a + (n - 1)d]
a = first term,
n = number of terms of the series,
d = common difference and 50 = concluding term
Question i :
Detect the sum of the following.
(i) three, 7, 11,… up to 40 terms.
Solution :
Number of terms (n) = 40
Outset term (a) = 3
Common difference (d) = 7 - 3 = iv
Sn = (n/2) [2a + (n - 1)d]
= (40/2) [2(3) + (40 - 1)iv]
= xx [6 + 156]
= 3240
Hence the sum of 40 terms of the given serial is 3240.
(2) 102, 97, 92,… up to 27 terms.
Solution :
Number of terms (n) = 27
First term (a) = 102
Mutual divergence (d) = 97 - 102 = -5
Sn = (n/2) [2a + (n - 1)d]
= (27/2) [2(102) + (27 - one)(-5)]
= (27/two) [204 + 26(-v)]
= (27/2) [204 - 130]
= (27/2) (74)
= 27 (37)
= 999
Hence the sum of 27 terms of the given series is 999.
(three) 6 + 13 + 20 +...........+ 97
Solution :
l = 97, a = 6, d = 13 - 6 = 7
From this, nosotros have to notice the number of terms.
due north = [(l - a)/d] + i
n = [(97 - 6)/vii] + 1
= (91/7) + one
n = 13 + 1
n = xiv
Sn = (n/ii) [a + 50]
= (14/2)[6 + 97]
= vii[103]
= 721
Hence the sum of given serial is 721.
Question 2 :
How many consecutive odd integers beginning with five will sum to 480?
Solution :
Consecutive integers starting from 5 are 5, vii, nine, ............
v + 7 + ix + eleven + ............. = 480
Southn = 480
(n/ii)[2a + (northward - 1)d] = 480
a = v, d = 7 - 5 = 2
(north/2)[2(5) + (n - 1)(2)] = 480
(northward/ii)[x + 2n - 2] = 480
(north/2)[2n + 8] = 480
2n2 + 8n = 960
Dividing by 2, we go
n ii + 4n - 480 = 0
(n + 24) (n - 20) = 0
n = -24 or northward = 20
Hence by finding sum of 20 terms we will get 480.
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