We have to consider the given sum as Sn. Instead of Sn, we may use one of the formulas given below.
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
a = first term, d = common difference and n = number of terms.
Question 1 :
How many terms of the AP
9, 17, 25,..........
must be taken to give a sum of 636?
Solution :
Sn = 636
a = 9, d = 17 - 9 = 6
Sn = (n/2) [2a + (n - 1)d]
(n/2) [2(9) + (n - 1)8] = 636
(n/2) [18 + 8n - 8] = 636
(n/2) [10 + 8n] = 636
n[5 + 4n] = 636
5n + 4n2 = 636
4n2 + 5n - 636 = 0
4n2 - 48n + 53n - 636 = 0
4n(n - 12) + 53(n - 12) = 0
(n - 12) (4n + 53) = 0
n = 12
By solving other factor (4n + 53), we get negative value for n, which is not admissible.
So, 12 terms to be added to get the sum 636.
Question 2 :
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.
Solution :
Given that :
First term (a) = 5, last term (l) = 45 and Sn = 400
Sn = (n/2) [a + l]
(n/2) [5 + 45] = 400
(n/2) [50] = 400
25n = 400
n = 400/25
n = 16
Hence the given series consist of 16 terms.
Question 3 :
In an arithmetic sequence
60, 56, 52, 48,.......
starting from the first term, how many terms are needed so that their sum is 368 ?
Solution :
Given that :
Sn = 368
60, 56, 52, 48,.......
a = 60, d = 56 - 60 = -6
(n/2)[2a+(n-1)d] = 368
(n/2)[2(60)+(n-1)(-4)] = 368
(n/2) (120-4n+4) = 368
(n/2) (124 - 4n) = 368
n(124-4n) = 368(2)
124n - 4n2 = 736
4n2-124n+736 = 0
÷ by 4 = > n2-31n+184 = 0
(n-23)(n-8) = 0
n = 23, 8
Sum of 8 or 23 terms of arithmetic sequence is 368.
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