hey guys here is question for u......
how many terms of the AP :9, 17, 25 .....must be taken to give a sum of 636?
plz tell me answer with explanation...
Answers
Answered by
1
A.p= 9,17,25
first term (a)=9
common difference
(d)=17-9
d=8
Sum of n terms=n÷2[2a+(n-1)d
636=n÷2[2(9)+(n-1)8]
636=n÷2[18+8n-8]
636=n÷2[10+8n]
1272=10n+8n^2
8n^2+10n-1272=0
2[4n^2+5n-636]=0
4n^2+5n-636=0
4n^2+53n-48n-636=0
n(4n+53)-12(4n+53)=0
(n-12)(4n+53)=0
n-12=0
n=12
12 terms must be given to sum of 636.
I hope it is helped you
first term (a)=9
common difference
(d)=17-9
d=8
Sum of n terms=n÷2[2a+(n-1)d
636=n÷2[2(9)+(n-1)8]
636=n÷2[18+8n-8]
636=n÷2[10+8n]
1272=10n+8n^2
8n^2+10n-1272=0
2[4n^2+5n-636]=0
4n^2+5n-636=0
4n^2+53n-48n-636=0
n(4n+53)-12(4n+53)=0
(n-12)(4n+53)=0
n-12=0
n=12
12 terms must be given to sum of 636.
I hope it is helped you
aman3495:
ur welcome
Answered by
0
a = 9
d = 8
Sn = (n/2)(2a + (n-1)d)
636 = (n/2)(18 + (n-1) × 8)
1272 = n ( 18 + 8 n - 8)
1272 = 10 n + 8 n²
4 n² + 5 n - 636 = 0
(n-12) (4n-53) = 0
Therefore n = 12 or 53/4
Therefore number of terms = 12.
d = 8
Sn = (n/2)(2a + (n-1)d)
636 = (n/2)(18 + (n-1) × 8)
1272 = n ( 18 + 8 n - 8)
1272 = 10 n + 8 n²
4 n² + 5 n - 636 = 0
(n-12) (4n-53) = 0
Therefore n = 12 or 53/4
Therefore number of terms = 12.
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