find the number of terms of A.P. 64, 60, 56...... so that their sum is 544
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HERE IS YOUR ANSWER
Sn = 544
a = 64
d = 60 - 64 = -4
n = ?
Sn = n/2 ( 2a + (n-1) d)
544 = n/2 ( 2 x 64) + (n-1) ( -4)
544 x 2 = n [ 128 + (n-1) (-4)
1088 = n [ 128 - 4n + 4 ]
1088 = n( 132 - 4n)
272 = n ( 33 - n) ...[taking 4 common ]
272 = 33n -n2
n2 - 33n + 272 = 0
n2 - 17n - 16n + 272 = 0
n ( n -17) - 16 ( n - 17) = 0
(n - 17) (n-16) = 0
n-17 =0
n=17
n -16 =0
n=16
value of n = 16, 17.
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HERE IS YOUR ANSWER
Sn = 544
a = 64
d = 60 - 64 = -4
n = ?
Sn = n/2 ( 2a + (n-1) d)
544 = n/2 ( 2 x 64) + (n-1) ( -4)
544 x 2 = n [ 128 + (n-1) (-4)
1088 = n [ 128 - 4n + 4 ]
1088 = n( 132 - 4n)
272 = n ( 33 - n) ...[taking 4 common ]
272 = 33n -n2
n2 - 33n + 272 = 0
n2 - 17n - 16n + 272 = 0
n ( n -17) - 16 ( n - 17) = 0
(n - 17) (n-16) = 0
n-17 =0
n=17
n -16 =0
n=16
value of n = 16, 17.
HOPE IT HELPS YOU
PLZ MARK AS BRAINLIEST IF IT HELPED
^_^
kumarsuresh237p9qz6n:
both factors are positive
thanks
Answered by
11
no of terms are 16 or 17
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