find the middle term of the sequence formed by all three digits numbers which leave a remainder 3, when divided by 4.also find the sum of all numbers on both side of the middle terms separately
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A.P. = 103, 107, 111,........,999
a = 103 ; d = 107 - 103 = 4
l = a + (n - 1) d
999 = 103 + (n - 1)(4)
999 - 103 /4 = n - 1
224 + 1 = n
n = 225
so, the middle term is 113
a113 = a + 112 d
= 103 + 112(4)
= 551.
Thus, the answer is 551.
mark it as brainlist please
a = 103 ; d = 107 - 103 = 4
l = a + (n - 1) d
999 = 103 + (n - 1)(4)
999 - 103 /4 = n - 1
224 + 1 = n
n = 225
so, the middle term is 113
a113 = a + 112 d
= 103 + 112(4)
= 551.
Thus, the answer is 551.
mark it as brainlist please
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