Question

Find the middle term of the sequence formed by all three-digit numbers which leave
a remainder 3. When divided by 4. Also find the sum of all numbers on both sides of
the middle term separately.
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Answer 1

Step-by-step explanation:

The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :

103 , 107 , 111 , 115 , .... 999

The above list is in AP with first term, a = 103 and common difference, d = 4

Let n be the number of terms in the AP.

Now, an = 999

103 + ( n - 1 ) 4 = 999

103 + 4n - 4 = 999

4n + 99 = 999

4n = 900

n = 225

Since, the number of terms is odd, so there will be only one middle term.

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