Question

iii) Find the middle term of sequence formed by all three digit number which leave a remainder 3 when divided by 4. Also find sum of all number on both sides of middle term.​

Answer 1

Answer:

Now, all three digit-numbers that leave a remainder 3 when divided by 4 are 103, 107, 111, 115 , ..., 999, which forms an AP. Here, a = 103 and d = 4. Thus, the middle term of the sequence is 551

Step-by-step explanation:

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Answer 2

Answer:

Middle term sequence = 551

the sum of all numbers on both sides of the middle term =

Before: 36400

After: 87024

All: 123975

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 the 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.

Middle term = (n+12)th term = 113th term = a + 112d = 103 + 112×4 = 551

We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]

Now, Sum = 112/2 × [2×103 + 111×4] = 36400

Sum of all terms before  middle term  = 36400

sum of all numbers=  225/2 × [2×103+224×4] = 123975

Now, sum of terms after  middle term = S225 − (S112+551) = 123975−(36400+551) = 87024