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

Answer 1

Hello mate !!
Here's ur answer

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.


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


Hope it helpful

Answer 2

107,111...999.
a=107,d=4,tn=999
tn=a+(n-1)d
999=107+(n-1)4
892=(n-1)4
n-1=223
n=224.
sn=n/2(2a+(n-1)d)
sn=112(214+892)
sn=112×1086
sn=121632.
middle term = sn+1÷2
=121633÷2
=60816.5