Question

Find the middle term of ap formed by all three digit no. Which leaves the remaider 5 when divided by 7 also find the sum of all no. On both sides of the middle term separated

Answer 1

List of three digit numbers when divided by 7 lives remainder 5 are 

110,117,124,131...887

a = 110 

d = 7 

let n be the number of terms of AP

a(n) = 887

110 + (n-1)7 = 887

110 + 7n - 7 = 887

7n + 103 = 887

7n = 784 

n = 784/7 

n = 112

Answer 2

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