Question : Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 3 when divided by 5. Also find the sum of all numbers.
Doubt by Gurvinder
Solution :
List of three digit numbers which leave a remainder 3 when divided by 5 are :
103, 108, 113 . . . 998
a = 103
d = 5
an = 998
an = a+(n-1)d
998 = 103 + (n-1)5
998-103 = (n-1)5
895/5 = n-1
179=n-1
179+1=n
n = 180 (even)
Since, n is even.
Therefore, there are two middle terms i.e.
(n/2)th and (n/2+1)th
(180/2)th and (180/2 + 1)th
90th and 9th term
an = a+(n-1)d
a90 = 103+(90-1)5
= 103 + 89×5
= 103 + 445
= 548
a91 = 103+(91-1)5
= 103+90×5
= 103+450
= 553
Now Sum of all terms
Sn = {n[a+an]}/2
sn = {180[103+998]}/2
sn = {90×1101}
sn = 99090