Saved Bookmarks
| 1. |
Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately. |
|
Answer» The sequence is 103, 110, ......, 999 ∴ 999 = 103 + (n – 1) × 7 ⇒ n = 129 Therefore (129+1)/2 = 65th term is the middle term Middle term = 103 + (64) × 7 = 551 Sum of first 64 terms = 32 [206 + 63 × 7] = 20704 Sum of last 64 terms= 32 [1116 + 63 × 7] = 49824 |
|