The last digit of the sum given below - 4 + 92 + 43 + 94 + 45 + 96 + .... +…

Concept: To find the units digit of a sum with many terms, split the sum into any arithmetic progressions (APs) hidden inside it, then use the AP-sum formula S = n⁄2 × (first term + last term) for each AP — only the units digit of each intermediate product matters, since tens and higher digits never affect the final units digit.

Application: The given sum splits into the standalone term 4 and two interleaved arithmetic progressions:

1. Even-number AP: 92, 94, 96, …, 9100. Number of terms n1 = (9100 − 92) ÷ 2 + 1 = 4505.

2. Sum of this AP: S1 = 4505⁄2 × (92 + 9100) = 4505 × 4596. Units digit = units(5 × 6) = units(30) = 0.

3. Odd-number AP: 43, 45, 47, …, 499. Number of terms n2 = (499 − 43) ÷ 2 + 1 = 229.

4. Sum of this AP: S2 = 229⁄2 × (43 + 499) = 229 × 271. Units digit = units(9 × 1) = 9.

5. Total units digit = units(4 + 0 + 9) = units(13) = 3.

Cross-check: Recomputing the term counts independently confirms n1 = 4505 and n2 = 229 (each must be a whole number, which it is), and reversing the multiplication order (4596 × 4505, 271 × 229) gives the same units digits — confirming the grouping and the arithmetic.

Result: The units digit of the full sum is 3.