AP sums with exclusions: An AP has 10 terms. The sum of all terms except the 1st is 99, and the sum of all terms except the 6th is 89. If t1 + t5 = 10, find the 3rd term.

Introduction / Context:
This problem combines AP sum identities with partial exclusions. Using S_10 − t_1 and S_10 − t_6 yields two equations. Together with t_1 + t_5 = 10, they allow solving for the first term and common difference, then t_3.

Given Data / Assumptions:

• S_10 − t_1 = 99.
• S_10 − t_6 = 89.
• t_1 + t_5 = 10.

Concept / Approach:
Let t_1 = a; t_k = a + (k − 1)d. Compute S_10 = 10/2[2a + 9d] = 5(2a + 9d). Use exclusions to form a linear system in a, d; then find t_3 = a + 2d.

Step-by-Step Solution:

Verification / Alternative check:
t_1 + t_5 = a + (a + 4d) = 2a + 4d = 2*1 + 8 = 10, as given. Consistent.

Why Other Options Are Wrong:

• 15, 8, 10: Do not align with the solved a = 1 and d = 2.

Common Pitfalls:
Dropping signs or mixing indices when excluding terms. Keep S_10 explicit and carefully subtract the appropriate term.

Final Answer: