Arithmetic Progression

Question

If the 3rd term of an AP is 7 and the 7th term is 23, then the 10th term is

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1. **Create Equations**:
   - $T_3 = a + 2d = 7$
   - $T_7 = a + 6d = 23$
2. **Solve for $d$**: Subtract the first from the second:
   - $(a + 6d) - (a + 2d) = 23 - 7$
   - $4d = 16 \implies d = 4$.
3. **Solve for $a$**:
   - $a + 2(4) = 7 \implies a = 7 - 8 = -1$.
4. **Find 10th Term**:
   - $T_{10} = a + 9d = -1 + 9(4) = -1 + 36 = 35$.

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