The 4th term of an arithmetic progression is 15, 15th term is -29, find the 10th term?

The $$n^{th}$$ term of an A.P. = $$a + (n - 1) d$$, where 'a' is the first term , 'n' is the number of terms and 'd' is the common difference.

4th term, $$A_4 = a + (4 - 1) d = 15$$

=> $$a + 3d = 15$$ -----------------(i)

Similarly, 15th term, $$A_{15} = a + 14d = -29$$ ------------------(ii)

Subtracting equation (i) from (ii), we get :

=> $$(14d - 3d) = -29 - 15$$

=> $$d = \frac{-44}{11} = -4$$

Substituting it in equation (i), => $$a - 12 = 15$$

=> $$a = 15 + 12 = 27$$

$$\therefore$$ 10th term, $$A_{10} = a + (10 - 1)d$$

= $$27 + (9 \times -4) = 27 - 36 = -9$$

=> Ans - (D)