1. **State the problem:** We are given an arithmetic progression (A.P.) with first term $A_1 = -17$, fifth term $A_5 = 2$, and the sum of all terms is 40. We need to find the total number of terms $n$ in the A.P. 2. **Recall formulas:** - The $n$th term of an A.P. is given by $$A_n = A_1 + (n-1)d$$ where $d$ is the common difference. - The sum of $n$ terms of an A.P. is $$S_n = \frac{n}{2}(A_1 + A_n)$$ 3. **Find the common difference $d$:** Given $A_5 = 2$, use the formula: $$2 = -17 + (5-1)d = -17 + 4d$$ Solve for $d$: $$4d = 2 + 17 = 19$$ $$d = \frac{19}{4} = 4.75$$ 4. **Express $A_n$ in terms of $n$:** $$A_n = -17 + (n-1) \times 4.75 = -17 + 4.75n - 4.75 = 4.75n - 21.75$$ 5. **Use the sum formula to find $n$:** Given $S_n = 40$, substitute: $$40 = \frac{n}{2}(-17 + A_n) = \frac{n}{2}(-17 + 4.75n - 21.75) = \frac{n}{2}(4.75n - 38.75)$$ Multiply both sides by 2: $$80 = n(4.75n - 38.75) = 4.75n^2 - 38.75n$$ Rewrite as quadratic equation: $$4.75n^2 - 38.75n - 80 = 0$$ 6. **Solve the quadratic equation:** Divide entire equation by 4.75 to simplify: $$n^2 - 8.1579n - 16.8421 = 0$$ Use quadratic formula: $$n = \frac{8.1579 \pm \sqrt{8.1579^2 + 4 \times 16.8421}}{2}$$ Calculate discriminant: $$8.1579^2 = 66.56, \quad 4 \times 16.8421 = 67.37$$ $$\sqrt{66.56 + 67.37} = \sqrt{133.93} \approx 11.57$$ So, $$n = \frac{8.1579 \pm 11.57}{2}$$ Two possible values: - $$n = \frac{8.1579 + 11.57}{2} = \frac{19.7279}{2} = 9.86$$ (not an integer) - $$n = \frac{8.1579 - 11.57}{2} = \frac{-3.4121}{2} = -1.71$$ (not valid) Since $n$ must be a positive integer, check if rounding $n=10$ works. 7. **Check sum for $n=10$:** Calculate $A_{10}$: $$A_{10} = 4.75 \times 10 - 21.75 = 47.5 - 21.75 = 25.75$$ Sum: $$S_{10} = \frac{10}{2}(-17 + 25.75) = 5 \times 8.75 = 43.75$$ (not 40) Try $n=9$: $$A_9 = 4.75 \times 9 - 21.75 = 42.75 - 21.75 = 21$$ $$S_9 = \frac{9}{2}(-17 + 21) = 4.5 \times 4 = 18$$ (too small) Try $n=8$: $$A_8 = 4.75 \times 8 - 21.75 = 38 - 21.75 = 16.25$$ $$S_8 = \frac{8}{2}(-17 + 16.25) = 4 \times (-0.75) = -3$$ (too small) Try $n=11$: $$A_{11} = 4.75 \times 11 - 21.75 = 52.25 - 21.75 = 30.5$$ $$S_{11} = \frac{11}{2}(-17 + 30.5) = 5.5 \times 13.5 = 74.25$$ (too large) 8. **Conclusion:** The exact $n$ is not an integer, but closest integer $n=10$ gives sum near 40. The problem likely expects $n=10$ as the number of terms. **Final answer:** There are approximately **10 terms** in the A.P.