Question 1: The sum of three terms of an A.P. is $\displaystyle 21$ and the product of the first and the third terms exceeds the second term by $\displaystyle 6$, find three terms. $\displaystyle \text{Let the three terms of the A.P. be } a-d, a, a+d$ $\displaystyle \therefore (a-d) + a + ( a+d) = 21$ $\displaystyle \Rightarrow 3a = 21$ $\displaystyle \Rightarrow a = 7$ $\displaystyle \text{Also } (a-d)(a+d) - a = 6$ $\displaystyle \Rightarrow a^2 - d^2 - a = 6$ $\displaystyle \Rightarrow 49 - d^2-7 = 6$ $\displaystyle \Rightarrow d^2 = 36$ $\displaystyle \Rightarrow d = \pm 6$ $\displaystyle \text{When } a = 7, d = 6 , \text{the three terms are } 1, 7, 13$ $\displaystyle \text{When } a = 7, d = -6 , \text{the three terms are } 13, 7, 1$ $\displaystyle \\$

Question 2: Three numbers are in A.P. If the sum of these numbers be $\displaystyle 27$ and the product $\displaystyle 548$, find the numbers. $\displaystyle \text{Let the three terms of the A.P. be } a-d, a, a+d$ $\displaystyle \therefore (a-d) + a + ( a+d) = 27$ $\displaystyle \Rightarrow 3a = 27$ $\displaystyle \Rightarrow a = 9$ $\displaystyle \text{Also } (a-d) \cdot a \cdot (a+d) = 648$ $\displaystyle \Rightarrow a(a^2 - d^2) = 648$ $\displaystyle \Rightarrow 9(81-d^2) = 648$ $\displaystyle \Rightarrow (81-d^2) = 72$ $\displaystyle \Rightarrow d^2 = 9$ $\displaystyle \Rightarrow d = \pm 3$ $\displaystyle \text{When } a = 9, d = 3 , \text{the three terms are } 6, 9, 12$ $\displaystyle \text{When } a = 9, d = -3 , \text{the three terms are } 12, 9, 6$ $\displaystyle \\$

Question 3: Find the four numbers in A.P., whose sum is $\displaystyle 50$ and in which the greatest number is $\displaystyle 4$ times the least. $\displaystyle \text{Let the four numbers be } a-3d, a-d, a+d, a+3d$ $\displaystyle \therefore (a-3d) + ( a-d) + (a+d) + ( a+3d) = 50$ $\displaystyle \Rightarrow 4a = 50$ $\displaystyle \Rightarrow a = \frac{25}{2}$ $\displaystyle \text{Also } a+3d = 4 ( a - 3d)$ $\displaystyle \Rightarrow a+3d = 4a - 12 d$ $\displaystyle \Rightarrow 15 d = 3a$ $\displaystyle \Rightarrow a = 5d$ $\displaystyle \Rightarrow d = \frac{25}{2} \times \frac{1}{5} = \frac{5}{2}$ $\displaystyle \therefore$ the terms are $\displaystyle \Big( \frac{25}{2}- 3 \times \frac{5}{2} \Big), \Big( \frac{25}{2}- \frac{5}{2} \Big) , \Big( \frac{25}{2}+ \frac{5}{2} \Big) , \Big( \frac{25}{2}+ 3 \times \frac{5}{2} \Big) \text{ or } 5, 10, 15, 20$ $\displaystyle \\$

Question 4: The sum of three numbers in A.P. is $\displaystyle 12$ and the sum of their cubes is $\displaystyle 288$. Find the numbers. $\displaystyle \text{Let the three terms of the A.P. be } a-d, a, a+d$ $\displaystyle \therefore (a-d) + a + ( a+d) = 12$ $\displaystyle \Rightarrow 3a = 12$ $\displaystyle \Rightarrow a = 4$ $\displaystyle \text{Also } (a-d)^3 + a^3 + ( a+d)^3 = 288$ $\displaystyle \Rightarrow a^3 - d^3 -3a^2d + 3ad^2 + a^3 +a^3 +d^3 + 3a^2d+ 3ad^2 = 288$ $\displaystyle \Rightarrow 3a^3 + 6ad^2 = 288$ $\displaystyle \Rightarrow 3(4)^3 + 6(4) d^2 = 288$ $\displaystyle \Rightarrow 192 + 24d^2 = 288$ $\displaystyle \Rightarrow 24d^2 = 96$ $\displaystyle \Rightarrow d= \pm 2$ $\displaystyle \text{When } a=4, d = 2 , \text{the three terms are } 2,4, 6$ $\displaystyle \text{When } a = 4, d = -2 , \text{the three terms are } 6, 4, 2$ $\displaystyle \\$

Question 5: If the sum of three numbers in A.P. is $\displaystyle 24$ and their product $\displaystyle 440$, find the numbers. $\displaystyle \text{Let the three terms of the A.P. be } a-d, a, a+d$ $\displaystyle \therefore (a-d) + a + ( a+d) = 24$ $\displaystyle \Rightarrow 3a = 24$ $\displaystyle \Rightarrow a = 8$ $\displaystyle \text{Also, } a(a-d)(a+d) = 440$ $\displaystyle \Rightarrow a(a^2 - d^2) = 440$ $\displaystyle \Rightarrow 8(64-d^2) = 440$ $\displaystyle \Rightarrow d^2 = 9$ $\displaystyle \Rightarrow d = \pm 3$ $\displaystyle \text{When } a = 8, d = 3 , \text{the three terms are } 5, 8, 11$ $\displaystyle \text{When } a = 8, d = -3 , \text{the three terms are } 11, 8, 5$ $\displaystyle \\$

Question 6: The angles of a quadrilateral are in A.P. whose common difference is $\displaystyle 10^{\circ}$. Find the angles. $\displaystyle \text{Let the angles be } A, (A+d), (A+2d), (A+3d)$ $\displaystyle \text{Given } d = 10^{\circ}$ $\displaystyle \text{Sum of the angles } A+ (A+d)+ (A+2d)+ (A+3d) = 360^{\circ}$ $\displaystyle \Rightarrow 4A + 60^{\circ} = 360^{\circ}$ $\displaystyle \Rightarrow A = 75^{\circ}$ $\displaystyle \text{Therefore the angles are } 75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$.