Question 1: The sum of three terms of an A.P. is $21$ and the product of the first and the third terms exceeds the second term by $6$, find three terms.

Let the three terms of the A.P. be $a-d, a, a+d$

$\therefore (a-d) + a + ( a+d) = 21$

$\Rightarrow 3a = 21$

$\Rightarrow a = 7$

Also $(a-d)(a+d) - a = 6$

$\Rightarrow a^2 - d^2 - a = 6$

$\Rightarrow 49 - d^2-7 = 6$

$\Rightarrow d^2 = 36$

$\Rightarrow d = \pm 6$

When $a = 7, d = 6$,  the three terms are $1, 7, 13$

When $a = 7, d = -6$, the three terms are $13, 7, 1$

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Question 2: Three numbers are in A.P. If the sum of these numbers be $27$ and the product $548$, find the numbers.

Let the three terms of the A.P. be $a-d, a, a+d$

$\therefore (a-d) + a + ( a+d) = 27$

$\Rightarrow 3a = 27$

$\Rightarrow a = 9$

Also $(a-d) \cdot a \cdot (a+d) = 648$

$\Rightarrow a(a^2 - d^2) = 648$

$\Rightarrow 9(81-d^2) = 648$

$\Rightarrow (81-d^2) = 72$

$\Rightarrow d^2 = 9$

$\Rightarrow d = \pm 3$

When $a = 9, d = 3$,  the three terms are $6, 9, 12$

When $a = 9, d = -3$, the three terms are $12, 9, 6$

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Question 3: Find the four numbers in A.P., whose sum is $50$ and in which the greatest number is $4$ times the least.

Let the four numbers be $a-3d, a-d, a+d, a+3d$

$\therefore (a-3d) + ( a-d) + (a+d) + ( a+3d) = 50$

$\Rightarrow 4a = 50$

$\Rightarrow a =$ $\frac{25}{2}$

Also $a+3d = 4 ( a - 3d)$

$\Rightarrow a+3d = 4a - 12 d$

$\Rightarrow 15 d = 3a$

$\Rightarrow a = 5d$

$\Rightarrow d =$ $\frac{25}{2}$ $\times$ $\frac{1}{5}$ $=$ $\frac{5}{2}$

$\therefore$ the terms are $\Big( \frac{25}{2}- 3 \times \frac{5}{2} \Big), \Big( \frac{25}{2}- \frac{5}{2} \Big) , \Big( \frac{25}{2}+ \frac{5}{2} \Big) , \Big( \frac{25}{2}+ 3 \times \frac{5}{2} \Big)$ or $5, 10, 15, 20$

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Question 4: The sum of three numbers in A.P. is $12$ and the sum of their cubes is $288$. Find the numbers.

Let the three terms of the A.P. be $a-d, a, a+d$

$\therefore (a-d) + a + ( a+d) = 12$

$\Rightarrow 3a = 12$

$\Rightarrow a = 4$

Also $(a-d)^3 + a^3 + ( a+d)^3 = 288$

$\Rightarrow a^3 - d^3 -3a^2d + 3ad^2 + a^3 +a^3 +d^3 + 3a^2d+ 3ad^2 = 288$

$\Rightarrow 3a^3 + 6ad^2 = 288$

$\Rightarrow 3(4)^3 + 6(4) d^2 = 288$

$\Rightarrow 192 + 24d^2 = 288$

$\Rightarrow 24d^2 = 96$

$\Rightarrow d= \pm 2$

When $a=4, d = 2$,  the three terms are $2,4, 6$

When $a = 4, d = -2$, the three terms are $6, 4, 2$

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Question 5: If the sum of three numbers in A.P. is $24$ and their product $440$, find the numbers.

Let the three terms of the A.P. be $a-d, a, a+d$

$\therefore (a-d) + a + ( a+d) = 24$

$\Rightarrow 3a = 24$

$\Rightarrow a = 8$

Also, $a(a-d)(a+d) = 440$

$\Rightarrow a(a^2 - d^2) = 440$

$\Rightarrow 8(64-d^2) = 440$

$\Rightarrow d^2 = 9$

$\Rightarrow d = \pm 3$

When $a = 8, d = 3$,  the three terms are $5, 8, 11$

When $a = 8, d = -3$, the three terms are $11, 8, 5$

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Question 6: The angles of a quadrilateral are in A.P. whose common difference is $10^{\circ}$. Find the angles.

Let the angles be $A, (A+d), (A+2d), (A+3d)$
Given $d = 10^{\circ}$
Sum of the angles $A+ (A+d)+ (A+2d)+ (A+3d) = 360^{\circ}$
$\Rightarrow 4A + 60^{\circ} = 360^{\circ}$
$\Rightarrow A = 75^{\circ}$
Therefore the angles are $75^{\circ}, 85^{\circ}, 95^{\circ}, 105^{\circ}$.