Class 11: Arithmetic Progressions

$\displaystyle \textbf{Question 1: } \text{The sum of three terms of an A.P. is }21\text{ and the product of the first and}$
$\displaystyle \text{third terms exceeds the second term by }6.\text{ Find the three terms.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three terms of the A.P. be }a-d,\;a,\;a+d.$
$\displaystyle (a-d)+a+(a+d)=21.$
$\displaystyle 3a=21.$
$\displaystyle \therefore a=7.$
$\displaystyle \text{Also, }(a-d)(a+d)-a=6.$
$\displaystyle a^2-d^2-a=6.$
$\displaystyle 49-d^2-7=6.$
$\displaystyle d^2=36.$
$\displaystyle \therefore d=\pm6.$
$\displaystyle \text{When }d=6,\text{ the three terms are }1,7,13.$
$\displaystyle \text{When }d=-6,\text{ the three terms are }13,7,1.$
$\displaystyle \therefore \text{The three terms are }1,7,13\text{ or }13,7,1.$
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$\displaystyle \textbf{Question 2: } \text{Three numbers are in A.P. If their sum is }27\text{ and their product is }648,$
$\displaystyle \text{find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three terms of the A.P. be }a-d,\;a,\;a+d.$
$\displaystyle (a-d)+a+(a+d)=27.$
$\displaystyle 3a=27.$
$\displaystyle \therefore a=9.$
$\displaystyle \text{Also, }(a-d)\cdot a\cdot(a+d)=648.$
$\displaystyle a(a^2-d^2)=648.$
$\displaystyle 9(81-d^2)=648.$
$\displaystyle 81-d^2=72.$
$\displaystyle d^2=9.$
$\displaystyle \therefore d=\pm3.$
$\displaystyle \text{When }d=3,\text{ the three numbers are }6,9,12.$
$\displaystyle \text{When }d=-3,\text{ the three numbers are }12,9,6.$
$\displaystyle \therefore \text{The three numbers are }6,9,12\text{ in either order.}$
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$\displaystyle \textbf{Question 3: } \text{Find four numbers in A.P. whose sum is }50\text{ and in which the greatest}$
$\displaystyle \text{number is four times the least number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the four numbers be }a-3d,\;a-d,\;a+d,\;a+3d.$
$\displaystyle (a-3d)+(a-d)+(a+d)+(a+3d)=50.$
$\displaystyle 4a=50.$
$\displaystyle \therefore a=\frac{25}{2}.$
$\displaystyle \text{Also, the greatest number is four times the least number.}$
$\displaystyle a+3d=4(a-3d).$
$\displaystyle a+3d=4a-12d.$
$\displaystyle 15d=3a.$
$\displaystyle a=5d.$
$\displaystyle \therefore d=\frac{a}{5}=\frac{25}{2}\times\frac{1}{5}=\frac{5}{2}.$
$\displaystyle a-3d=\frac{25}{2}-3\left(\frac{5}{2}\right)=5.$
$\displaystyle a-d=\frac{25}{2}-\frac{5}{2}=10.$
$\displaystyle a+d=\frac{25}{2}+\frac{5}{2}=15.$
$\displaystyle a+3d=\frac{25}{2}+3\left(\frac{5}{2}\right)=20.$
$\displaystyle \therefore \text{The four numbers are }5,10,15\text{ and }20.$
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$\displaystyle \textbf{Question 4: } \text{The sum of three numbers in A.P. is }12\text{ and the sum of their cubes}$
$\displaystyle \text{is }288.\text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three terms of the A.P. be }a-d,\;a,\;a+d.$
$\displaystyle (a-d)+a+(a+d)=12.$
$\displaystyle 3a=12.$
$\displaystyle \therefore a=4.$
$\displaystyle \text{Also, }(a-d)^3+a^3+(a+d)^3=288.$
$\displaystyle (a-d)^3+(a+d)^3=2a^3+6ad^2.$
$\displaystyle \therefore 3a^3+6ad^2=288.$
$\displaystyle 3(4)^3+6(4)d^2=288.$
$\displaystyle 192+24d^2=288.$
$\displaystyle 24d^2=96.$
$\displaystyle d^2=4.$
$\displaystyle \therefore d=\pm2.$
$\displaystyle \text{When }d=2,\text{ the three numbers are }2,4,6.$
$\displaystyle \text{When }d=-2,\text{ the three numbers are }6,4,2.$
$\displaystyle \therefore \text{The three numbers are }2,4,6\text{ in either order.}$
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$\displaystyle \textbf{Question 5: } \text{If the sum of three numbers in A.P. is }24\text{ and their product is }440,$
$\displaystyle \text{find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three terms of the A.P. be }a-d,\;a,\;a+d.$
$\displaystyle (a-d)+a+(a+d)=24.$
$\displaystyle 3a=24.$
$\displaystyle \therefore a=8.$
$\displaystyle \text{Also, }a(a-d)(a+d)=440.$
$\displaystyle a(a^2-d^2)=440.$
$\displaystyle 8(64-d^2)=440.$
$\displaystyle 64-d^2=55.$
$\displaystyle d^2=9.$
$\displaystyle \therefore d=\pm3.$
$\displaystyle \text{When }d=3,\text{ the three numbers are }5,8,11.$
$\displaystyle \text{When }d=-3,\text{ the three numbers are }11,8,5.$
$\displaystyle \therefore \text{The three numbers are }5,8,11\text{ in either order.}$
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$\displaystyle \textbf{Question 6: } \text{The angles of a quadrilateral are in A.P. with common difference }10^\circ.$
$\displaystyle \text{Find the angles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the angles be }A,\;(A+d),\;(A+2d)\text{ and }(A+3d).$
$\displaystyle \text{Given, }d=10^\circ.$
$\displaystyle \text{The sum of the angles of a quadrilateral is }360^\circ.$
$\displaystyle A+(A+d)+(A+2d)+(A+3d)=360^\circ.$
$\displaystyle 4A+6d=360^\circ.$
$\displaystyle 4A+60^\circ=360^\circ.$
$\displaystyle 4A=300^\circ.$
$\displaystyle \therefore A=75^\circ.$
$\displaystyle \therefore \text{The angles are }75^\circ,\;85^\circ,\;95^\circ\text{ and }105^\circ.$
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