Class 10: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{For the following arithmetic progressions write the first term }$
$\displaystyle a\text{ and the common} \ \text{difference }d:$
$\displaystyle \text{(i) }-5,-1,3,7,\ldots\ \qquad\text{(ii) }\frac15,\frac35,\frac55,\frac75,\ldots$
$\displaystyle \text{(iii) }0.3,0.55,0.80,1.05,\ldots\qquad\text{(iv) }-1.1,-3.1,-5.1,-7.1,\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }a=-5$
$\displaystyle d=-1-(-5)=4$
$\displaystyle \therefore a=-5,\quad d=4.$

$\displaystyle \text{(ii) }a=\frac15$
$\displaystyle d=\frac35-\frac15=\frac25$
$\displaystyle \therefore a=\frac15,\quad d=\frac25.$

$\displaystyle \text{(iii) }a=0.3$
$\displaystyle d=0.55-0.3=0.25$
$\displaystyle \therefore a=0.3,\quad d=0.25.$

$\displaystyle \text{(iv) }a=-1.1$
$\displaystyle d=-3.1-(-1.1)=-2$
$\displaystyle \therefore a=-1.1,\quad d=-2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Write the arithmetic progression when first term } a\text{ and common}$
$\displaystyle \text{difference } d \ \text{are as follows:}$
$\displaystyle \text{(i) }a=4,\ d=-3\ \qquad\text{(ii) }a=-1,\ d=\frac12\$
$\displaystyle \text{(iii) }a=-1.5,\ d=-0.5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }a=4,\quad d=-3$
$\displaystyle \text{A.P.}=a,\ a+d,\ a+2d,\ a+3d,\ldots$
$\displaystyle =4,\ 4-3,\ 4-6,\ 4-9,\ldots$
$\displaystyle \therefore \text{The A.P. is }4,1,-2,-5,\ldots$

$\displaystyle \text{(ii) Given, }a=-1,\quad d=\frac12$
$\displaystyle \text{A.P.}=a,\ a+d,\ a+2d,\ a+3d,\ldots$
$\displaystyle =-1,\ -1+\frac12,\ -1+1,\ -1+\frac32,\ldots$
$\displaystyle \therefore \text{The A.P. is }-1,-\frac12,0,\frac12,\ldots$

$\displaystyle \text{(iii) Given, }a=-1.5,\quad d=-0.5$
$\displaystyle \text{A.P.}=a,\ a+d,\ a+2d,\ a+3d,\ldots$
$\displaystyle =-1.5,\ -1.5-0.5,\ -1.5-1,\ -1.5-1.5,\ldots$
$\displaystyle \therefore \text{The A.P. is }-1.5,-2,-2.5,-3,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{In which of the following situations, the sequence of numbers formed} \\ \text{will form an A.P.?}$
$\displaystyle \text{(i) The cost of digging a well for the first metre is Rs. 150 and rises by Rs. 20 for each}$
$\displaystyle \text{succeeding metre.}$
$\displaystyle \text{(ii) The amount of air present in the cylinder when a vacuum pump removes each time }\frac14$
$\displaystyle \text{of the air remaining in the cylinder.}$
$\displaystyle \text{(iii) Divya deposited Rs. 1000 at compound interest at the rate of }10\%\text{ per annum. The}$
$\displaystyle \text{amount at the end of first year, second year, third year, and so on.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The costs of digging successive metres are }150,170,190,210,\ldots$
$\displaystyle 170-150=20,\quad190-170=20,\quad210-190=20$
$\displaystyle \therefore \text{The common difference is constant. Hence, the sequence forms an A.P.}$

$\displaystyle \text{(ii) Let the initial amount of air in the cylinder be }A.$
$\displaystyle \text{Since }\frac14\text{ of the remaining air is removed each time, }\frac34\text{ of it remains.}$
$\displaystyle \text{The amounts of air are }A,\frac34A,\left(\frac34\right)^2A,\left(\frac34\right)^3A,\ldots$
$\displaystyle \frac34A-A=-\frac14A,\quad\frac9{16}A-\frac34A=-\frac3{16}A$
$\displaystyle \therefore \text{The differences are not constant. Hence, the sequence does not form an A.P.}$

$\displaystyle \text{(iii) Amount after the first year}=1000\left(1+\frac{10}{100}\right)=1100$
$\displaystyle \text{Amount after the second year}=1100(1.1)=1210$
$\displaystyle \text{Amount after the third year}=1210(1.1)=1331$
$\displaystyle \text{Thus, the sequence is }1100,1210,1331,\ldots$
$\displaystyle 1210-1100=110,\quad1331-1210=121$
$\displaystyle \therefore \text{The differences are not constant. Hence, the sequence does not form an A.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the common difference and write the next four terms of each of}$
$\displaystyle \text{the following arithmetic progressions:}$
$\displaystyle \text{(i) }1,-2,-5,-8,\ldots\qquad\text{(ii) }0,-3,-6,-9,\ldots$
$\displaystyle \text{(iii) }-1,\frac14,\frac32,\ldots\qquad\text{(iv) }-1,-\frac56,-\frac23,\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }d=-2-1=-3$
$\displaystyle \text{Next four terms are }-8-3=-11,\ -11-3=-14,\ -14-3=-17,\ -17-3=-20.$
$\displaystyle \therefore d=-3;\quad\text{next four terms are }-11,-14,-17,-20.$

$\displaystyle \text{(ii) }d=-3-0=-3$
$\displaystyle \text{Next four terms are }-9-3=-12,\ -12-3=-15,\ -15-3=-18,\ -18-3=-21.$
$\displaystyle \therefore d=-3;\quad\text{next four terms are }-12,-15,-18,-21.$

$\displaystyle \text{(iii) }d=\frac14-(-1)=\frac54$
$\displaystyle \text{Also, }\frac32-\frac14=\frac54$
$\displaystyle \text{Next four terms are }\frac32+\frac54=\frac{11}{4},\ \frac{11}{4}+\frac54=4,\ 4+\frac54=\frac{21}{4},\ \frac{21}{4}+\frac54=\frac{13}{2}.$
$\displaystyle \therefore d=\frac54;\quad\text{next four terms are }\frac{11}{4},4,\frac{21}{4},\frac{13}{2}.$

$\displaystyle \text{(iv) }d=-\frac56-(-1)=\frac16$
$\displaystyle \text{Also, }-\frac23-\left(-\frac56\right)=\frac16$
$\displaystyle \text{Next four terms are }-\frac23+\frac16=-\frac12,\ -\frac12+\frac16=-\frac13,\ -\frac13+\frac16=-\frac16,\ -\frac16+\frac16=0.$
$\displaystyle \therefore d=\frac16;\quad\text{next four terms are }-\frac12,-\frac13,-\frac16,0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find out which of the following sequences are arithmetic progressions.}$
$\displaystyle \text{For those which are arithmetic progressions, find out the common difference.}$
$\displaystyle \text{(i) }3,3,3,3,\ldots\qquad\text{(ii) }p,p+90,p+180,p+270,\ldots,\text{ where }p=(999)^{999}$
$\displaystyle \text{(iii) }1.0,1.7,2.4,3.1,\ldots\qquad\text{(iv) }-225,-425,-625,-825,\ldots$
$\displaystyle \text{(v) }10,10+2^5,10+2^6,10+2^7,\ldots\qquad\text{(vi) }1^2,3^2,5^2,7^2,\ldots$
$\displaystyle \text{(vii) }a+b,(a+1)+b,(a+1)+(b+1),(a+2)+(b+1),(a+2)+(b+2),\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }3-3=0,\quad3-3=0,\quad3-3=0$
$\displaystyle \therefore \text{The sequence is an A.P. with common difference }d=0.$

$\displaystyle \text{(ii) }(p+90)-p=90$
$\displaystyle (p+180)-(p+90)=90,\quad(p+270)-(p+180)=90$
$\displaystyle \therefore \text{The sequence is an A.P. with common difference }d=90.$

$\displaystyle \text{(iii) }1.7-1.0=0.7,\quad2.4-1.7=0.7,\quad3.1-2.4=0.7$
$\displaystyle \therefore \text{The sequence is an A.P. with common difference }d=0.7.$

$\displaystyle \text{(iv) }-425-(-225)=-200,\quad-625-(-425)=-200$
$\displaystyle -825-(-625)=-200$
$\displaystyle \therefore \text{The sequence is an A.P. with common difference }d=-200.$

$\displaystyle \text{(v) The sequence is }10,42,74,138,\ldots$
$\displaystyle 42-10=32,\quad74-42=32,\quad138-74=64$
$\displaystyle \therefore \text{The differences are not constant. Hence, the sequence is not an A.P.}$

$\displaystyle \text{(vi) The sequence is }1,9,25,49,\ldots$
$\displaystyle 9-1=8,\quad25-9=16,\quad49-25=24$
$\displaystyle \therefore \text{The differences are not constant. Hence, the sequence is not an A.P.}$

$\displaystyle \text{(vii) Simplifying the terms, we get}$
$\displaystyle a+b,\ a+b+1,\ a+b+2,\ a+b+3,\ a+b+4,\ldots$
$\displaystyle (a+b+1)-(a+b)=1,\quad(a+b+2)-(a+b+1)=1$
$\displaystyle (a+b+3)-(a+b+2)=1$
$\displaystyle \therefore \text{The sequence is an A.P. with common difference }d=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Prove that no matter what the real numbers }a\text{ and }b\text{ are, the}$
$\displaystyle \text{sequence with }n\text{th}\text{term }a+nb\text{ is always an A.P. What is the common difference?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=a+nb$
$\displaystyle a_{n+1}=a+(n+1)b=a+nb+b$
$\displaystyle a_{n+1}-a_n=(a+nb+b)-(a+nb)=b$
$\displaystyle \text{Since }a_{n+1}-a_n=b\text{ is constant, the given sequence is always an A.P.}$
$\displaystyle \therefore \text{The common difference is }b.$
$\displaystyle \\$