Class 10: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{Write the sequence with }n\text{th term:}$
$\displaystyle \text{(i) }a_n=3+4n\qquad\text{(ii) }a_n=5+2n\qquad\text{(iii) }a_n=6-n\qquad\text{(iv) }a_n=9-5n$
$\displaystyle \text{Show that all of the above sequences form A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }a_n=3+4n$
$\displaystyle a_1=7,\quad a_2=11,\quad a_3=15,\quad a_4=19,\quad a_5=23$
$\displaystyle \therefore \text{The sequence is }7,11,15,19,23,\ldots$
$\displaystyle a_2-a_1=11-7=4,\quad a_3-a_2=15-11=4,\quad a_4-a_3=19-15=4$
$\displaystyle \therefore \text{The common difference is constant. Hence, the sequence forms an A.P.}$

$\displaystyle \text{(ii) Given, }a_n=5+2n$
$\displaystyle a_1=7,\quad a_2=9,\quad a_3=11,\quad a_4=13,\quad a_5=15$
$\displaystyle \therefore \text{The sequence is }7,9,11,13,15,\ldots$
$\displaystyle a_2-a_1=9-7=2,\quad a_3-a_2=11-9=2,\quad a_4-a_3=13-11=2$
$\displaystyle \therefore \text{The common difference is constant. Hence, the sequence forms an A.P.}$

$\displaystyle \text{(iii) Given, }a_n=6-n$
$\displaystyle a_1=5,\quad a_2=4,\quad a_3=3,\quad a_4=2,\quad a_5=1$
$\displaystyle \therefore \text{The sequence is }5,4,3,2,1,\ldots$
$\displaystyle a_2-a_1=4-5=-1,\quad a_3-a_2=3-4=-1,\quad a_4-a_3=2-3=-1$
$\displaystyle \therefore \text{The common difference is constant. Hence, the sequence forms an A.P.}$

$\displaystyle \text{(iv) Given, }a_n=9-5n$
$\displaystyle a_1=4,\quad a_2=-1,\quad a_3=-6,\quad a_4=-11,\quad a_5=-16$
$\displaystyle \therefore \text{The sequence is }4,-1,-6,-11,-16,\ldots$
$\displaystyle a_2-a_1=-1-4=-5,\quad a_3-a_2=-6-(-1)=-5,\quad a_4-a_3=-11-(-6)=-5$
$\displaystyle \therefore \text{The common difference is constant. Hence, the sequence forms an A.P.}$
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$\displaystyle \textbf{Question 2: }\text{The }n\text{th term of an A.P. is }6n+2.\text{ Find the common difference.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=6n+2$
$\displaystyle a_{n+1}=6(n+1)+2=6n+8$
$\displaystyle d=a_{n+1}-a_n=(6n+8)-(6n+2)=6$
$\displaystyle \therefore \text{The common difference is }6.$
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$\displaystyle \textbf{Question 3: }\text{Show that the sequence defined by }a_n=5n-7\text{ is an A.P.; find its}$
$\displaystyle \text{common difference.}\hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=5n-7$
$\displaystyle a_{n+1}=5(n+1)-7=5n-2$
$\displaystyle a_{n+1}-a_n=(5n-2)-(5n-7)=5$
$\displaystyle \text{Since }a_{n+1}-a_n\text{ is constant, the given sequence is an A.P.}$
$\displaystyle \therefore \text{The common difference is }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Show that the sequence defined by }a_n=3n^2-5\text{ is not an A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=3n^2-5$
$\displaystyle a_{n+1}=3(n+1)^2-5=3n^2+6n-2$
$\displaystyle a_{n+1}-a_n=(3n^2+6n-2)-(3n^2-5)=6n+3$
$\displaystyle \text{Since }a_{n+1}-a_n\text{ depends on }n,\text{ it is not constant.}$
$\displaystyle \therefore \text{The given sequence is not an A.P.}$
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$\displaystyle \textbf{Question 5: }\text{The general term of a sequence is given by }a_n=-4n+15.\text{ Is the sequence}$
$\displaystyle \text{an A.P.? If so, find its }15\text{th term and the common difference.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=-4n+15$
$\displaystyle a_{n+1}=-4(n+1)+15=-4n+11$
$\displaystyle a_{n+1}-a_n=(-4n+11)-(-4n+15)=-4$
$\displaystyle \text{Since }a_{n+1}-a_n\text{ is constant, the given sequence is an A.P.}$
$\displaystyle \therefore d=-4$
$\displaystyle a_{15}=-4(15)+15=-60+15=-45$
$\displaystyle \therefore \text{The }15\text{th term is }-45\text{ and the common difference is }-4.$
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$\displaystyle \textbf{Question 6: }\text{Justify whether it is true to say that the sequence having the following }n\text{th}$
$\displaystyle \text{term is an A.P.}$
$\displaystyle \text{(i) }a_n=2n-1\qquad\text{(ii) }a_n=3n^2+5\qquad\text{(iii) }a_n=1+n+n^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }a_n=2n-1$
$\displaystyle a_{n+1}=2(n+1)-1=2n+1$
$\displaystyle a_{n+1}-a_n=(2n+1)-(2n-1)=2$
$\displaystyle \therefore \text{The difference is constant. Hence, the sequence is an A.P. with }d=2.$

$\displaystyle \text{(ii) Given, }a_n=3n^2+5$
$\displaystyle a_{n+1}=3(n+1)^2+5=3n^2+6n+8$
$\displaystyle a_{n+1}-a_n=(3n^2+6n+8)-(3n^2+5)=6n+3$
$\displaystyle \therefore \text{The difference depends on }n.\text{ Hence, the sequence is not an A.P.}$

$\displaystyle \text{(iii) Given, }a_n=1+n+n^2$
$\displaystyle a_{n+1}=1+(n+1)+(n+1)^2=n^2+3n+3$
$\displaystyle a_{n+1}-a_n=(n^2+3n+3)-(n^2+n+1)=2n+2$
$\displaystyle \therefore \text{The difference depends on }n.\text{ Hence, the sequence is not an A.P.}$
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$\displaystyle \textbf{Question 7: }\text{The }n\text{th term of an A.P. cannot be }n^2+1.\text{ Justify your answer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a_n=n^2+1$
$\displaystyle a_{n+1}=(n+1)^2+1=n^2+2n+2$
$\displaystyle a_{n+1}-a_n=(n^2+2n+2)-(n^2+1)=2n+1$
$\displaystyle \text{Since }a_{n+1}-a_n\text{ depends on }n,\text{ it is not constant.}$
$\displaystyle \therefore n^2+1\text{ cannot be the }n\text{th term of an A.P.}$
$\displaystyle \\$