\displaystyle \textbf{Question 1: } \text{If the }n^{\text{th}}\text{ term }a_n\text{ of a sequence is given by}
\displaystyle a_n=n^2-n+1,\text{ write down the first five terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=n^2-n+1.
\displaystyle \text{The first five terms are }a_1,a_2,a_3,a_4\text{ and }a_5.
\displaystyle a_1=(1)^2-1+1=1.
\displaystyle a_2=(2)^2-2+1=3.
\displaystyle a_3=(3)^2-3+1=7.
\displaystyle a_4=(4)^2-4+1=13.
\displaystyle a_5=(5)^2-5+1=21.
\displaystyle \therefore \text{The first five terms are }1,\,3,\,7,\,13\text{ and }21.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{A sequence is defined by }a_n=n^3-6n^2+11n-6,\;n\in N.
\displaystyle \text{Show that the first three terms of the sequence are zero and all other terms are positive.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=n^3-6n^2+11n-6,\;n\in N.
\displaystyle a_1=(1)^3-6(1)^2+11(1)-6=0.
\displaystyle a_2=(2)^3-6(2)^2+11(2)-6=0.
\displaystyle a_3=(3)^3-6(3)^2+11(3)-6=0.
\displaystyle \therefore \text{The first three terms of the sequence are zero.}
\displaystyle \text{Now, }a_n=n^3-6n^2+11n-6=(n-1)(n-2)(n-3).
\displaystyle \text{For }n\ge4,\;(n-1)>0,\;(n-2)>0\text{ and }(n-3)>0.
\displaystyle \therefore a_n=(n-1)(n-2)(n-3)>0\text{ for all }n\ge4.
\displaystyle \therefore \text{All terms after the first three are positive.}
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Find the first four terms of the sequence defined by }a_1=3\text{ and}
\displaystyle a_n=3a_{n-1}+2\text{ for all }n>1.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_1=3\text{ and }a_n=3a_{n-1}+2.
\displaystyle a_2=3a_{2-1}+2=3a_1+2=3(3)+2=11.
\displaystyle a_3=3a_{3-1}+2=3a_2+2=3(11)+2=35.
\displaystyle a_4=3a_{4-1}+2=3a_3+2=3(35)+2=107.
\displaystyle \therefore \text{The first four terms are }3,\,11,\,35\text{ and }107.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Write the first five terms of each of the following sequences:}
\displaystyle \text{(i) }a_1=1,\quad a_n=a_{n-1}+2,\quad n>1
\displaystyle \text{(ii) }a_1=a_2=1,\quad a_n=a_{n-1}+a_{n-2},\quad n>2
\displaystyle \text{(iii) }a_1=a_2=2,\quad a_n=a_{n-1}-1,\quad n>2
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given, }a_1=1\text{ and }a_n=a_{n-1}+2.
\displaystyle a_1=1.
\displaystyle a_2=a_{2-1}+2=a_1+2=1+2=3.
\displaystyle a_3=a_{3-1}+2=a_2+2=3+2=5.
\displaystyle a_4=a_{4-1}+2=a_3+2=5+2=7.
\displaystyle a_5=a_{5-1}+2=a_4+2=7+2=9.
\displaystyle \therefore \text{The first five terms are }1,\,3,\,5,\,7\text{ and }9.

\displaystyle \text{(ii) Given, }a_1=a_2=1\text{ and }a_n=a_{n-1}+a_{n-2}.
\displaystyle a_1=1.
\displaystyle a_2=1.
\displaystyle a_3=a_{3-1}+a_{3-2}=a_2+a_1=1+1=2.
\displaystyle a_4=a_{4-1}+a_{4-2}=a_3+a_2=2+1=3.
\displaystyle a_5=a_{5-1}+a_{5-2}=a_4+a_3=3+2=5.
\displaystyle \therefore \text{The first five terms are }1,\,1,\,2,\,3\text{ and }5.

\displaystyle \text{(iii) Given, }a_1=a_2=2\text{ and }a_n=a_{n-1}-1.
\displaystyle a_1=2.
\displaystyle a_2=2.
\displaystyle a_3=a_{3-1}-1=a_2-1=2-1=1.
\displaystyle a_4=a_{4-1}-1=a_3-1=1-1=0.
\displaystyle a_5=a_{5-1}-1=a_4-1=0-1=-1.
\displaystyle \therefore \text{The first five terms are }2,\,2,\,1,\,0\text{ and }-1.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{The Fibonacci sequence is defined by }a_1=a_2=1,\;a_n=a_{n-1}+a_{n-2}
\displaystyle \text{for }n>2.\text{ Find }\frac{a_{n+1}}{a_n}\text{ for }n=1,2,3,4,5.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_1=1,\;a_2=1.
\displaystyle a_3=a_{3-1}+a_{3-2}=a_2+a_1=1+1=2.
\displaystyle a_4=a_{4-1}+a_{4-2}=a_3+a_2=2+1=3.
\displaystyle a_5=a_{5-1}+a_{5-2}=a_4+a_3=3+2=5.
\displaystyle a_6=a_{6-1}+a_{6-2}=a_5+a_4=5+3=8.
\displaystyle \text{Hence,}
\displaystyle \text{For }n=1,\;\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{1}{1}=1.
\displaystyle \text{For }n=2,\;\frac{a_{n+1}}{a_n}=\frac{a_3}{a_2}=\frac{2}{1}=2.
\displaystyle \text{For }n=3,\;\frac{a_{n+1}}{a_n}=\frac{a_4}{a_3}=\frac{3}{2}.
\displaystyle \text{For }n=4,\;\frac{a_{n+1}}{a_n}=\frac{a_5}{a_4}=\frac{5}{3}.
\displaystyle \text{For }n=5,\;\frac{a_{n+1}}{a_n}=\frac{a_6}{a_5}=\frac{8}{5}.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Show that each of the following sequences is in A.P. Also find the common}
\displaystyle \text{difference and write three more terms in each case.}
\displaystyle \text{(i) }3,-1,-5,-9,\ldots
\displaystyle \text{(ii) }-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},\ldots
\displaystyle \text{(iii) }\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\ldots
\displaystyle \text{(iv) }9,7,5,3,\ldots
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given sequence: }3,-1,-5,-9,\ldots
\displaystyle a_1=3,\quad a_2=-1,\quad a_3=-5,\quad a_4=-9.
\displaystyle a_2-a_1=-1-3=-4.
\displaystyle a_3-a_2=-5-(-1)=-4.
\displaystyle a_4-a_3=-9-(-5)=-4.
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \therefore d=-4.
\displaystyle a_5=a_4+d=-9+(-4)=-13.
\displaystyle a_6=a_5+d=-13+(-4)=-17.
\displaystyle a_7=a_6+d=-17+(-4)=-21.
\displaystyle \therefore \text{The next three terms are }-13,-17\text{ and }-21.

\displaystyle \text{(ii) Given sequence: }-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},\ldots
\displaystyle a_1=-1,\quad a_2=\frac{1}{4},\quad a_3=\frac{3}{2},\quad a_4=\frac{11}{4}.
\displaystyle a_2-a_1=\frac{1}{4}-(-1)=\frac{5}{4}.
\displaystyle a_3-a_2=\frac{3}{2}-\frac{1}{4}=\frac{6-1}{4}=\frac{5}{4}.
\displaystyle a_4-a_3=\frac{11}{4}-\frac{3}{2}=\frac{11-6}{4}=\frac{5}{4}.
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \therefore d=\frac{5}{4}.
\displaystyle a_5=a_4+d=\frac{11}{4}+\frac{5}{4}=\frac{16}{4}=4.
\displaystyle a_6=a_5+d=4+\frac{5}{4}=\frac{21}{4}.
\displaystyle a_7=a_6+d=\frac{21}{4}+\frac{5}{4}=\frac{26}{4}=\frac{13}{2}.
\displaystyle \therefore \text{The next three terms are }4,\frac{21}{4}\text{ and }\frac{13}{2}.

\displaystyle \text{(iii) Given sequence: }\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},\ldots
\displaystyle a_1=\sqrt{2},\quad a_2=3\sqrt{2},\quad a_3=5\sqrt{2},\quad a_4=7\sqrt{2}.
\displaystyle a_2-a_1=3\sqrt{2}-\sqrt{2}=2\sqrt{2}.
\displaystyle a_3-a_2=5\sqrt{2}-3\sqrt{2}=2\sqrt{2}.
\displaystyle a_4-a_3=7\sqrt{2}-5\sqrt{2}=2\sqrt{2}.
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \therefore d=2\sqrt{2}.
\displaystyle a_5=a_4+d=7\sqrt{2}+2\sqrt{2}=9\sqrt{2}.
\displaystyle a_6=a_5+d=9\sqrt{2}+2\sqrt{2}=11\sqrt{2}.
\displaystyle a_7=a_6+d=11\sqrt{2}+2\sqrt{2}=13\sqrt{2}.
\displaystyle \therefore \text{The next three terms are }9\sqrt{2},11\sqrt{2}\text{ and }13\sqrt{2}.

\displaystyle \text{(iv) Given sequence: }9,7,5,3,\ldots
\displaystyle a_1=9,\quad a_2=7,\quad a_3=5,\quad a_4=3.
\displaystyle a_2-a_1=7-9=-2.
\displaystyle a_3-a_2=5-7=-2.
\displaystyle a_4-a_3=3-5=-2.
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \therefore d=-2.
\displaystyle a_5=a_4+d=3+(-2)=1.
\displaystyle a_6=a_5+d=1+(-2)=-1.
\displaystyle a_7=a_6+d=-1+(-2)=-3.
\displaystyle \therefore \text{The next three terms are }1,-1\text{ and }-3.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{The }n^{\text{th}}\text{ term of a sequence is given by }a_n=2n+7.
\displaystyle \text{Show that it is an A.P. Also, find its }7^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=2n+7.
\displaystyle a_1=2(1)+7=9.
\displaystyle a_2=2(2)+7=11.
\displaystyle a_3=2(3)+7=13.
\displaystyle a_2-a_1=11-9=2.
\displaystyle a_3-a_2=13-11=2.
\displaystyle \text{Since the difference between consecutive terms is constant, the sequence is an A.P.}
\displaystyle \therefore d=2.
\displaystyle a_7=2(7)+7=21.
\displaystyle \therefore \text{The }7^{\text{th}}\text{ term is }21.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{The }n^{\text{th}}\text{ term of a sequence is given by }a_n=2n^2+n+1.
\displaystyle \text{Show that it is not an A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=2n^2+n+1.
\displaystyle a_1=2(1)^2+1+1=4.
\displaystyle a_2=2(2)^2+2+1=11.
\displaystyle a_3=2(3)^2+3+1=22.
\displaystyle a_2-a_1=11-4=7.
\displaystyle a_3-a_2=22-11=11.
\displaystyle \text{Since }a_3-a_2\neq a_2-a_1,\text{ the sequence is not an A.P.}
\displaystyle \\

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