\displaystyle \textbf{Question 1: } \text{Find:}
\displaystyle \text{(i) }10^{\text{th}}\text{ term of the A.P. }1,4,7,10,\ldots
\displaystyle \text{(ii) }18^{\text{th}}\text{ term of the A.P. }\sqrt{2},3\sqrt{2},5\sqrt{2},\ldots
\displaystyle \text{(iii) }n^{\text{th}}\text{ term of the A.P. }13,8,3,-2,\ldots
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }1,4,7,10,\ldots
\displaystyle a=1,\qquad d=4-1=3.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a_{10}=1+(10-1)(3)=1+27=28.
\displaystyle \therefore \text{The }10^{\text{th}}\text{ term is }28.
\displaystyle \text{(ii) Given A.P. }\sqrt{2},3\sqrt{2},5\sqrt{2},\ldots
\displaystyle a=\sqrt{2},\qquad d=3\sqrt{2}-\sqrt{2}=2\sqrt{2}.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a_{18}=\sqrt{2}+(18-1)(2\sqrt{2})=\sqrt{2}+34\sqrt{2}=35\sqrt{2}.
\displaystyle \therefore \text{The }18^{\text{th}}\text{ term is }35\sqrt{2}.
\displaystyle \text{(iii) Given A.P. }13,8,3,-2,\ldots
\displaystyle a=13,\qquad d=8-13=-5.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a_n=13+(n-1)(-5)=13-5n+5=18-5n.
\displaystyle \therefore a_n=18-5n.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{In an A.P., show that }a_{m+n}+a_{m-n}=2a_m.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common difference be }d.
\displaystyle \text{L.H.S.}=a_{m+n}+a_{m-n}.
\displaystyle =[a+(m+n-1)d]+[a+(m-n-1)d].
\displaystyle =2a+(m+n-1+m-n-1)d.
\displaystyle =2a+2(m-1)d.
\displaystyle =2[a+(m-1)d].
\displaystyle =2a_m=\text{R.H.S.}
\displaystyle \therefore a_{m+n}+a_{m-n}=2a_m.
\displaystyle \\

\displaystyle \textbf{Question 3:} \text{Find which term of each A.P. is the given number:}
\displaystyle \text{(i) }3,8,13,\ldots\text{ is }248.
\displaystyle \text{(ii) }84,80,76,\ldots\text{ is }0.
\displaystyle \text{(iii) }4,9,14,\ldots\text{ is }254.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }3,8,13,\ldots
\displaystyle a=3,\qquad d=8-3=5,\qquad a_n=248.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 248=3+(n-1)(5).
\displaystyle 245=5(n-1).
\displaystyle 49=n-1.
\displaystyle \therefore n=50.
\displaystyle \therefore 248\text{ is the }50^{\text{th}}\text{ term.}
\displaystyle \text{(ii) Given A.P. }84,80,76,\ldots
\displaystyle a=84,\qquad d=80-84=-4,\qquad a_n=0.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 0=84+(n-1)(-4).
\displaystyle 84=4(n-1).
\displaystyle 21=n-1.
\displaystyle \therefore n=22.
\displaystyle \therefore 0\text{ is the }22^{\text{nd}}\text{ term.}
\displaystyle \text{(iii) Given A.P. }4,9,14,\ldots
\displaystyle a=4,\qquad d=9-4=5,\qquad a_n=254.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 254=4+(n-1)(5).
\displaystyle 250=5(n-1).
\displaystyle 50=n-1.
\displaystyle \therefore n=51.
\displaystyle \therefore 254\text{ is the }51^{\text{st}}\text{ term.}
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{(i) Is }68\text{ a term of the A.P. }7,10,13,\ldots\text{?}
\displaystyle \text{(ii) Is }302\text{ a term of the A.P. }3,8,13,\ldots\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }7,10,13,\ldots
\displaystyle a=7,\qquad d=10-7=3,\qquad a_n=68.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 68=7+(n-1)(3).
\displaystyle 61=3(n-1).
\displaystyle n-1=\frac{61}{3}.
\displaystyle n=\frac{61}{3}+1=\frac{64}{3}.
\displaystyle \text{Since }n\text{ is not a natural number, }68\text{ is not a term of the given A.P.}
\displaystyle \text{(ii) Given A.P. }3,8,13,\ldots
\displaystyle a=3,\qquad d=8-3=5,\qquad a_n=302.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 302=3+(n-1)(5).
\displaystyle 299=5(n-1).
\displaystyle n-1=\frac{299}{5}.
\displaystyle n=\frac{299}{5}+1=\frac{304}{5}.
\displaystyle \text{Since }n\text{ is not a natural number, }302\text{ is not a term of the given A.P.}
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{(i) Which term of the sequence }24,23\frac{1}{4},22\frac{1}{2},21\frac{3}{4},\ldots
\displaystyle \text{is the first negative term?}
\displaystyle \text{(ii) Which term of the sequence }12+8i,11+6i,10+4i,\ldots\text{ is}
\displaystyle \text{(a) purely real and (b) purely imaginary?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }24,23\frac{1}{4},22\frac{1}{2},21\frac{3}{4},\ldots
\displaystyle a=24,\qquad d=23\frac{1}{4}-24=-\frac{3}{4}.
\displaystyle \text{Let }a_n\text{ be the first negative term.}
\displaystyle a_n<0.
\displaystyle a+(n-1)d<0.
\displaystyle 24+(n-1)\left(-\frac{3}{4}\right)<0.
\displaystyle 24-\frac{3n}{4}+\frac{3}{4}<0.
\displaystyle \frac{99}{4}<\frac{3n}{4}.
\displaystyle 99<3n.
\displaystyle n>33.
\displaystyle \text{The least natural number greater than }33\text{ is }34.
\displaystyle \therefore \text{The }34^{\text{th}}\text{ term is the first negative term.}
\displaystyle \text{(ii) Given A.P. }12+8i,11+6i,10+4i,\ldots
\displaystyle a=12+8i.
\displaystyle d=(11+6i)-(12+8i)=-1-2i.
\displaystyle a_n=a+(n-1)d.
\displaystyle =(12+8i)+(n-1)(-1-2i).
\displaystyle =12+8i-n+1-2in+2i.
\displaystyle =(13-n)+i(10-2n).
\displaystyle \text{(a) For }a_n\text{ to be purely real, its imaginary part must be zero.}
\displaystyle 10-2n=0.
\displaystyle \therefore n=5.
\displaystyle a_5=(13-5)+i(10-10)=8.
\displaystyle \therefore \text{The }5^{\text{th}}\text{ term is purely real.}
\displaystyle \text{(b) For }a_n\text{ to be purely imaginary, its real part must be zero.}
\displaystyle 13-n=0.
\displaystyle \therefore n=13.
\displaystyle a_{13}=(13-13)+i(10-26)=-16i.
\displaystyle \therefore \text{The }13^{\text{th}}\text{ term is purely imaginary.}
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{(i) How many terms are there in the A.P. }7,10,13,\ldots,43\text{?}
\displaystyle \text{(ii) How many terms are there in the A.P. }-1,\frac{-5}{6},\frac{-2}{3},\frac{-1}{2},\ldots,\frac{10}{3}\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }7,10,13,\ldots,43.
\displaystyle a=7,\qquad d=10-7=3,\qquad a_n=43.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 43=7+(n-1)(3).
\displaystyle 36=3(n-1).
\displaystyle n-1=12.
\displaystyle \therefore n=13.
\displaystyle \therefore \text{There are }13\text{ terms in the given A.P.}
\displaystyle \text{(ii) Given A.P. }-1,\frac{-5}{6},\frac{-2}{3},\frac{-1}{2},\ldots,\frac{10}{3}.
\displaystyle a=-1,\qquad d=\frac{-5}{6}-(-1)=\frac{1}{6},\qquad a_n=\frac{10}{3}.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle \frac{10}{3}=-1+(n-1)\left(\frac{1}{6}\right).
\displaystyle \frac{13}{3}=\frac{n-1}{6}.
\displaystyle n-1=26.
\displaystyle \therefore n=27.
\displaystyle \therefore \text{There are }27\text{ terms in the given A.P.}
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{The first term of an A.P. is }5,\text{ the common difference is }3\text{ and the}
\displaystyle \text{last term is }80.\text{ Find the number of terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=5,\qquad d=3,\qquad a_n=80.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 80=5+(n-1)(3).
\displaystyle 75=3(n-1).
\displaystyle n-1=25.
\displaystyle \therefore n=26.
\displaystyle \therefore \text{There are }26\text{ terms in the given A.P.}
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{The }6^{\text{th}}\text{ and }17^{\text{th}}\text{ terms of an A.P. are }19\text{ and }41
\displaystyle \text{respectively. Find the }40^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_6=19\text{ and }a_{17}=41.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a+5d=19 \qquad \ldots\text{(i)}
\displaystyle a+16d=41 \qquad \ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 11d=22.
\displaystyle \therefore d=2.
\displaystyle a=19-5(2)=9.
\displaystyle a_{40}=a+(40-1)d=9+39(2)=87.
\displaystyle \therefore \text{The }40^{\text{th}}\text{ term is }87.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{If the }9^{\text{th}}\text{ term of an A.P. is zero, prove that its }29^{\text{th}}\text{ term}
\displaystyle \text{is double the }19^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_9=0.
\displaystyle a+(9-1)d=0.
\displaystyle a+8d=0.
\displaystyle \therefore a=-8d.
\displaystyle a_{19}=a+(19-1)d.
\displaystyle =-8d+18d=10d.
\displaystyle a_{29}=a+(29-1)d.
\displaystyle =-8d+28d=20d.
\displaystyle =2(10d)=2a_{19}.
\displaystyle \therefore a_{29}=2a_{19}.
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{If }10\text{ times the }10^{\text{th}}\text{ term of an A.P. is equal to }15\text{ times the}
\displaystyle 15^{\text{th}}\text{ term, show that the }25^{\text{th}}\text{ term of the A.P. is zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }10a_{10}=15a_{15}.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 10[a+(10-1)d]=15[a+(15-1)d].
\displaystyle 10(a+9d)=15(a+14d).
\displaystyle 10a+90d=15a+210d.
\displaystyle 5a+120d=0.
\displaystyle a+24d=0.
\displaystyle \therefore a=-24d.
\displaystyle a_{25}=a+(25-1)d.
\displaystyle =-24d+24d=0.
\displaystyle \therefore \text{The }25^{\text{th}}\text{ term of the A.P. is zero.}
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{The }10^{\text{th}}\text{ and }18^{\text{th}}\text{ terms of an A.P. are }41\text{ and }73
\displaystyle \text{respectively. Find the }26^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_{10}=41\text{ and }a_{18}=73.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a+9d=41 \qquad \ldots\text{(i)}
\displaystyle a+17d=73 \qquad \ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 8d=32.
\displaystyle \therefore d=4.
\displaystyle \text{Substituting }d=4\text{ in (i),}
\displaystyle a+9(4)=41.
\displaystyle \therefore a=5.
\displaystyle a_{26}=a+(26-1)d.
\displaystyle =5+25(4)=105.
\displaystyle \therefore \text{The }26^{\text{th}}\text{ term is }105.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{In a certain A.P., the }24^{\text{th}}\text{ term is twice the }10^{\text{th}}\text{ term. Prove}
\displaystyle \text{that the }72^{\text{nd}}\text{ term is twice the }34^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_{24}=2a_{10}.
\displaystyle a+23d=2(a+9d).
\displaystyle a+23d=2a+18d.
\displaystyle \therefore a=5d.
\displaystyle a_{34}=a+(34-1)d.
\displaystyle =5d+33d=38d.
\displaystyle a_{72}=a+(72-1)d.
\displaystyle =5d+71d=76d.
\displaystyle =2(38d)=2a_{34}.
\displaystyle \therefore a_{72}=2a_{34}.
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{If the }(m+1)^{\text{th}}\text{ term of an A.P. is twice the }(n+1)^{\text{th}}\text{ term,}
\displaystyle \text{prove that the }(3m+1)^{\text{th}}\text{ term is twice the }(m+n+1)^{\text{th}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_{m+1}=2a_{n+1}.
\displaystyle a+[(m+1)-1]d=2\{a+[(n+1)-1]d\}.
\displaystyle a+md=2(a+nd).
\displaystyle a+md=2a+2nd.
\displaystyle \therefore a=(m-2n)d.\qquad\ldots\text{(i)}
\displaystyle a_{3m+1}=a+[(3m+1)-1]d.
\displaystyle =a+3md.
\displaystyle =(m-2n)d+3md.
\displaystyle =(4m-2n)d.
\displaystyle a_{m+n+1}=a+[(m+n+1)-1]d.
\displaystyle =a+(m+n)d.
\displaystyle =(m-2n)d+(m+n)d.
\displaystyle =(2m-n)d.
\displaystyle 2a_{m+n+1}=2(2m-n)d=(4m-2n)d.
\displaystyle \therefore a_{3m+1}=2a_{m+n+1}.
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{If the }n^{\text{th}}\text{ term of the A.P. }9,7,5,\ldots\text{ is the same as the}
\displaystyle \text{ }n^{\text{th}}\text{ term of the A.P. }15,12,9,\ldots,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \text{For the A.P. }9,7,5,\ldots
\displaystyle a=9,\qquad d=7-9=-2.
\displaystyle a_n=a+(n-1)d=9+(n-1)(-2)=11-2n.
\displaystyle \text{For the A.P. }15,12,9,\ldots
\displaystyle a=15,\qquad d=12-15=-3.
\displaystyle a_n=a+(n-1)d=15+(n-1)(-3)=18-3n.
\displaystyle \text{Since the }n^{\text{th}}\text{ terms are equal,}
\displaystyle 11-2n=18-3n.
\displaystyle \therefore n=7.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Find the }12^{\text{th}}\text{ term from the end of each of the following A.P.s:}
\displaystyle \text{(i) }3,5,7,9,\ldots,201
\displaystyle \text{(ii) }3,8,13,\ldots,253
\displaystyle \text{(iii) }1,4,7,10,\ldots,88
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }3,5,7,9,\ldots,201.
\displaystyle a=3,\qquad d=5-3=2,\qquad l=201.
\displaystyle \text{The }n^{\text{th}}\text{ term from the end}=l-(n-1)d.
\displaystyle \therefore \text{The }12^{\text{th}}\text{ term from the end}=201-(12-1)(2)=201-22=179.
\displaystyle \text{(ii) Given A.P. }3,8,13,\ldots,253.
\displaystyle a=3,\qquad d=8-3=5,\qquad l=253.
\displaystyle \text{The }n^{\text{th}}\text{ term from the end}=l-(n-1)d.
\displaystyle \therefore \text{The }12^{\text{th}}\text{ term from the end}=253-(12-1)(5)=253-55=198.
\displaystyle \text{(iii) Given A.P. }1,4,7,10,\ldots,88.
\displaystyle a=1,\qquad d=4-1=3,\qquad l=88.
\displaystyle \text{The }n^{\text{th}}\text{ term from the end}=l-(n-1)d.
\displaystyle \therefore \text{The }12^{\text{th}}\text{ term from the end}=88-(12-1)(3)=88-33=55.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{The }4^{\text{th}}\text{ term of an A.P. is three times its first term and the}
\displaystyle 7^{\text{th}}\text{ term exceeds twice the }3^{\text{rd}}\text{ term by }1.\text{ Find the first term and}
\displaystyle \text{the common difference.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_4=3a.
\displaystyle a+(4-1)d=3a.
\displaystyle a+3d=3a.
\displaystyle 3d=2a.\qquad\ldots\text{(i)}
\displaystyle \text{Also, }a_7-2a_3=1.
\displaystyle [a+(7-1)d]-2[a+(3-1)d]=1.
\displaystyle a+6d-2(a+2d)=1.
\displaystyle a+6d-2a-4d=1.
\displaystyle 2d-a=1.
\displaystyle \therefore a=2d-1.\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting (ii) in (i),}
\displaystyle 3d=2(2d-1).
\displaystyle 3d=4d-2.
\displaystyle \therefore d=2.
\displaystyle a=2(2)-1=3.
\displaystyle \therefore \text{The first term is }3\text{ and the common difference is }2.
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{Find the second term and the }n^{\text{th}}\text{ term of an A.P. whose }6^{\text{th}}\text{ term}
\displaystyle \text{is }12\text{ and the }8^{\text{th}}\text{ term is }22.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_6=12\text{ and }a_8=22.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a+5d=12.\qquad\ldots\text{(i)}
\displaystyle a+7d=22.\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 2d=10.
\displaystyle \therefore d=5.
\displaystyle \text{Substituting }d=5\text{ in (i),}
\displaystyle a+5(5)=12.
\displaystyle \therefore a=-13.
\displaystyle a_2=a+(2-1)d=-13+5=-8.
\displaystyle a_n=a+(n-1)d.
\displaystyle =-13+(n-1)(5).
\displaystyle =-13+5n-5=5n-18.
\displaystyle \therefore \text{The second term is }-8\text{ and }a_n=5n-18.
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{How many two-digit numbers are divisible by }3\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{The two-digit numbers divisible by }3\text{ form the A.P.}
\displaystyle 12,15,18,\ldots,96,99.
\displaystyle a=12,\qquad d=15-12=3,\qquad a_n=99.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 99=12+(n-1)(3).
\displaystyle 87=3(n-1).
\displaystyle n-1=29.
\displaystyle \therefore n=30.
\displaystyle \therefore \text{There are }30\text{ two-digit numbers divisible by }3.
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{An A.P. consists of }60\text{ terms. If the first and last terms are }7\text{ and }125
\displaystyle \text{respectively, find the }32^{\text{nd}}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=7,\qquad n=60,\qquad l=125.
\displaystyle \text{Using }l=a+(n-1)d,
\displaystyle 125=7+(60-1)d.
\displaystyle 118=59d.
\displaystyle \therefore d=2.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a_{32}=7+(32-1)(2).
\displaystyle =7+62=69.
\displaystyle \therefore \text{The }32^{\text{nd}}\text{ term is }69.
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{The sum of the }4^{\text{th}}\text{ and }8^{\text{th}}\text{ terms of an A.P. is }24\text{ and the}
\displaystyle \text{sum of the }6^{\text{th}}\text{ and }10^{\text{th}}\text{ terms is }34.\text{ Find the first term and the}
\displaystyle \text{common difference of the A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_4+a_8=24.
\displaystyle [a+(4-1)d]+[a+(8-1)d]=24.
\displaystyle 2a+10d=24.
\displaystyle a+5d=12.\qquad\ldots\text{(i)}
\displaystyle \text{Also, }a_6+a_{10}=34.
\displaystyle [a+(6-1)d]+[a+(10-1)d]=34.
\displaystyle 2a+14d=34.
\displaystyle a+7d=17.\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 2d=5.
\displaystyle \therefore d=\frac{5}{2}.
\displaystyle \text{Substituting }d=\frac{5}{2}\text{ in (i),}
\displaystyle a+5\left(\frac{5}{2}\right)=12.
\displaystyle a=12-\frac{25}{2}.
\displaystyle a=\frac{24-25}{2}=-\frac{1}{2}.
\displaystyle \therefore \text{The first term is }-\frac{1}{2}\text{ and the common difference is }\frac{5}{2}.
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{How many numbers are there between }1\text{ and }1000\text{ which when divided by }7
\displaystyle \text{leave remainder }4\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{If a number }N\text{ leaves remainder }4\text{ when divided by }7,\text{ then }N=7q+4,
\displaystyle \text{where }q\text{ is a non-negative integer.}
\displaystyle \text{Hence the required sequence is }4,11,18,\ldots,998.
\displaystyle a=4,\qquad d=11-4=7,\qquad l=998.
\displaystyle \text{Using }l=a+(n-1)d,
\displaystyle 998=4+(n-1)(7).
\displaystyle 994=7(n-1).
\displaystyle n-1=142.
\displaystyle \therefore n=143.
\displaystyle \therefore \text{There are }143\text{ numbers between }1\text{ and }1000\text{ which leave remainder }4
\displaystyle \text{when divided by }7.
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{The first and the last terms of an A.P. are }a\text{ and }l\text{ respectively. Show}
\displaystyle \text{that the sum of the }n^{\text{th}}\text{ term from the beginning and the }n^{\text{th}}\text{ term from}
\displaystyle \text{the end is }a+l.
\displaystyle \text{Answer:}
\displaystyle \text{Let the common difference be }d.
\displaystyle \text{The }n^{\text{th}}\text{ term from the beginning is }a_n=a+(n-1)d.
\displaystyle \text{The }n^{\text{th}}\text{ term from the end is }l-(n-1)d.
\displaystyle \therefore \text{Their sum}=[a+(n-1)d]+[l-(n-1)d].
\displaystyle =a+l.
\displaystyle \therefore \text{The sum of the }n^{\text{th}}\text{ term from the beginning and the }n^{\text{th}}\text{ term}
\displaystyle \text{from the end is }a+l.
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{If an A.P. is such that }\frac{a_4}{a_7}=\frac{2}{3},\text{ find }\frac{a_6}{a_8}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{a_4}{a_7}=\frac{2}{3}.
\displaystyle \therefore \frac{a+(4-1)d}{a+(7-1)d}=\frac{2}{3}.
\displaystyle \frac{a+3d}{a+6d}=\frac{2}{3}.
\displaystyle 3(a+3d)=2(a+6d).
\displaystyle 3a+9d=2a+12d.
\displaystyle \therefore a=3d.
\displaystyle \therefore \frac{a_6}{a_8}=\frac{a+(6-1)d}{a+(8-1)d}.
\displaystyle =\frac{a+5d}{a+7d}=\frac{3d+5d}{3d+7d}.
\displaystyle =\frac{8d}{10d}=\frac{4}{5}.
\displaystyle \therefore \frac{a_6}{a_8}=\frac{4}{5}.
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{If }\theta_1,\theta_2,\theta_3,\ldots,\theta_n\text{ are in A.P. with common difference }d,
\displaystyle \text{show that }\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+\cdots+\sec\theta_{n-1}\sec\theta_n
\displaystyle =\frac{\tan\theta_n-\tan\theta_1}{\sin d}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\theta_1,\theta_2,\theta_3,\ldots,\theta_n\text{ are in A.P.,}
\displaystyle d=\theta_2-\theta_1=\theta_3-\theta_2=\cdots=\theta_n-\theta_{n-1}.
\displaystyle \text{L.H.S.}=\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+\cdots+\sec\theta_{n-1}\sec\theta_n.
\displaystyle =\frac{1}{\cos\theta_1\cos\theta_2}+\frac{1}{\cos\theta_2\cos\theta_3}+\cdots
\displaystyle \qquad+\frac{1}{\cos\theta_{n-1}\cos\theta_n}.
\displaystyle =\frac{1}{\sin d}\left[\frac{\sin d}{\cos\theta_1\cos\theta_2}+\frac{\sin d}{\cos\theta_2\cos\theta_3}+\cdots\right.
\displaystyle \left.\qquad+\frac{\sin d}{\cos\theta_{n-1}\cos\theta_n}\right].
\displaystyle =\frac{1}{\sin d}\left[\frac{\sin(\theta_2-\theta_1)}{\cos\theta_1\cos\theta_2}+\frac{\sin(\theta_3-\theta_2)}{\cos\theta_2\cos\theta_3}+\cdots\right.
\displaystyle \left.\qquad+\frac{\sin(\theta_n-\theta_{n-1})}{\cos\theta_{n-1}\cos\theta_n}\right].
\displaystyle \text{Using }\sin(A-B)=\sin A\cos B-\cos A\sin B,
\displaystyle \frac{\sin(\theta_{r+1}-\theta_r)}{\cos\theta_r\cos\theta_{r+1}}
\displaystyle =\frac{\sin\theta_{r+1}\cos\theta_r-\cos\theta_{r+1}\sin\theta_r}{\cos\theta_r\cos\theta_{r+1}}
\displaystyle =\tan\theta_{r+1}-\tan\theta_r.
\displaystyle \therefore \text{L.H.S.}=\frac{1}{\sin d}\left[(\tan\theta_2-\tan\theta_1)+(\tan\theta_3-\tan\theta_2)+\cdots\right.
\displaystyle \left.\qquad+(\tan\theta_n-\tan\theta_{n-1})\right].
\displaystyle =\frac{\tan\theta_n-\tan\theta_1}{\sin d}.
\displaystyle =\text{R.H.S.}
\displaystyle \therefore \sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+\cdots+\sec\theta_{n-1}\sec\theta_n
\displaystyle =\frac{\tan\theta_n-\tan\theta_1}{\sin d}.
\displaystyle \\

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