\displaystyle \textbf{Question 1: } \text{Find the sum of each of the following arithmetic progressions:}
\displaystyle \text{(i) }50,46,42,\ldots\text{ to }10\text{ terms}
\displaystyle \text{(ii) }1,3,5,7,\ldots\text{ to }12\text{ terms}
\displaystyle \text{(iii) }3,\frac{9}{2},6,\frac{15}{2},\ldots\text{ to }25\text{ terms}
\displaystyle \text{(iv) }41,36,31,\ldots\text{ to }12\text{ terms}
\displaystyle \text{(v) }a+b,a-b,a-3b,\ldots\text{ to }22\text{ terms}
\displaystyle \text{(vi) }(x-y)^2,x^2+y^2,(x+y)^2,\ldots\text{ to }n\text{ terms}
\displaystyle \text{(vii) }\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y},\ldots\text{ to }n\text{ terms}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given A.P. }50,46,42,\ldots
\displaystyle a=50,\qquad d=46-50=-4,\qquad n=10.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{10}=\frac{10}{2}[2(50)+(10-1)(-4)].
\displaystyle =5(100-36)=5(64)=320.
\displaystyle \therefore \text{The required sum is }320.

\displaystyle \text{(ii) Given A.P. }1,3,5,7,\ldots
\displaystyle a=1,\qquad d=3-1=2,\qquad n=12.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{12}=\frac{12}{2}[2(1)+(12-1)(2)].
\displaystyle =6(2+22)=6(24)=144.
\displaystyle \therefore \text{The required sum is }144.

\displaystyle \text{(iii) Given A.P. }3,\frac{9}{2},6,\frac{15}{2},\ldots
\displaystyle a=3,\qquad d=\frac{9}{2}-3=\frac{3}{2},\qquad n=25.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{25}=\frac{25}{2}\left[2(3)+(25-1)\left(\frac{3}{2}\right)\right].
\displaystyle =\frac{25}{2}(6+36)=\frac{25}{2}(42)=525.
\displaystyle \therefore \text{The required sum is }525.

\displaystyle \text{(iv) Given A.P. }41,36,31,\ldots
\displaystyle a=41,\qquad d=36-41=-5,\qquad n=12.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{12}=\frac{12}{2}[2(41)+(12-1)(-5)].
\displaystyle =6(82-55)=6(27)=162.
\displaystyle \therefore \text{The required sum is }162.

\displaystyle \text{(v) Given A.P. }a+b,a-b,a-3b,\ldots
\displaystyle \text{First term}=a+b,\qquad d=(a-b)-(a+b)=-2b,\qquad n=22.
\displaystyle \text{Using }S_n=\frac{n}{2}[2(\text{first term})+(n-1)d],
\displaystyle S_{22}=\frac{22}{2}[2(a+b)+(22-1)(-2b)].
\displaystyle =11[2a+2b-42b].
\displaystyle =11(2a-40b)=22a-440b.
\displaystyle \therefore \text{The required sum is }22a-440b.

\displaystyle \text{(vi) Given A.P. }(x-y)^2,x^2+y^2,(x+y)^2,\ldots
\displaystyle a=(x-y)^2,\qquad d=(x^2+y^2)-(x-y)^2=2xy.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_n=\frac{n}{2}[2(x-y)^2+(n-1)(2xy)].
\displaystyle =n[(x-y)^2+(n-1)xy].
\displaystyle \therefore \text{The required sum is }n[(x-y)^2+(n-1)xy].

\displaystyle \text{(vii) Given A.P. }\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y},\ldots
\displaystyle a=\frac{x-y}{x+y},\qquad d=\frac{3x-2y}{x+y}-\frac{x-y}{x+y}.
\displaystyle d=\frac{2x-y}{x+y},\qquad x+y\neq0.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_n=\frac{n}{2}\left[2\left(\frac{x-y}{x+y}\right)+(n-1)\left(\frac{2x-y}{x+y}\right)\right].
\displaystyle =\frac{n}{2(x+y)}[2(x-y)+(n-1)(2x-y)].
\displaystyle =\frac{n}{2(x+y)}[2nx-ny-y].
\displaystyle =\frac{n[n(2x-y)-y]}{2(x+y)}.
\displaystyle \therefore \text{The required sum is }\frac{n[n(2x-y)-y]}{2(x+y)}.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Find the sum of each of the following series:}
\displaystyle \text{(i) }2+5+8+\cdots+182
\displaystyle \text{(ii) }101+99+97+\cdots+47
\displaystyle \text{(iii) }(a-b)^2+(a^2+b^2)+(a+b)^2+\cdots+[(a+b)^2+6ab]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given A.P. }2,5,8,\ldots,182.
\displaystyle a=2,\qquad d=5-2=3,\qquad a_n=182.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 182=2+(n-1)(3).
\displaystyle 182=3n-1.
\displaystyle 3n=183.
\displaystyle \therefore n=61.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{61}=\frac{61}{2}[2(2)+(61-1)(3)].
\displaystyle =\frac{61}{2}(4+180)=\frac{61}{2}(184)=5612.
\displaystyle \therefore \text{The required sum is }5612.
\displaystyle \text{(ii) Given A.P. }101,99,97,\ldots,47.
\displaystyle a=101,\qquad d=99-101=-2,\qquad a_n=47.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 47=101+(n-1)(-2).
\displaystyle 47=103-2n.
\displaystyle 2n=56.
\displaystyle \therefore n=28.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{28}=\frac{28}{2}[2(101)+(28-1)(-2)].
\displaystyle =14(202-54)=14(148)=2072.
\displaystyle \therefore \text{The required sum is }2072.
\displaystyle \text{(iii) Given A.P. }(a-b)^2,\;a^2+b^2,\;(a+b)^2,\ldots,[(a+b)^2+6ab].
\displaystyle \text{First term}=(a-b)^2.
\displaystyle d=(a^2+b^2)-(a-b)^2=2ab.
\displaystyle a_n=(a+b)^2+6ab.
\displaystyle \text{Using }a_n=\text{first term}+(n-1)d,
\displaystyle (a+b)^2+6ab=(a-b)^2+(n-1)(2ab).
\displaystyle a^2+b^2+8ab=a^2+b^2-2ab+(n-1)(2ab).
\displaystyle (n-1)(2ab)=10ab.
\displaystyle n-1=5.
\displaystyle \therefore n=6.
\displaystyle \text{Using }S_n=\frac{n}{2}[2(\text{first term})+(n-1)d],
\displaystyle S_6=\frac{6}{2}[2(a-b)^2+(6-1)(2ab)].
\displaystyle =3[2a^2+2b^2-4ab+10ab].
\displaystyle =3(2a^2+2b^2+6ab).
\displaystyle =6a^2+18ab+6b^2.
\displaystyle \therefore \text{The required sum is }6a^2+18ab+6b^2.
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Find the sum of the first }n\text{ natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first }n\text{ natural numbers form the A.P. }1,2,3,\ldots,n.
\displaystyle a=1,\qquad d=2-1=1.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_n=\frac{n}{2}[2(1)+(n-1)(1)].
\displaystyle =\frac{n}{2}(n+1).
\displaystyle \therefore \text{The sum of the first }n\text{ natural numbers is }\frac{n(n+1)}{2}.
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Find the sum of all natural numbers between }1\text{ and }100\text{ which are}
\displaystyle \text{divisible by }2\text{ or }5.
\displaystyle \text{Answer:}
\displaystyle \text{The numbers divisible by }2\text{ are }2,4,6,\ldots,98,100.
\displaystyle a=2,\qquad d=2,\qquad a_n=100.
\displaystyle 100=2+(n-1)(2).
\displaystyle \therefore n=50.
\displaystyle S_{50}=\frac{50}{2}[2(2)+(50-1)(2)].
\displaystyle =25(4+98)=2550.
\displaystyle \text{The numbers divisible by }5\text{ are }5,10,15,\ldots,95,100.
\displaystyle a=5,\qquad d=5,\qquad a_n=100.
\displaystyle 100=5+(n-1)(5).
\displaystyle \therefore n=20.
\displaystyle S_{20}=\frac{20}{2}[2(5)+(20-1)(5)].
\displaystyle =10(10+95)=1050.
\displaystyle \text{The numbers divisible by both }2\text{ and }5\text{ are multiples of }10:
\displaystyle 10,20,30,\ldots,90,100.
\displaystyle a=10,\qquad d=10,\qquad a_n=100.
\displaystyle 100=10+(n-1)(10).
\displaystyle \therefore n=10.
\displaystyle S_{10}=\frac{10}{2}[2(10)+(10-1)(10)].
\displaystyle =5(20+90)=550.
\displaystyle \therefore \text{Required sum}=2550+1050-550=3050.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Find the sum of the first }n\text{ odd natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{The first }n\text{ odd natural numbers form the A.P. }1,3,5,7,\ldots
\displaystyle a=1,\qquad d=3-1=2.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_n=\frac{n}{2}[2(1)+(n-1)(2)].
\displaystyle =\frac{n}{2}(2n)=n^2.
\displaystyle \therefore \text{The sum of the first }n\text{ odd natural numbers is }n^2.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Find the sum of all odd numbers between }100\text{ and }200.
\displaystyle \text{Answer:}
\displaystyle \text{The required A.P. is }101,103,105,\ldots,199.
\displaystyle a=101,\qquad d=103-101=2,\qquad a_n=199.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 199=101+(n-1)(2).
\displaystyle 98=2(n-1).
\displaystyle n-1=49.
\displaystyle \therefore n=50.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{50}=\frac{50}{2}[2(101)+(50-1)(2)].
\displaystyle =25(202+98)=25(300)=7500.
\displaystyle \therefore \text{The required sum is }7500.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{Show that the sum of all odd integers between }1\text{ and }1000\text{ which are}
\displaystyle \text{divisible by }3\text{ is }83667.
\displaystyle \text{Answer:}
\displaystyle \text{The required odd integers form the A.P. }3,9,15,\ldots,999.
\displaystyle a=3,\qquad d=9-3=6,\qquad a_n=999.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 999=3+(n-1)(6).
\displaystyle 996=6(n-1).
\displaystyle n-1=166.
\displaystyle \therefore n=167.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{167}=\frac{167}{2}[2(3)+(167-1)(6)].
\displaystyle =\frac{167}{2}(6+996).
\displaystyle =\frac{167}{2}(1002)=167(501)=83667.
\displaystyle \therefore \text{The sum of all such odd integers is }83667.
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{Find the sum of all integers between }84\text{ and }219\text{ which are multiples}
\displaystyle \text{of }5.
\displaystyle \text{Answer:}
\displaystyle \text{The required integers form the A.P. }85,90,95,\ldots,210,215.
\displaystyle a=85,\qquad d=90-85=5,\qquad a_n=215.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 215=85+(n-1)(5).
\displaystyle 130=5(n-1).
\displaystyle n-1=26.
\displaystyle \therefore n=27.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{27}=\frac{27}{2}[2(85)+(27-1)(5)].
\displaystyle =\frac{27}{2}(170+130).
\displaystyle =\frac{27}{2}(300)=4050.
\displaystyle \therefore \text{The required sum is }4050.
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{Find the sum of all integers between }50\text{ and }500\text{ which are divisible}
\displaystyle \text{by }7.
\displaystyle \text{Answer:}
\displaystyle \text{The required integers form the A.P. }56,63,70,\ldots,497.
\displaystyle a=56,\qquad d=63-56=7,\qquad a_n=497.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 497=56+(n-1)(7).
\displaystyle 441=7(n-1).
\displaystyle n-1=63.
\displaystyle \therefore n=64.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{64}=\frac{64}{2}[2(56)+(64-1)(7)].
\displaystyle =32(112+441)=32(553)=17696.
\displaystyle \therefore \text{The required sum is }17696.
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{Find the sum of all even integers between }101\text{ and }999.
\displaystyle \text{Answer:}
\displaystyle \text{The required integers form the A.P. }102,104,106,\ldots,998.
\displaystyle a=102,\qquad d=104-102=2,\qquad a_n=998.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 998=102+(n-1)(2).
\displaystyle 896=2(n-1).
\displaystyle n-1=448.
\displaystyle \therefore n=449.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{449}=\frac{449}{2}[2(102)+(449-1)(2)].
\displaystyle =\frac{449}{2}(204+896).
\displaystyle =\frac{449}{2}(1100)=449(550)=246950.
\displaystyle \therefore \text{The required sum is }246950.
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Find the sum of all integers between }100\text{ and }550\text{ which are divisible}
\displaystyle \text{by }9.
\displaystyle \text{Answer:}
\displaystyle \text{The required integers form the A.P. }108,117,126,\ldots,549.
\displaystyle a=108,\qquad d=117-108=9,\qquad a_n=549.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 549=108+(n-1)(9).
\displaystyle 441=9(n-1).
\displaystyle n-1=49.
\displaystyle \therefore n=50.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{50}=\frac{50}{2}[2(108)+(50-1)(9)].
\displaystyle =25(216+441)=25(657)=16425.
\displaystyle \therefore \text{The required sum is }16425.
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Find the sum of the series }3+5+7+6+9+12+9+13+17+\cdots
\displaystyle \text{to }3n\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{The given series is }3,5,7,6,9,12,9,13,17,\ldots
\displaystyle \text{It can be separated into the following three A.P.s:}
\displaystyle 3,6,9,\ldots;\qquad 5,9,13,\ldots;\qquad 7,12,17,\ldots
\displaystyle \text{Since the original series has }3n\text{ terms, each A.P. has }n\text{ terms.}
\displaystyle \text{For the A.P. }3,6,9,\ldots,\quad a=3,\quad d=3.
\displaystyle \text{For the A.P. }5,9,13,\ldots,\quad a=5,\quad d=4.
\displaystyle \text{For the A.P. }7,12,17,\ldots,\quad a=7,\quad d=5.
\displaystyle \therefore \text{Required sum}
\displaystyle =\frac{n}{2}[2(3)+(n-1)(3)]+\frac{n}{2}[2(5)+(n-1)(4)]
\displaystyle \qquad+\frac{n}{2}[2(7)+(n-1)(5)].
\displaystyle =\frac{n}{2}[6+3(n-1)+10+4(n-1)+14+5(n-1)].
\displaystyle =\frac{n}{2}[30+12(n-1)].
\displaystyle =\frac{n}{2}(12n+18).
\displaystyle =3n(2n+3).
\displaystyle \therefore \text{The required sum is }3n(2n+3).
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{Find the sum of all integers between }100\text{ and }800\text{ which leave}
\displaystyle \text{remainder }7\text{ when divided by }16.
\displaystyle \text{Answer:}
\displaystyle \text{Such numbers are of the form }N=16q+7,\text{ where }q\text{ is an integer.}
\displaystyle \text{The required integers form the A.P. }103,119,135,\ldots,791.
\displaystyle a=103,\qquad d=119-103=16,\qquad a_n=791.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle 791=103+(n-1)(16).
\displaystyle 688=16(n-1).
\displaystyle n-1=43.
\displaystyle \therefore n=44.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{44}=\frac{44}{2}[2(103)+(44-1)(16)].
\displaystyle =22(206+688).
\displaystyle =22(894)=19668.
\displaystyle \therefore \text{The required sum is }19668.
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{Solve:}
\displaystyle \text{(i) }25+22+19+16+\cdots+x=115
\displaystyle \text{(ii) }1+4+7+10+\cdots+x=590
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }25+22+19+16+\cdots+x=115.
\displaystyle S_n=115,\qquad a=25,\qquad d=22-25=-3.
\displaystyle \text{Let the number of terms be }n.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle 115=\frac{n}{2}[2(25)+(n-1)(-3)].
\displaystyle 230=n[50-3(n-1)].
\displaystyle 230=n(53-3n).
\displaystyle 3n^2-53n+230=0.
\displaystyle (3n-23)(n-10)=0.
\displaystyle n=\frac{23}{3}\text{ or }n=10.
\displaystyle \text{Since }n\text{ must be a natural number, }n=10.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle x=25+(10-1)(-3).
\displaystyle =25-27=-2.
\displaystyle \therefore x=-2.
\displaystyle \text{(ii) Given, }1+4+7+10+\cdots+x=590.
\displaystyle S_n=590,\qquad a=1,\qquad d=4-1=3.
\displaystyle \text{Let the number of terms be }n.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle 590=\frac{n}{2}[2(1)+(n-1)(3)].
\displaystyle 1180=n(3n-1).
\displaystyle 3n^2-n-1180=0.
\displaystyle (3n+59)(n-20)=0.
\displaystyle n=-\frac{59}{3}\text{ or }n=20.
\displaystyle \text{Since }n\text{ must be a natural number, }n=20.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle x=1+(20-1)(3).
\displaystyle =1+57=58.
\displaystyle \therefore x=58.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Find the }r^{\text{th}}\text{ term of an A.P. whose sum of the first }n\text{ terms is}
\displaystyle 3n^2+2n.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common difference be }d.
\displaystyle \text{Given, }S_n=3n^2+2n.
\displaystyle S_1=3(1)^2+2(1)=5.
\displaystyle \therefore a=5.
\displaystyle S_2=3(2)^2+2(2)=12+4=16.
\displaystyle \therefore a+(a+d)=16.
\displaystyle 5+5+d=16.
\displaystyle \therefore d=6.
\displaystyle \text{Using }a_r=a+(r-1)d,
\displaystyle a_r=5+(r-1)(6).
\displaystyle =6r-1.
\displaystyle \therefore \text{The }r^{\text{th}}\text{ term is }6r-1.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{How many terms are there in an A.P. whose first and fifth terms are }-14
\displaystyle \text{and }2\text{ respectively, and the sum of the terms is }40\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common difference be }d.
\displaystyle \text{Given, }a=-14\text{ and }a_5=2.
\displaystyle a+(5-1)d=2.
\displaystyle -14+4d=2.
\displaystyle 4d=16.
\displaystyle \therefore d=4.
\displaystyle \text{Also, }S_n=40.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle 40=\frac{n}{2}[-28+(n-1)(4)].
\displaystyle 80=n(-28+4n-4).
\displaystyle 80=n(4n-32).
\displaystyle 4n^2-32n-80=0.
\displaystyle n^2-8n-20=0.
\displaystyle (n-10)(n+2)=0.
\displaystyle n=10\text{ or }n=-2.
\displaystyle \text{Since the number of terms cannot be negative, }n=10.
\displaystyle \therefore \text{There are }10\text{ terms in the A.P.}
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{The sum of the first }7\text{ terms of an A.P. is }10\text{ and the sum of the}
\displaystyle \text{next }7\text{ terms is }17.\text{ Find the progression.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S_7=10.
\displaystyle \frac{7}{2}[2a+(7-1)d]=10.
\displaystyle \frac{7}{2}(2a+6d)=10.
\displaystyle 2a+6d=\frac{20}{7}.
\displaystyle a+3d=\frac{10}{7}.\qquad\ldots\text{(i)}
\displaystyle \text{Also, the sum of the next }7\text{ terms is }17.
\displaystyle S_{14}-S_7=17.
\displaystyle \frac{14}{2}[2a+(14-1)d]-10=17.
\displaystyle 7(2a+13d)=27.
\displaystyle 14a+91d=27.\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying (i) by }14,
\displaystyle 14a+42d=20.\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting (iii) from (ii),}
\displaystyle 49d=7.
\displaystyle \therefore d=\frac{1}{7}.
\displaystyle \text{Substituting }d=\frac{1}{7}\text{ in (i),}
\displaystyle a+3\left(\frac{1}{7}\right)=\frac{10}{7}.
\displaystyle a=\frac{7}{7}=1.
\displaystyle \therefore \text{The progression is }1,\frac{8}{7},\frac{9}{7},\frac{10}{7},\ldots
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{The third term of an A.P. is }7\text{ and the seventh term exceeds three}
\displaystyle \text{times the third term by }2.\text{ Find the first term, the common difference and the sum}
\displaystyle \text{of the first }20\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_3=7.
\displaystyle \text{Using }a_n=a+(n-1)d,
\displaystyle a+2d=7.\qquad\ldots\text{(i)}
\displaystyle \text{Also, }a_7-3a_3=2.
\displaystyle a_7-3(7)=2.
\displaystyle a_7=23.
\displaystyle \therefore a+6d=23.\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 4d=16.
\displaystyle \therefore d=4.
\displaystyle \text{Substituting }d=4\text{ in (i),}
\displaystyle a+2(4)=7.
\displaystyle \therefore a=-1.
\displaystyle \text{Using }S_n=\frac{n}{2}[2a+(n-1)d],
\displaystyle S_{20}=\frac{20}{2}[2(-1)+(20-1)(4)].
\displaystyle =10(-2+76)=10(74)=740.
\displaystyle \therefore \text{The first term is }-1,\text{ the common difference is }4\text{ and }S_{20}=740.
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{The first term of an A.P. is }2,\text{ the last term is }50\text{ and the sum of}
\displaystyle \text{all the terms is }442.\text{ Find the common difference.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=2,\qquad l=50,\qquad S_n=442.
\displaystyle \text{Using }l=a+(n-1)d,
\displaystyle 50=2+(n-1)d.
\displaystyle (n-1)d=48.\qquad\ldots\text{(i)}
\displaystyle \text{Also, }S_n=\frac{n}{2}(a+l).
\displaystyle 442=\frac{n}{2}(2+50).
\displaystyle 442=26n.
\displaystyle \therefore n=17.
\displaystyle \text{From (i),}
\displaystyle d=\frac{48}{17-1}=\frac{48}{16}=3.
\displaystyle \therefore \text{The common difference is }3.
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{The number of terms of an A.P. is even. The sum of the odd-positioned}
\displaystyle \text{terms is }24,\text{ the sum of the even-positioned terms is }30,\text{ and the last term}
\displaystyle \text{exceeds the first term by }10\frac{1}{2}.\text{ Find the number of terms and the A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total number of terms be }2n.
\displaystyle a_1+a_3+\cdots+a_{2n-1}=24.\qquad\ldots\text{(i)}
\displaystyle a_2+a_4+\cdots+a_{2n}=30.\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle (a_2-a_1)+(a_4-a_3)+\cdots+(a_{2n}-a_{2n-1})=6.
\displaystyle d+d+\cdots+d=6.
\displaystyle nd=6.\qquad\ldots\text{(iii)}
\displaystyle \text{Also, the last term exceeds the first term by }10\frac{1}{2}=\frac{21}{2}.
\displaystyle a_{2n}-a_1=\frac{21}{2}.
\displaystyle [a+(2n-1)d]-a=\frac{21}{2}.
\displaystyle (2n-1)d=\frac{21}{2}.
\displaystyle 2nd-d=\frac{21}{2}.
\displaystyle \text{Using }nd=6,
\displaystyle 12-d=\frac{21}{2}.
\displaystyle \therefore d=\frac{3}{2}.
\displaystyle \text{From (iii),}
\displaystyle n\left(\frac{3}{2}\right)=6.
\displaystyle \therefore n=4.
\displaystyle \therefore 2n=8.
\displaystyle \text{Hence, the A.P. contains }8\text{ terms.}
\displaystyle \text{Now, the even-positioned terms form the A.P.}
\displaystyle a+d,\;a+3d,\;a+5d,\ldots,a+(2n-1)d.
\displaystyle 30=\frac{n}{2}[(a+d)+\{a+(2n-1)d\}].
\displaystyle 30=\frac{4}{2}\left[\left(a+\frac{3}{2}\right)+\left(a+7\cdot\frac{3}{2}\right)\right].
\displaystyle 30=2(2a+12).
\displaystyle 15=2a+12.
\displaystyle 2a=3.
\displaystyle \therefore a=\frac{3}{2}.
\displaystyle \therefore \text{The A.P. is }\frac{3}{2},3,\frac{9}{2},6,\frac{15}{2},9,\frac{21}{2},12.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }S_n=n^2p\text{ and }S_m=m^2p,\ m\neq n,\text{ in an A.P., prove that}
\displaystyle S_p=p^3.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S_n=n^2p
\displaystyle \frac{n}{2}\left[2a+(n-1)d\right]=n^2p
\displaystyle 2a+(n-1)d=2np\qquad\ldots\text{(i)}
\displaystyle \text{Also, }S_m=m^2p
\displaystyle \frac{m}{2}\left[2a+(m-1)d\right]=m^2p
\displaystyle 2a+(m-1)d=2mp\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle (n-m)d=2p(n-m)
\displaystyle \because m\neq n,\quad n-m\neq0
\displaystyle \therefore d=2p\qquad\ldots\text{(iii)}
\displaystyle \text{Substituting }d=2p\text{ in (i),}
\displaystyle 2a+(n-1)(2p)=2np
\displaystyle 2a=2np-2np+2p
\displaystyle 2a=2p
\displaystyle \therefore a=p
\displaystyle S_p=\frac{p}{2}\left[2a+(p-1)d\right]
\displaystyle =\frac{p}{2}\left[2p+(p-1)(2p)\right]
\displaystyle =\frac{p}{2}\left(2p+2p^2-2p\right)
\displaystyle =\frac{p}{2}\times2p^2
\displaystyle \therefore S_p=p^3
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If the }12^{\text{th}}\text{ term of an A.P. is }-13\text{ and the sum of the first}
\displaystyle \text{four terms is }24,\text{ what is the sum of the first }10\text{ terms?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_{12}=-13\text{ and }S_4=24
\displaystyle a_n=a+(n-1)d
\displaystyle -13=a+(12-1)d
\displaystyle a+11d=-13\qquad\ldots\text{(i)}
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle 24=\frac{4}{2}\left[2a+(4-1)d\right]
\displaystyle 2a+3d=12\qquad\ldots\text{(ii)}
\displaystyle \text{From (i), }a=-13-11d
\displaystyle \text{Substituting in (ii),}
\displaystyle 2(-13-11d)+3d=12
\displaystyle -26-22d+3d=12
\displaystyle -19d=38
\displaystyle d=-2
\displaystyle \therefore a=-13-11(-2)=9
\displaystyle S_{10}=\frac{10}{2}\left[2(9)+(10-1)(-2)\right]
\displaystyle =5(18-18)
\displaystyle \therefore S_{10}=0
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If the }5^{\text{th}}\text{ and }12^{\text{th}}\text{ terms of an A.P. are }30\text{ and }65
\displaystyle \text{respectively, what is the sum of the first }20\text{ terms?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_5=30\text{ and }a_{12}=65
\displaystyle a_n=a+(n-1)d
\displaystyle 30=a+(5-1)d
\displaystyle a+4d=30\qquad\ldots\text{(i)}
\displaystyle 65=a+(12-1)d
\displaystyle a+11d=65\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (i) from (ii),}
\displaystyle 7d=35
\displaystyle d=5
\displaystyle \text{Substituting }d=5\text{ in (i),}
\displaystyle a+4(5)=30
\displaystyle a=10
\displaystyle S_{20}=\frac{20}{2}\left[2a+(20-1)d\right]
\displaystyle =10\left[2(10)+19(5)\right]
\displaystyle =10(20+95)
\displaystyle \therefore S_{20}=1150
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Find the sum of }n\text{ terms of the A.P. whose }k^{\text{th}}\text{ term is }5k+1.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_k=5k+1
\displaystyle \text{For }k=1,\quad a_1=5(1)+1=6
\displaystyle \text{For }k=2,\quad a_2=5(2)+1=11
\displaystyle \therefore d=a_2-a_1=11-6=5
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle =\frac{n}{2}\left[2(6)+(n-1)(5)\right]
\displaystyle =\frac{n}{2}\left(12+5n-5\right)
\displaystyle \therefore S_n=\frac{n}{2}(5n+7)
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Find the sum of all two-digit numbers which, when divided by }4,\text{ leave}
\displaystyle \text{a remainder of }1.
\displaystyle \text{Answer:}
\displaystyle \text{The required numbers are of the form }4q+1.
\displaystyle \text{Hence the A.P. is }13,17,21,\ldots,97.
\displaystyle a=13,\qquad d=4,\qquad a_n=97
\displaystyle a_n=a+(n-1)d
\displaystyle 97=13+(n-1)(4)
\displaystyle 84=4(n-1)
\displaystyle n-1=21
\displaystyle \therefore n=22
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle S_{22}=\frac{22}{2}\left[2(13)+(22-1)(4)\right]
\displaystyle =11(26+84)
\displaystyle =11\times110
\displaystyle \therefore S_{22}=1210
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If the sum of a certain number of terms of the A.P. }25,22,19,\ldots\text{ is}
\displaystyle 116,\text{ find the last term.}
\displaystyle \text{Answer:}
\displaystyle \text{The given A.P. is }25,22,19,\ldots
\displaystyle a=25,\qquad d=22-25=-3,\qquad S_n=116
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle 116=\frac{n}{2}\left[2(25)+(n-1)(-3)\right]
\displaystyle 232=n\left[50-3(n-1)\right]
\displaystyle 232=n(53-3n)
\displaystyle 3n^2-53n+232=0
\displaystyle (3n-29)(n-8)=0
\displaystyle n=\frac{29}{3}\quad\text{or}\quad n=8
\displaystyle \text{Since the number of terms must be a positive integer, }n=8.
\displaystyle a_8=a+(8-1)d
\displaystyle =25+7(-3)
\displaystyle =25-21
\displaystyle \therefore \text{The last term is }4.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Find the sum of all odd integers from }1\text{ to }2001.
\displaystyle \text{Answer:}
\displaystyle \text{The required A.P. is }1,3,5,\ldots,2001.
\displaystyle a=1,\qquad d=3-1=2,\qquad a_n=2001
\displaystyle a_n=a+(n-1)d
\displaystyle 2001=1+(n-1)(2)
\displaystyle 2000=2(n-1)
\displaystyle n-1=1000
\displaystyle n=1001
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle S_{1001}=\frac{1001}{2}\left[2(1)+(1001-1)(2)\right]
\displaystyle =\frac{1001}{2}(2+2000)
\displaystyle =\frac{1001}{2}\times2002
\displaystyle \therefore S_{1001}=1002001
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{How many terms of the A.P. }-6,-\frac{11}{2},-5,\ldots\text{ are needed to give}
\displaystyle \text{the sum }-25?
\displaystyle \text{Answer:}
\displaystyle \text{The given A.P. is }-6,-\frac{11}{2},-5,\ldots
\displaystyle a=-6,\qquad d=-\frac{11}{2}-(-6)=\frac{1}{2},\qquad S_n=-25
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle -25=\frac{n}{2}\left[2(-6)+(n-1)\left(\frac{1}{2}\right)\right]
\displaystyle -25=\frac{n}{2}\left(-12+\frac{n-1}{2}\right)
\displaystyle -25=\frac{n}{2}\left(\frac{n-25}{2}\right)
\displaystyle -100=n(n-25)
\displaystyle n^2-25n+100=0
\displaystyle (n-20)(n-5)=0
\displaystyle n=20\quad\text{or}\quad n=5
\displaystyle \therefore \text{The required number of terms is }5\text{ or }20.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In an A.P., the first term is }2\text{ and the sum of the first five terms is}
\displaystyle \text{one-fourth of the sum of the next five terms. Show that the }20^{\text{th}}\text{ term is }-112.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=2\text{ and }S_5=\frac{1}{4}(S_{10}-S_5)
\displaystyle S_n=\frac{n}{2}\left[2a+(n-1)d\right]
\displaystyle S_5=\frac{5}{2}\left[2(2)+(5-1)d\right]
\displaystyle =\frac{5}{2}(4+4d)
\displaystyle =5(2+2d)\qquad\ldots\text{(i)}
\displaystyle S_{10}=\frac{10}{2}\left[2(2)+(10-1)d\right]
\displaystyle =5(4+9d)\qquad\ldots\text{(ii)}
\displaystyle \text{Substituting (i) and (ii) in }S_5=\frac{1}{4}(S_{10}-S_5),
\displaystyle 5(2+2d)=\frac{1}{4}\left[5(4+9d)-5(2+2d)\right]
\displaystyle 4(2+2d)=(4+9d)-(2+2d)
\displaystyle 8+8d=2+7d
\displaystyle d=-6
\displaystyle a_n=a+(n-1)d
\displaystyle a_{20}=2+(20-1)(-6)
\displaystyle =2-114
\displaystyle \therefore a_{20}=-112
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }S_1\text{ is the sum of }(2n+1)\text{ terms of an A.P. and }S_2\text{ is the sum}
\displaystyle \text{of its odd-numbered terms, prove that }S_1:S_2=(2n+1):(n+1).
\displaystyle \text{Answer:}
\displaystyle \text{Let the A.P. be }a,a+d,a+2d,\ldots
\displaystyle S_1=\frac{2n+1}{2}\left[2a+\{(2n+1)-1\}d\right]
\displaystyle =\frac{2n+1}{2}(2a+2nd)
\displaystyle =(2n+1)(a+nd)\qquad\ldots\text{(i)}
\displaystyle \text{The odd-numbered terms are }a,a+2d,a+4d,\ldots,a+2nd.
\displaystyle \text{They form an A.P. containing }n+1\text{ terms with common difference }2d.
\displaystyle S_2=\frac{n+1}{2}\left[2a+\{(n+1)-1\}(2d)\right]
\displaystyle =\frac{n+1}{2}(2a+2nd)
\displaystyle =(n+1)(a+nd)\qquad\ldots\text{(ii)}
\displaystyle \text{Provided }a+nd\neq0,\text{ dividing (i) by (ii),}
\displaystyle \frac{S_1}{S_2}=\frac{(2n+1)(a+nd)}{(n+1)(a+nd)}
\displaystyle =\frac{2n+1}{n+1}
\displaystyle \therefore S_1:S_2=(2n+1):(n+1).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Find an A.P. in which the sum of any number of terms is three times the}
\displaystyle \text{square of the number of terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S_n=3n^2
\displaystyle a_n=S_n-S_{n-1}
\displaystyle =3n^2-3(n-1)^2
\displaystyle =3\left[n^2-(n^2-2n+1)\right]
\displaystyle =3(2n-1)
\displaystyle =6n-3
\displaystyle \text{For }n=1,\quad a_1=6(1)-3=3
\displaystyle \text{For }n=2,\quad a_2=6(2)-3=9
\displaystyle \text{For }n=3,\quad a_3=6(3)-3=15
\displaystyle \therefore d=a_2-a_1=9-3=6
\displaystyle \therefore \text{The required A.P. is }3,9,15,21,\ldots
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If the sum of }n\text{ terms of an A.P. is }nP+\frac{1}{2}n(n-1)Q,\text{ where}
\displaystyle P\text{ and }Q\text{ are constants, find the common difference.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S_n=nP+\frac{1}{2}n(n-1)Q
\displaystyle \text{For }n=1,\quad S_1=(1)P+\frac{1}{2}(1)(1-1)Q=P
\displaystyle \therefore a_1=P
\displaystyle \text{For }n=2,\quad S_2=(2)P+\frac{1}{2}(2)(2-1)Q=2P+Q
\displaystyle a_2=S_2-S_1=2P+Q-P=P+Q
\displaystyle \therefore d=a_2-a_1=(P+Q)-P=Q
\displaystyle \therefore \text{The common difference is }Q.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The sums of }n\text{ terms of two arithmetic progressions are in the ratio}
\displaystyle (5n+4):(9n+6).\text{ Find the ratio of their }18^{\text{th}}\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first terms of the two A.P.s be }a_1\text{ and }a_2,\text{ and their common differences}
\displaystyle \text{be }d_1\text{ and }d_2\text{ respectively.}
\displaystyle \text{Given, }\frac{S_n^{(1)}}{S_n^{(2)}}=\frac{5n+4}{9n+6}
\displaystyle \frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}=\frac{5n+4}{9n+6}
\displaystyle \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}
\displaystyle \text{To obtain the }18^{\text{th}}\text{ terms, take }n=2(18)-1=35.
\displaystyle \frac{2a_1+34d_1}{2a_2+34d_2}=\frac{5(35)+4}{9(35)+6}
\displaystyle \frac{2(a_1+17d_1)}{2(a_2+17d_2)}=\frac{179}{321}
\displaystyle \frac{a_1+(18-1)d_1}{a_2+(18-1)d_2}=\frac{179}{321}
\displaystyle \therefore \frac{\text{18}^{\text{th}}\text{ term of the first A.P.}}{\text{18}^{\text{th}}\text{ term of the second A.P.}}=\frac{179}{321}
\displaystyle \therefore \text{The required ratio is }179:321.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The sums of the first }n\text{ terms of two A.P.s are in the ratio}
\displaystyle (7n+2):(n+4).\text{ Find the ratio of their }5^{\text{th}}\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first terms of the two A.P.s be }a_1\text{ and }a_2,\text{ and their common differences}
\displaystyle \text{be }d_1\text{ and }d_2\text{ respectively.}
\displaystyle \text{Given, }\frac{S_n^{(1)}}{S_n^{(2)}}=\frac{7n+2}{n+4}
\displaystyle \frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}=\frac{7n+2}{n+4}
\displaystyle \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+2}{n+4}
\displaystyle \text{To obtain the }5^{\text{th}}\text{ terms, take }n=2(5)-1=9.
\displaystyle \frac{2a_1+8d_1}{2a_2+8d_2}=\frac{7(9)+2}{9+4}
\displaystyle \frac{2(a_1+4d_1)}{2(a_2+4d_2)}=\frac{65}{13}
\displaystyle \frac{a_1+(5-1)d_1}{a_2+(5-1)d_2}=\frac{65}{13}
\displaystyle \therefore \frac{5^{\text{th}}\text{ term of the first A.P.}}{5^{\text{th}}\text{ term of the second A.P.}}=\frac{65}{13}
\displaystyle \therefore \text{The required ratio is }5:1.
\displaystyle \\

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