\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{If the }n^{th}\text{ term of an A.P. is }3n-4,\text{ the }10^{th}\text{ term of A.P. is}
\displaystyle \text{(a) }12\qquad\text{(b) }22\qquad\text{(c) }26\qquad\text{(d) }30
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=3n-4.
\displaystyle \text{For the }10^{th}\text{ term, put }n=10.
\displaystyle a_{10}=3(10)-4=30-4=26.
\displaystyle \therefore \text{The }10^{th}\text{ term is }26.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\frac{2}{3},\ k,\ \frac{5}{8}\text{ are in A.P. then the value of }k\text{ is}
\displaystyle \text{(a) }\frac{31}{24}\qquad\text{(b) }\frac{31}{48}\qquad\text{(c) }\frac{24}{31}\qquad\text{(d) }\frac{48}{31}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\frac{2}{3},\ k,\ \frac{5}{8}\text{ are in A.P., the middle term is the arithmetic mean.}
\displaystyle 2k=\frac{2}{3}+\frac{5}{8}.
\displaystyle =\frac{16+15}{24}=\frac{31}{24}.
\displaystyle \therefore k=\frac{31}{48}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the third term of an A.P. is }7\text{ and its }7^{th}\text{ term is }2\text{ more than three}
\displaystyle \text{times of its third term, then the sum of its first }20\text{ terms is}
\displaystyle \text{(a) }228\qquad\text{(b) }74\qquad\text{(c) }740\qquad\text{(d) }1090
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle a+2d=7.\qquad ...(1)
\displaystyle \text{The }7^{th}\text{ term is }2\text{ more than three times the third term.}
\displaystyle a+6d=3(7)+2=23.\qquad ...(2)
\displaystyle \text{Subtracting (1) from (2), we get}
\displaystyle 4d=16\Rightarrow d=4.
\displaystyle \text{From (1), }a+8=7\Rightarrow a=-1.
\displaystyle S_{20}=\frac{20}{2}[2a+(20-1)d].
\displaystyle =10[2(-1)+19(4)].
\displaystyle =10[-2+76]=740.
\displaystyle \therefore \text{The sum of the first }20\text{ terms is }740.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{For a G.P. the ratio of the }7^{th}\text{ and the third terms is }16.\text{ What is the}
\displaystyle \text{common ratio?}
\displaystyle \text{(a) }2\qquad\text{(b) }\pm2\qquad\text{(c) }4\qquad\text{(d) }\pm4
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r.
\displaystyle T_7=ar^6,\qquad T_3=ar^2.
\displaystyle \text{Given, }\frac{T_7}{T_3}=16.
\displaystyle \frac{ar^6}{ar^2}=16.
\displaystyle r^4=16.
\displaystyle \therefore r=\pm2.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the sum of the first }2n\text{ terms of the A.P. }2,5,8,\ldots\text{ is equal to}
\displaystyle \text{the sum of the first }n\text{ terms of the A.P. }57,59,61,\ldots,\text{ then }n\text{ equals}
\displaystyle \text{(a) }10\qquad\text{(b) }12\qquad\text{(c) }11\qquad\text{(d) }13
\displaystyle \text{Answer:}
\displaystyle \text{For the first A.P., }a=2,\ d=3.
\displaystyle S_{2n}=\frac{2n}{2}[2(2)+(2n-1)3].
\displaystyle =n[4+6n-3]=n(6n+1).
\displaystyle \text{For the second A.P., }a=57,\ d=2.
\displaystyle S_n=\frac{n}{2}[2(57)+(n-1)2].
\displaystyle =\frac{n}{2}[114+2n-2]=n(n+56).
\displaystyle \text{Given, }n(6n+1)=n(n+56).
\displaystyle \text{Since }n>0,\quad6n+1=n+56.
\displaystyle 5n=55.
\displaystyle \therefore n=11.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{How many terms of G.P. }3,\ 3^2,\ 3^3,\ldots\text{ are needed to give the sum }120?
\displaystyle \text{(a) }3\qquad\text{(b) }4\qquad\text{(c) }5\qquad\text{(d) }6
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=3,\quad r=3,\quad S_n=120.
\displaystyle \text{Using }S_n=\frac{a(r^n-1)}{r-1},
\displaystyle 120=\frac{3(3^n-1)}{3-1}.
\displaystyle 240=3(3^n-1).
\displaystyle 80=3^n-1.
\displaystyle 3^n=81=3^4.
\displaystyle \therefore n=4.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Next term of the sequence }0.02,\ 0.006,\ 0.0018,\ldots\text{ is}
\displaystyle \text{(a) }0.000054\qquad\text{(b) }0.0054\qquad\text{(c) }0.00054\qquad\text{(d) }0.00036
\displaystyle \text{Answer:}
\displaystyle \frac{0.006}{0.02}=0.3,\qquad\frac{0.0018}{0.006}=0.3.
\displaystyle \therefore \text{the sequence is a G.P. with common ratio }r=0.3.
\displaystyle \text{Next term}=0.0018\times0.3=0.00054.
\displaystyle \therefore \text{The next term is }0.00054.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A man saves Rs. }135\text{ in the first year, Rs. }150\text{ in the second year}
\displaystyle \text{and in this way he increases his savings by Rs. }15\text{ every year. In what time will his total}
\displaystyle \text{savings be Rs. }5550\text{?}
\displaystyle \text{(a) }20\text{ years}\qquad\text{(b) }25\text{ years}\qquad\text{(c) }30\text{ years}\qquad\text{(d) }35\text{ years}
\displaystyle \text{Answer:}
\displaystyle \text{The yearly savings form an A.P. with }a=135,\quad d=15.
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d].
\displaystyle 5550=\frac{n}{2}[2(135)+(n-1)15].
\displaystyle 11100=n[270+15n-15].
\displaystyle 11100=n(15n+255).
\displaystyle 740=n(n+17).
\displaystyle n^2+17n-740=0.
\displaystyle (n+37)(n-20)=0.
\displaystyle \text{Since }n\text{ is positive, }n=20.
\displaystyle \therefore \text{The total savings will be Rs. }5550\text{ in }20\text{ years.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The fourth, seventh and tenth terms of a G.P. are }p,\ q,\ r
\displaystyle \text{respectively, then:}
\displaystyle \text{(a) }p^2=q^2+r^2\qquad\text{(b) }q^2=pr
\displaystyle \text{(c) }p^2=qr\qquad\text{(d) }pqr+pq+1=0
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }R.
\displaystyle p=T_4=aR^3,\qquad q=T_7=aR^6,\qquad r=T_{10}=aR^9.
\displaystyle q^2=(aR^6)^2=a^2R^{12}.
\displaystyle pr=(aR^3)(aR^9)=a^2R^{12}.
\displaystyle \therefore q^2=pr.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The }4^{th}\text{ term from the end of the G.P. }3,6,12,24,\ldots,3072\text{ is}
\displaystyle \text{(a) }348\qquad\text{(b) }843\qquad\text{(c) }438\qquad\text{(d) }384
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=3,\quad R=2.
\displaystyle \text{Let }3072\text{ be the }n^{th}\text{ term.}
\displaystyle 3072=3(2)^{n-1}.
\displaystyle 2^{n-1}=\frac{3072}{3}=1024=2^{10}.
\displaystyle \therefore n=11.
\displaystyle \text{The }4^{th}\text{ term from the end is the }(11-4+1)^{th}=8^{th}\text{ term from the beginning.}
\displaystyle T_8=3(2)^7=3\times128=384.
\displaystyle \therefore \text{The }4^{th}\text{ term from the end is }384.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED QUESTIONS}

\displaystyle \textbf{Question 11: }\text{150 workers were engaged to finish a job in a certain number of}
\displaystyle \text{days. }4\text{ workers dropped out on second day, }4\text{ more workers dropped out on third day}
\displaystyle \text{and so on. It took }8\text{ more days to finish the work.}
\displaystyle \text{(i) Find the A.P. representing the above situation.}
\displaystyle \text{(ii) In how many days was the work completed?}
\displaystyle \text{Answer:}

\displaystyle \text{(i) The number of workers on successive days is}
\displaystyle 150,\ 146,\ 142,\ 138,\ldots
\displaystyle \therefore \text{the required A.P. is }150,\ 146,\ 142,\ 138,\ldots
\displaystyle \text{with first term }a=150\text{ and common difference }d=-4.

\displaystyle \text{(ii) Let the work be originally planned to be completed in }n\text{ days.}
\displaystyle \text{Then total work}=150n\text{ worker-days.}
\displaystyle \text{Since the work actually took }8\text{ more days, the actual number of days is }n+8.
\displaystyle \text{The number of workers each day forms the A.P. }150,146,142,\ldots
\displaystyle \therefore \text{actual total work}=\frac{n+8}{2}[2(150)+(n+8-1)(-4)].
\displaystyle 150n=\frac{n+8}{2}[300-4(n+7)].
\displaystyle 150n=\frac{n+8}{2}(272-4n).
\displaystyle 300n=(n+8)(272-4n).
\displaystyle 300n=272n+2176-4n^2-32n.
\displaystyle 4n^2+60n-2176=0.
\displaystyle n^2+15n-544=0.
\displaystyle (n+32)(n-17)=0.
\displaystyle \text{Since }n\text{ is positive, }n=17.
\displaystyle \therefore \text{actual number of days}=17+8=25.
\displaystyle \therefore \text{The work was completed in }25\text{ days.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A company produces }500\text{ computers in the third year and }600
\displaystyle \text{computers in the seventh year. Assuming that the production increases uniformly by a constant}
\displaystyle \text{number every year, answer the following questions.}

\displaystyle \text{(i) How many computers were produced in the first year?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first-year production be }a\text{ and the annual increase be }d.
\displaystyle a+2d=500.\qquad ...(1)
\displaystyle a+6d=600.\qquad ...(2)
\displaystyle \text{Subtracting (1) from (2), we get}
\displaystyle 4d=100.
\displaystyle \therefore d=25.
\displaystyle \text{From (1), }a+2(25)=500.
\displaystyle \therefore a=450.
\displaystyle \therefore \text{450 computers were produced in the first year.}
\displaystyle \\

\displaystyle \text{(ii) By what number does the production increase every year?}
\displaystyle \text{Answer:}
\displaystyle d=25.
\displaystyle \therefore \text{The production increases by }25\text{ computers every year.}
\displaystyle \\

\displaystyle \text{(iii) How many computers will be produced in the }21^{st}\text{ year?}
\displaystyle \text{Answer:}
\displaystyle a_{21}=a+20d.
\displaystyle =450+20(25).
\displaystyle =450+500=950.
\displaystyle \therefore 950\text{ computers will be produced in the }21^{st}\text{ year.}
\displaystyle \\

\displaystyle \text{(iv) Find the total production in }10\text{ years.}
\displaystyle \text{Answer:}
\displaystyle S_{10}=\frac{10}{2}[2(450)+(10-1)(25)].
\displaystyle =5[900+225].
\displaystyle =5(1125)=5625.
\displaystyle \therefore \text{The total production in }10\text{ years is }5625\text{ computers.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Each side of an equilateral triangle is }24\text{ cm. The mid-points of its sides are}
\displaystyle \text{joined to form another triangle. This process is going on continuously.} \displaystyle \text{Based on the above information, answer the following questions.}

\displaystyle \text{(i) What is the length of the side of the fifth triangle?}
\displaystyle \text{Answer:}
\displaystyle \text{The sides of the successive triangles are}
\displaystyle 24,\ 12,\ 6,\ 3,\ldots
\displaystyle \text{This forms a G.P. with }a=24\text{ and }r=\frac{1}{2}.
\displaystyle T_n=ar^{n-1}.
\displaystyle \therefore T_5=24\left(\frac{1}{2}\right)^4.
\displaystyle =24\times\frac{1}{16}=\frac{3}{2}\text{ cm}.
\displaystyle \therefore \text{The length of the side of the fifth triangle is }\frac{3}{2}\text{ cm}.
\displaystyle \\

\displaystyle \text{(ii) What is the sum of the perimeters of the first six triangles?}
\displaystyle \text{Answer:}
\displaystyle \text{The perimeters of the successive triangles are}
\displaystyle 72,\ 36,\ 18,\ 9,\ldots
\displaystyle \text{This forms a G.P. with }a=72\text{ and }r=\frac{1}{2}.
\displaystyle S_6=\frac{a(1-r^6)}{1-r}.
\displaystyle =\frac{72\left[1-\left(\frac{1}{2}\right)^6\right]}{1-\frac{1}{2}}.
\displaystyle =144\left(1-\frac{1}{64}\right).
\displaystyle =144\times\frac{63}{64}=\frac{567}{4}\text{ cm}.
\displaystyle \therefore \text{The sum of the perimeters of the first six triangles is }\frac{567}{4}\text{ cm}.
\displaystyle \therefore \text{The required sum is }141.75\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Rahul being a plant lover decides to open a nursery and he bought few plants}
\displaystyle \text{with pots. He wants to place pots in such a way that number of pots in first row is }2,\text{ in second}
\displaystyle \text{row is }4\text{ and in third row is }8\text{ and so on.}
\displaystyle \text{Answer the following questions based on the above information.}

\displaystyle \text{(i) Find the number of pots in the }8^{th}\text{ row.}
\displaystyle \text{Answer:}
\displaystyle \text{The number of pots in successive rows is }2,\ 4,\ 8,\ 16,\ldots
\displaystyle \text{This forms a G.P. with }a=2\text{ and }r=2.
\displaystyle T_n=ar^{n-1}.
\displaystyle \therefore T_8=2(2)^7=2^8=256.
\displaystyle \therefore \text{The number of pots in the }8^{th}\text{ row is }256.
\displaystyle \\

\displaystyle \text{(ii) Find the total number of pots in }10\text{ rows.}
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{a(r^n-1)}{r-1}.
\displaystyle \therefore S_{10}=\frac{2(2^{10}-1)}{2-1}.
\displaystyle =2(1024-1)=2(1023)=2046.
\displaystyle \therefore \text{The total number of pots in }10\text{ rows is }2046.
\displaystyle \\

\displaystyle \text{(iii) If Rahul wants to place }510\text{ pots in all, how many rows will be formed?}
\displaystyle \text{Answer:}
\displaystyle S_n=510.
\displaystyle \frac{2(2^n-1)}{2-1}=510.
\displaystyle 2(2^n-1)=510.
\displaystyle 2^n-1=255.
\displaystyle 2^n=256=2^8.
\displaystyle \therefore n=8.
\displaystyle \therefore \text{Rahul will form }8\text{ rows.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Asif buys a scooter for Rs. }22000.\text{ He pays Rs. }4000\text{ in cash and}
\displaystyle \text{agrees to pay the balance in annual installments of Rs. }1000\text{ plus }10\%\text{ interest on the unpaid amount.}
\displaystyle \text{Based on the above information, answer the following questions.}

\displaystyle \text{(i) What will be the first installment?}
\displaystyle \text{Answer:}
\displaystyle \text{Amount remaining after the cash payment}=22000-4000=18000.
\displaystyle \text{Interest for the first year}=10\%\text{ of }18000=1800.
\displaystyle \text{First installment}=1000+1800=2800.
\displaystyle \therefore \text{The first installment is Rs. }2800.
\displaystyle \\

\displaystyle \text{(ii) In how many installments will he be able to repay the loan?}
\displaystyle \text{Answer:}
\displaystyle \text{The unpaid principal is Rs. }18000\text{ and Rs. }1000\text{ of principal is repaid in each installment.}
\displaystyle \therefore \text{Number of installments}=\frac{18000}{1000}=18.
\displaystyle \therefore \text{He will repay the loan in }18\text{ installments.}
\displaystyle \\

\displaystyle \text{(iii) Find the total amount paid as installments.}
\displaystyle \text{Answer:}
\displaystyle \text{The interest amounts are }1800,\ 1700,\ 1600,\ldots,100.
\displaystyle \text{Therefore, the installments are }2800,\ 2700,\ 2600,\ldots,1100.
\displaystyle \text{These installments form an A.P. with }a=2800,\ d=-100,\ n=18.
\displaystyle S_{18}=\frac{18}{2}[2(2800)+(18-1)(-100)].
\displaystyle =9[5600-1700].
\displaystyle =9(3900)=35100.
\displaystyle \therefore \text{The total amount paid as installments is Rs. }35100.
\displaystyle \\

\displaystyle \text{VERY SHORT QUESTIONS}

\displaystyle \textbf{Question 16: }\text{If }S_n=3n^2+2n,\text{ then write }a_2.
\displaystyle \text{Answer:}
\displaystyle a_2=S_2-S_1.
\displaystyle S_2=3(2)^2+2(2)=16,\qquad S_1=3(1)^2+2(1)=5.
\displaystyle \therefore a_2=16-5=11.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If sum of first }n\text{ terms of an A.P. is }2n^2+7n,\text{ write its }n^{th}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle =2n^2+7n-\left[2(n-1)^2+7(n-1)\right].
\displaystyle =2n^2+7n-(2n^2+3n-5).
\displaystyle \therefore a_n=4n+5.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If in a G.P., }a_3+a_5=90\text{ and if }r=2,\text{ find the first term of the G.P.}
\displaystyle \text{Answer:}
\displaystyle a_3=ar^2,\qquad a_5=ar^4.
\displaystyle \therefore ar^2+ar^4=90.
\displaystyle a(2^2+2^4)=90.
\displaystyle 20a=90.
\displaystyle \therefore a=\frac{9}{2}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If the product of }3\text{ consecutive terms of G.P. is }27,\text{ find the middle term.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive terms be }\frac{a}{r},\ a,\ ar.
\displaystyle \frac{a}{r}\times a\times ar=a^3=27.
\displaystyle \therefore a=3.
\displaystyle \therefore \text{The middle term is }3.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In an A.P. }8,11,14,\ldots\text{ find }S_n-S_{n-1}.
\displaystyle \text{Answer:}
\displaystyle S_n-S_{n-1}=a_n.
\displaystyle a_n=a+(n-1)d.
\displaystyle =8+(n-1)3.
\displaystyle \therefore S_n-S_{n-1}=3n+5.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The first term of a G.P. is }2\text{ and sum to infinity is }6,\text{ find common ratio.}
\displaystyle \text{Answer:}
\displaystyle S_{\infty}=\frac{a}{1-r}.
\displaystyle 6=\frac{2}{1-r}.
\displaystyle 1-r=\frac{1}{3}.
\displaystyle \therefore r=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The A.M. of two numbers is }34\text{ and G.M. is }16,\text{ the numbers are}
\displaystyle \text{\_\_\_\_\_\_\_ and \_\_\_\_\_\_\_.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numbers be }x\text{ and }y.
\displaystyle \frac{x+y}{2}=34\Rightarrow x+y=68.
\displaystyle \sqrt{xy}=16\Rightarrow xy=256.
\displaystyle \therefore t^2-68t+256=0.
\displaystyle (t-64)(t-4)=0.
\displaystyle \therefore \text{The numbers are }64\text{ and }4.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The sum of first }10\text{ terms of G.P. is equal to }244\text{ times the sum of}
\displaystyle \text{first five terms. Then the common ratio is \_\_\_\_\_\_\_.}
\displaystyle \text{Answer:}
\displaystyle S_{10}=244S_5.
\displaystyle \frac{a(1-r^{10})}{1-r}=244\frac{a(1-r^5)}{1-r}.
\displaystyle 1-r^{10}=244(1-r^5).
\displaystyle (1-r^5)(1+r^5)=244(1-r^5).
\displaystyle \therefore 1+r^5=244.
\displaystyle r^5=243=3^5.
\displaystyle \therefore r=3.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The third term of a geometric progression is }4.\text{ Find the product of}
\displaystyle \text{the first five terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r.
\displaystyle T_3=ar^2=4.
\displaystyle \text{Product of the first five terms}
\displaystyle =a(ar)(ar^2)(ar^3)(ar^4).
\displaystyle =a^5r^{10}=(ar^2)^5.
\displaystyle =4^5=1024.
\displaystyle \therefore \text{The product of the first five terms is }1024.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The first term of a G.P. is }1.\text{ The sum of the third term and fifth term}
\displaystyle \text{is }90.\text{ The common ratio of G.P. is \_\_\_\_\_\_\_.}
\displaystyle \text{Answer:}
\displaystyle a=1.
\displaystyle T_3=ar^2=r^2,\qquad T_5=ar^4=r^4.
\displaystyle r^2+r^4=90.
\displaystyle \text{Let }x=r^2.
\displaystyle x^2+x-90=0.
\displaystyle (x+10)(x-9)=0.
\displaystyle \text{Since }x=r^2\geq0,\quad x=9.
\displaystyle \therefore r^2=9.
\displaystyle \therefore r=\pm3.
\displaystyle \\

\displaystyle \text{ASSERTION REASON QUESTIONS}

\displaystyle \textbf{Question 26: }\text{Assertion (A): If }a_1,a_2,a_3,\ldots,a_n,\ldots\text{ is an A.P. such that}
\displaystyle a_1+a_4+a_7+\cdots+a_{16}=147,\text{ then }a_1+a_6+a_{11}+a_{16}=98.
\displaystyle \text{Reason (R): In an A.P., the sum of the terms equidistant from the beginning and the end}
\displaystyle \text{is always same and is equal to the sum of first and last term.}

\displaystyle \text{(a) Both A and R are true. R is a correct explanation for A.}
\displaystyle \text{(b) Both A and R are true. R is not a correct explanation for A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle a_1+a_4+a_7+a_{10}+a_{13}+a_{16}=147.
\displaystyle \text{The terms }a_1,a_4,a_7,a_{10},a_{13},a_{16}\text{ form an A.P. of }6\text{ terms.}
\displaystyle \text{Using the property of terms equidistant from the beginning and end,}
\displaystyle a_1+a_{16}=a_4+a_{13}=a_7+a_{10}.
\displaystyle \therefore 3(a_1+a_{16})=147.
\displaystyle \therefore a_1+a_{16}=49.
\displaystyle \text{Also, }a_1,a_6,a_{11},a_{16}\text{ form an A.P. of }4\text{ terms.}
\displaystyle a_1+a_{16}=a_6+a_{11}=49.
\displaystyle \therefore a_1+a_6+a_{11}+a_{16}=49+49=98.
\displaystyle \therefore \text{Assertion A is true.}
\displaystyle \text{Reason R is also true and correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion (A): If }n^{th}\text{ term of a sequence is }a_n=\frac{n^2}{2^n},\text{ then its }7^{th}\text{ term is}
\displaystyle \frac{49}{128}.
\displaystyle \text{Reason (R): If }n^{th}\text{ term of a sequence is }a_n=\frac{n(n-2)}{n+3},\text{ then its }20^{th}\text{ term is }\frac{323}{22}.

\displaystyle \text{(a) Both A and R are true. R is a correct explanation for A.}
\displaystyle \text{(b) Both A and R are true. R is not a correct explanation for A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For Assertion A, }a_n=\frac{n^2}{2^n}.
\displaystyle \therefore a_7=\frac{7^2}{2^7}=\frac{49}{128}.
\displaystyle \therefore \text{Assertion A is true.}

\displaystyle \text{For Reason R, }a_n=\frac{n(n-2)}{n+3}.
\displaystyle \therefore a_{20}=\frac{20(20-2)}{20+3}=\frac{360}{23}.
\displaystyle \frac{360}{23}\ne\frac{323}{22}.
\displaystyle \therefore \text{Reason R is false.}
\displaystyle \therefore \text{A is true but R is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion (A): If the numbers }\frac{-2}{7},\ k,\ \frac{-7}{2}\text{ are in G.P., then }k=\pm1.
\displaystyle \text{Reason (R): If }a,\ b,\ c\text{ are in G.P., then }ac=b^2.

\displaystyle \text{(a) Both A and R are true. R is a correct explanation for A.}
\displaystyle \text{(b) Both A and R are true. R is not a correct explanation for A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For three numbers }a,\ b,\ c\text{ in G.P., }b^2=ac.
\displaystyle \therefore k^2=\left(\frac{-2}{7}\right)\left(\frac{-7}{2}\right).
\displaystyle k^2=1.
\displaystyle \therefore k=\pm1.
\displaystyle \therefore \text{Assertion A is true.}
\displaystyle \text{Reason R is also true and directly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Let }a_1,a_2,a_3,\ldots,a_{n-1},a_n\text{ be an A.P.}
\displaystyle \text{Assertion (A): }a_1+a_2+a_3+\cdots+a_n=\frac{n}{2}(a_1+a_n).
\displaystyle \text{Reason (R): }a_k+a_{n-k+1}=a_1+a_n,\text{ for }k=1,2,3,\ldots,n.

\displaystyle \text{(a) Both A and R are true. R is a correct explanation for A.}
\displaystyle \text{(b) Both A and R are true. R is not a correct explanation for A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }S_n=a_1+a_2+a_3+\cdots+a_n.
\displaystyle \text{Writing the sum in reverse order,}
\displaystyle S_n=a_n+a_{n-1}+a_{n-2}+\cdots+a_1.
\displaystyle \text{In an A.P., terms equidistant from the beginning and end have the same sum.}
\displaystyle a_k+a_{n-k+1}=a_1+a_n.
\displaystyle \text{Adding the two expressions for }S_n,
\displaystyle 2S_n=n(a_1+a_n).
\displaystyle \therefore S_n=\frac{n}{2}(a_1+a_n).
\displaystyle \therefore \text{Assertion A and Reason R are both true.}
\displaystyle \text{Reason R correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion (A): The sum of the first }22\text{ terms of the A.P. }16,11,6,\ldots\text{ is }-803.
\displaystyle \text{Reason (R): The sum of the first }22\text{ terms of the A.P. }x+y,\ x-y,\ x-3y,\ldots\text{ is}
\displaystyle 22(x-20y).

\displaystyle \text{(a) Both A and R are true. R is a correct explanation for A.}
\displaystyle \text{(b) Both A and R are true. R is not a correct explanation for A.}
\displaystyle \text{(c) A is true but R is false.}
\displaystyle \text{(d) A is false but R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For Assertion A, }a=16,\quad d=11-16=-5,\quad n=22.
\displaystyle S_{22}=\frac{22}{2}[2(16)+(22-1)(-5)].
\displaystyle =11[32-105]=11(-73)=-803.
\displaystyle \therefore \text{Assertion A is true.}

\displaystyle \text{For Reason R, }a=x+y,\quad d=(x-y)-(x+y)=-2y.
\displaystyle S_{22}=\frac{22}{2}[2(x+y)+21(-2y)].
\displaystyle =11[2x+2y-42y].
\displaystyle =11(2x-40y)=22(x-20y).
\displaystyle \therefore \text{Reason R is true.}
\displaystyle \text{Also, putting }x=15\text{ and }y=1\text{ in Reason R gives the A.P. }16,11,6,\ldots
\displaystyle \text{and }S_{22}=22(15-20)=-110,\text{ which does not give }-803.
\displaystyle \text{Thus, Reason R does not correctly explain Assertion A.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{SHORT QUESTIONS}

\displaystyle \textbf{Question 31: }\text{Write the first negative term of the sequence }20,\ 19\frac{1}{4},\ 18\frac{1}{2},\ 17\frac{3}{4},\ldots
\displaystyle \text{Answer:}
\displaystyle a=20,\qquad d=-\frac{3}{4}.
\displaystyle a_n=20+(n-1)\left(-\frac{3}{4}\right).
\displaystyle \text{For the first negative term, }a_n<0.
\displaystyle 20-\frac{3}{4}(n-1)<0.
\displaystyle 80-3n+3<0\Rightarrow83<3n.
\displaystyle \therefore n>\frac{83}{3}=27\frac{2}{3}.
\displaystyle \therefore \text{the first negative term is the }28^{th}\text{ term.}
\displaystyle a_{28}=20-\frac{3}{4}(27)=-\frac{1}{4}.
\displaystyle \therefore \text{The first negative term is }-\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{How many numbers are there between }200\text{ and }500,\text{ which leave}
\displaystyle \text{remainder }7\text{ when divided by }9\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{The required numbers are }205,\ 214,\ 223,\ldots,493.
\displaystyle \text{This is an A.P. with }a=205,\quad d=9,\quad l=493.
\displaystyle 493=205+(n-1)9.
\displaystyle 288=9(n-1).
\displaystyle n-1=32.
\displaystyle \therefore n=33.
\displaystyle \therefore \text{There are }33\text{ such numbers.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{In an A.P., sum of first }4\text{ terms is }56\text{ and the sum of last }4\text{ terms is}
\displaystyle 112.\text{ If the first term is }11\text{ then find the number of terms.}
\displaystyle \text{Answer:}
\displaystyle a=11.
\displaystyle 11+(11+d)+(11+2d)+(11+3d)=56.
\displaystyle 44+6d=56.
\displaystyle \therefore d=2.
\displaystyle \text{Let the last term be }l.
\displaystyle l+(l-d)+(l-2d)+(l-3d)=112.
\displaystyle 4l-6d=112.
\displaystyle 4l-12=112\Rightarrow l=31.
\displaystyle l=a+(n-1)d.
\displaystyle 31=11+2(n-1).
\displaystyle 20=2(n-1)\Rightarrow n=11.
\displaystyle \therefore \text{The number of terms is }11.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The product of first three terms of a G.P. is }1000.\text{ If }6\text{ is added to its}
\displaystyle \text{second term and }7\text{ is added to its third term, the terms become in A.P. Find the G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive terms of the G.P. be }\frac{a}{r},\ a,\ ar.
\displaystyle \frac{a}{r}\times a\times ar=a^3=1000.
\displaystyle \therefore a=10.
\displaystyle \text{Thus, the terms are }\frac{10}{r},\ 10,\ 10r.
\displaystyle \text{After the given additions, }\frac{10}{r},\ 16,\ 10r+7\text{ are in A.P.}
\displaystyle \therefore 2(16)=\frac{10}{r}+10r+7.
\displaystyle 25=\frac{10}{r}+10r.
\displaystyle 25r=10+10r^2.
\displaystyle 2r^2-5r+2=0.
\displaystyle (2r-1)(r-2)=0.
\displaystyle \therefore r=\frac{1}{2}\text{ or }r=2.
\displaystyle \text{For }r=2,\text{ the G.P. is }5,\ 10,\ 20.
\displaystyle \text{For }r=\frac{1}{2},\text{ the G.P. is }20,\ 10,\ 5.
\displaystyle \therefore \text{The required G.P. is }5,10,20\text{ or }20,10,5.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Insert }5\text{ numbers between }7\text{ and }55,\text{ so that resulting series is A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{After inserting }5\text{ numbers, the A.P. contains }7\text{ terms.}
\displaystyle a=7,\qquad a_7=55.
\displaystyle a_7=a+6d.
\displaystyle 55=7+6d.
\displaystyle 6d=48.
\displaystyle \therefore d=8.
\displaystyle \text{Hence, the A.P. is }7,\ 15,\ 23,\ 31,\ 39,\ 47,\ 55.
\displaystyle \therefore \text{The five numbers are }15,\ 23,\ 31,\ 39,\ 47.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The sum of first three terms of a G.P. is }15\text{ and sum of next three}
\displaystyle \text{terms is }120.\text{ Find the sum of first }n\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r.
\displaystyle a+ar+ar^2=15.
\displaystyle a(1+r+r^2)=15.\qquad ...(1)
\displaystyle ar^3+ar^4+ar^5=120.
\displaystyle ar^3(1+r+r^2)=120.\qquad ...(2)
\displaystyle \text{Dividing (2) by (1), we get}
\displaystyle r^3=\frac{120}{15}=8.
\displaystyle \therefore r=2.
\displaystyle \text{From (1), }a(1+2+4)=15.
\displaystyle 7a=15.
\displaystyle \therefore a=\frac{15}{7}.
\displaystyle S_n=\frac{a(r^n-1)}{r-1}.
\displaystyle =\frac{\frac{15}{7}(2^n-1)}{2-1}.
\displaystyle \therefore S_n=\frac{15}{7}(2^n-1).
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Find the sum of all the natural numbers between }1\text{ and }200\text{ which}
\displaystyle \text{are neither divisible by }2\text{ nor by }5.
\displaystyle \text{Answer:}
\displaystyle \text{Sum of natural numbers from }1\text{ to }200=\frac{200(201)}{2}=20100.
\displaystyle \text{Sum of numbers divisible by }2=2(1+2+\cdots+100).
\displaystyle =2\left(\frac{100(101)}{2}\right)=10100.
\displaystyle \text{Sum of numbers divisible by }5=5(1+2+\cdots+40).
\displaystyle =5\left(\frac{40(41)}{2}\right)=4100.
\displaystyle \text{Numbers divisible by both }2\text{ and }5\text{ are divisible by }10.
\displaystyle \text{Their sum}=10(1+2+\cdots+20).
\displaystyle =10\left(\frac{20(21)}{2}\right)=2100.
\displaystyle \text{Therefore, sum of numbers divisible by }2\text{ or }5
\displaystyle =10100+4100-2100=12100.
\displaystyle \therefore \text{Required sum}=20100-12100=8000.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If in an A.P., }\frac{a_7}{a_{10}}=\frac{5}{7},\text{ find }\frac{a_4}{a_7}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle a_7=a+6d,\qquad a_{10}=a+9d.
\displaystyle \frac{a+6d}{a+9d}=\frac{5}{7}.
\displaystyle 7(a+6d)=5(a+9d).
\displaystyle 7a+42d=5a+45d.
\displaystyle \therefore 2a=3d.
\displaystyle \therefore a=\frac{3d}{2}.
\displaystyle \frac{a_4}{a_7}=\frac{a+3d}{a+6d}.
\displaystyle =\frac{\frac{3d}{2}+3d}{\frac{3d}{2}+6d}.
\displaystyle =\frac{\frac{9d}{2}}{\frac{15d}{2}}=\frac{3}{5}.
\displaystyle \therefore \frac{a_4}{a_7}=\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Using G.P. prove that }0.031111\ldots=\frac{7}{225}.
\displaystyle \text{Hint: }0.031111\ldots=0.03+0.001+0.0001+\cdots\text{. Now use infinite G.P.}
\displaystyle \text{Answer:}
\displaystyle 0.031111\ldots=0.03+(0.001+0.0001+0.00001+\cdots).
\displaystyle \text{The terms }0.001,\ 0.0001,\ 0.00001,\ldots\text{ form an infinite G.P.}
\displaystyle \text{Here, }a=0.001=\frac{1}{1000},\qquad r=\frac{1}{10}.
\displaystyle S_{\infty}=\frac{a}{1-r}.
\displaystyle =\frac{\frac{1}{1000}}{1-\frac{1}{10}}=\frac{1}{900}.
\displaystyle \therefore 0.031111\ldots=\frac{3}{100}+\frac{1}{900}.
\displaystyle =\frac{27+1}{900}=\frac{28}{900}=\frac{7}{225}.
\displaystyle \therefore 0.031111\ldots=\frac{7}{225}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Solve: }1+6+11+16+\cdots+x=148.
\displaystyle \text{Answer:}
\displaystyle \text{The terms form an A.P. with }a=1\text{ and }d=5.
\displaystyle \text{Let there be }n\text{ terms. Then}
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d].
\displaystyle 148=\frac{n}{2}[2+5(n-1)].
\displaystyle 296=n(5n-3).
\displaystyle 5n^2-3n-296=0.
\displaystyle (5n+37)(n-8)=0.
\displaystyle \text{Since }n\text{ is positive, }n=8.
\displaystyle x=a+(n-1)d.
\displaystyle =1+7(5)=36.
\displaystyle \therefore x=36.
\displaystyle \\

\displaystyle \text{LONG QUESTIONS}

\displaystyle \textbf{Question 41: }\text{A square is drawn by joining the mid points of the sides of a square.}
\displaystyle \text{A third square is drawn inside the second square in the same way and the process is continued}
\displaystyle \text{indefinitely. If the side of the first square is }15\text{ cm, then find the sum of the areas of all}
\displaystyle \text{the squares so formed.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of the first square}=15^2=225\text{ cm}^2.
\displaystyle \text{When the mid-points of the sides of a square are joined, the area of the new square is}
\displaystyle \frac{1}{2}\text{ of the area of the original square.}
\displaystyle \therefore \text{the areas form the G.P.}
\displaystyle 225,\ \frac{225}{2},\ \frac{225}{4},\ldots
\displaystyle \text{Here, }a=225,\qquad r=\frac{1}{2}.
\displaystyle S_{\infty}=\frac{a}{1-r}.
\displaystyle =\frac{225}{1-\frac{1}{2}}.
\displaystyle =\frac{225}{\frac{1}{2}}=450.
\displaystyle \therefore \text{The sum of the areas of all the squares is }450\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }a\text{ is arithmetic mean of }b\text{ and }c\text{ and }c,\ G_1,\ G_2,\ b\text{ are in G.P., then}
\displaystyle \text{prove that }G_1^3+G_2^3=2abc.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a\text{ is the arithmetic mean of }b\text{ and }c,
\displaystyle 2a=b+c.\qquad ...(1)
\displaystyle \text{Let the common ratio of the G.P. }c,\ G_1,\ G_2,\ b\text{ be }r.
\displaystyle \therefore G_1=cr,\qquad G_2=cr^2,\qquad b=cr^3.
\displaystyle G_1^3+G_2^3=(cr)^3+(cr^2)^3.
\displaystyle =c^3r^3+c^3r^6.
\displaystyle =c^3r^3(1+r^3).
\displaystyle =bc(c+b).
\displaystyle \text{Using (1), }b+c=2a.
\displaystyle \therefore G_1^3+G_2^3=bc(2a)=2abc.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If }p^{th},\ q^{th}\text{ and }r^{th}\text{ terms of an A.P. and G.P. both are }a,\ b\text{ and }c,
\displaystyle \text{respectively, then show that }a^{b-c}b^{c-a}c^{a-b}=1.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,\ b,\ c\text{ are the }p^{th},\ q^{th},\ r^{th}\text{ terms of an A.P.,}
\displaystyle a=A+(p-1)d,\qquad b=A+(q-1)d,\qquad c=A+(r-1)d.
\displaystyle \therefore \frac{b-a}{q-p}=\frac{c-a}{r-p}=d.
\displaystyle \therefore (b-a)(r-p)=(c-a)(q-p).
\displaystyle \therefore p(b-c)+q(c-a)+r(a-b)=0.\qquad ...(1)

\displaystyle \text{Now let the first term and common ratio of the G.P. be }B\text{ and }R.
\displaystyle a=BR^{p-1},\qquad b=BR^{q-1},\qquad c=BR^{r-1}.
\displaystyle a^{b-c}b^{c-a}c^{a-b}
\displaystyle =(BR^{p-1})^{b-c}(BR^{q-1})^{c-a}(BR^{r-1})^{a-b}.
\displaystyle =B^{(b-c)+(c-a)+(a-b)}
\displaystyle \qquad\times R^{(p-1)(b-c)+(q-1)(c-a)+(r-1)(a-b)}.
\displaystyle \text{The power of }B\text{ is }0.
\displaystyle \text{Also, by (1), the power of }R\text{ is }0.
\displaystyle \therefore a^{b-c}b^{c-a}c^{a-b}=B^0R^0=1.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Prove that the sum of }n\text{ numbers between }a\text{ and }b\text{ such that the}
\displaystyle \text{resulting series becomes A.P. is }\frac{n(a+b)}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }n\text{ arithmetic means }A_1,A_2,\ldots,A_n\text{ be inserted between }a\text{ and }b.
\displaystyle \therefore a,A_1,A_2,\ldots,A_n,b\text{ form an A.P. of }n+2\text{ terms.}
\displaystyle \text{In an A.P., the sum of terms equidistant from the beginning and end is constant.}
\displaystyle \therefore A_1+A_n=A_2+A_{n-1}=\cdots=a+b.
\displaystyle \text{Let }S=A_1+A_2+\cdots+A_n.
\displaystyle \text{Writing the sum in reverse order,}
\displaystyle S=A_n+A_{n-1}+\cdots+A_1.
\displaystyle \text{Adding the two expressions,}
\displaystyle 2S=n(a+b).
\displaystyle \therefore S=\frac{n(a+b)}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{If }a,b,c\text{ are in G.P., then prove that}
\displaystyle \frac{1}{a^2-b^2}=\frac{1}{b^2-c^2}-\frac{1}{b^2}.
\displaystyle \text{Hint: Put }b=ar,\ c=ar^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,b,c\text{ are in G.P., let the common ratio be }r.
\displaystyle \therefore b=ar,\qquad c=ar^2.
\displaystyle \text{Consider the right hand side:}
\displaystyle \frac{1}{b^2-c^2}-\frac{1}{b^2}.
\displaystyle =\frac{1}{a^2r^2-a^2r^4}-\frac{1}{a^2r^2}.
\displaystyle =\frac{1}{a^2r^2(1-r^2)}-\frac{1}{a^2r^2}.
\displaystyle =\frac{1-(1-r^2)}{a^2r^2(1-r^2)}.
\displaystyle =\frac{r^2}{a^2r^2(1-r^2)}.
\displaystyle =\frac{1}{a^2(1-r^2)}.
\displaystyle =\frac{1}{a^2-a^2r^2}.
\displaystyle =\frac{1}{a^2-b^2}.
\displaystyle \therefore \frac{1}{a^2-b^2}=\frac{1}{b^2-c^2}-\frac{1}{b^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{If the sum of }p\text{ terms of an A.P. is }q\text{ and the sum of }q\text{ terms is }p,
\displaystyle \text{then show that the sum of }p+q\text{ terms is }-(p+q).\text{ Also, find the sum of first}
\displaystyle p-q\text{ terms, where }p>q.
\displaystyle \text{Answer:}

\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d].

\displaystyle \text{Given, }S_p=q.
\displaystyle \frac{p}{2}[2a+(p-1)d]=q.
\displaystyle 2a+(p-1)d=\frac{2q}{p}.\qquad ...(1)

\displaystyle \text{Also, }S_q=p.
\displaystyle \frac{q}{2}[2a+(q-1)d]=p.
\displaystyle 2a+(q-1)d=\frac{2p}{q}.\qquad ...(2)

\displaystyle \text{Subtracting (2) from (1),}
\displaystyle (p-q)d=\frac{2q}{p}-\frac{2p}{q}.
\displaystyle =\frac{2(q^2-p^2)}{pq}.
\displaystyle =-\frac{2(p-q)(p+q)}{pq}.
\displaystyle \therefore d=-\frac{2(p+q)}{pq}.\qquad ...(3)

\displaystyle \text{Using (1),}
\displaystyle 2a=\frac{2q}{p}-(p-1)d.
\displaystyle =\frac{2q}{p}+\frac{2(p-1)(p+q)}{pq}.

\displaystyle \text{Now, }S_{p+q}=\frac{p+q}{2}[2a+(p+q-1)d].
\displaystyle =\frac{p+q}{2}\left[\frac{2q}{p}+qd\right].
\displaystyle =\frac{p+q}{2}\left[\frac{2q}{p}-\frac{2(p+q)}{p}\right].
\displaystyle =\frac{p+q}{2}\left(-2\right).
\displaystyle \therefore S_{p+q}=-(p+q).
\displaystyle \text{Hence proved.}

\displaystyle \text{Also, }S_{p-q}=\frac{p-q}{2}[2a+(p-q-1)d].
\displaystyle =\frac{p-q}{2}\left[\frac{2q}{p}-qd\right].
\displaystyle =\frac{p-q}{2}\left[\frac{2q}{p}+\frac{2(p+q)}{p}\right].
\displaystyle =\frac{p-q}{2}\left[\frac{2(p+2q)}{p}\right].
\displaystyle \therefore S_{p-q}=\frac{(p-q)(p+2q)}{p}.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }A\text{ is the arithmetic mean and }G_1,\ G_2\text{ are two geometric means}
\displaystyle \text{between any two numbers, then prove that }2A=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}.
\displaystyle \text{Answer:}

\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Since }A\text{ is the arithmetic mean of }a\text{ and }b,
\displaystyle A=\frac{a+b}{2}.
\displaystyle \therefore 2A=a+b.\qquad ...(1)

\displaystyle \text{Since }G_1\text{ and }G_2\text{ are two geometric means between }a\text{ and }b,
\displaystyle a,\ G_1,\ G_2,\ b\text{ are in G.P.}
\displaystyle \text{Let the common ratio be }r.
\displaystyle \therefore G_1=ar,\qquad G_2=ar^2,\qquad b=ar^3.

\displaystyle \frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}
\displaystyle =\frac{(ar)^2}{ar^2}+\frac{(ar^2)^2}{ar}.
\displaystyle =a+ar^3.
\displaystyle =a+b.

\displaystyle \text{Using (1), }a+b=2A.
\displaystyle \therefore \frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=2A.
\displaystyle \therefore 2A=\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Show that the sum of }(m+n)^{th}\text{ and }(m-n)^{th}\text{ terms of an A.P. is}
\displaystyle \text{equal to twice the }m^{th}\text{ term.}
\displaystyle \text{Answer:}

\displaystyle \text{Let the first term of the A.P. be }a\text{ and common difference be }d.
\displaystyle T_{m+n}=a+(m+n-1)d.
\displaystyle T_{m-n}=a+(m-n-1)d.
\displaystyle \therefore T_{m+n}+T_{m-n}
\displaystyle =[a+(m+n-1)d]+[a+(m-n-1)d].
\displaystyle =2a+(2m-2)d.
\displaystyle =2[a+(m-1)d].
\displaystyle =2T_m.
\displaystyle \therefore T_{m+n}+T_{m-n}=2T_m.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 49: }\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^{th}\text{ terms.}
\displaystyle \text{Answer:}

\displaystyle \text{Let the sums of }n\text{ terms of the two A.P.s be }S_n\text{ and }S'_n.
\displaystyle \frac{S_n}{S'_n}=\frac{5n+4}{9n+6}.
\displaystyle \text{Since the sum of an A.P. contains a factor }n,\text{ we may write}
\displaystyle S_n=kn(5n+4),\qquad S'_n=kn(9n+6),
\displaystyle \text{where }k\text{ is a constant.}

\displaystyle T_n=S_n-S_{n-1}.
\displaystyle =k[n(5n+4)-(n-1)\{5(n-1)+4\}].
\displaystyle =k(10n-1).

\displaystyle \text{Similarly, }T'_n=S'_n-S'_{n-1}.
\displaystyle =k[n(9n+6)-(n-1)\{9(n-1)+6\}].
\displaystyle =k(18n-3).

\displaystyle \therefore T_{18}:T'_{18}
\displaystyle =[10(18)-1]:[18(18)-3].
\displaystyle =179:321.
\displaystyle \therefore \text{The ratio of their }18^{th}\text{ terms is }179:321.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find two positive numbers whose difference is }12\text{ and whose}
\displaystyle \text{arithmetic mean exceeds the geometric mean by }2.
\displaystyle \text{Answer:}

\displaystyle \text{Let the two positive numbers be }x\text{ and }y,\text{ where }x>y.
\displaystyle x-y=12.\qquad ...(1)

\displaystyle \text{Also, arithmetic mean }-\text{ geometric mean}=2.
\displaystyle \frac{x+y}{2}-\sqrt{xy}=2.
\displaystyle x+y-2\sqrt{xy}=4.
\displaystyle (\sqrt{x}-\sqrt{y})^2=4.
\displaystyle \therefore \sqrt{x}-\sqrt{y}=2.\qquad ...(2)

\displaystyle x-y=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}).
\displaystyle 12=2(\sqrt{x}+\sqrt{y}).
\displaystyle \therefore \sqrt{x}+\sqrt{y}=6.\qquad ...(3)

\displaystyle \text{Adding (2) and (3),}
\displaystyle 2\sqrt{x}=8.
\displaystyle \therefore \sqrt{x}=4\Rightarrow x=16.

\displaystyle \text{Subtracting (2) from (3),}
\displaystyle 2\sqrt{y}=4.
\displaystyle \therefore \sqrt{y}=2\Rightarrow y=4.

\displaystyle \therefore \text{The two positive numbers are }16\text{ and }4.
\displaystyle \\

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