2. Choose the correct choice in the following and justify : (i) 30th term of the AP: 10, 7, 4, . . . , is (A) 97 (B) 77 (C) –77 (D) – 87Answer $a_n$ term of a series in AP is given as; $a_n=a_{n-1}+d=a+(n-1)d $ Here $a=10$ and $d=a_2-a_1=7-10=-3$ $a_{30}=10+(30-1)(-3)$ $a_{30}=-77$ Ans:(C) (ii) 11th term of the AP: – 3, $-1/2$, 2, . . ., is (A) 28 (B) 22 (C) –38 (D) -48$1/2$Answer Here $a=-3$ and $d=a_2-a_1=-1/2-(-3)=5/2$ $a_n=a+(n-1)d $ $a_{11}=-3+(11-1)(5/2) $ $=-3+50/2$ =$22$ Ans:(B) 3. In the following APs, find the missing terms in the blanks. (i) 2, ______ , 26Answer Since $a_n=a+(n-1)d $ Here $a=2$ and $a_3=26$ So $a_3=26=2+(3-1)d $ $d=24/2=12$ So, second term $a_2=2+12=14$ Ans:14 (ii) ______, 13,_________ , 3Answer Since $a_n=a+(n-1)d $ Here $a=?;a_2=13,a_3=?;a_4=3$ So, $a_2=a+(2-1)d$ $13=a+d ..........(1)$ and $a_4=a+(4-1)d$ $3=a+3d ...........(2)$ Solving (1) and (2) $10=-2d; d=-5$ Putting value of $d$ in equation (1); $13=a+(-5)$ $a=13+5=18$ Hence $a=18$ and $a_3=18+(3-1)(-5)=8$ Ans:Missing term: 18 & 8 (iii) 5,_____, ________, $-1/2$Answer Since $a_n=a+(n-1)d $ Here $a=5;a_2=?,a_3=?;a_4={-1/2}$ $a_4=a+(4-1)d $ ${-1/2}=5+(4-1)d $ ${-1/2-5}=(3)d $ $d=-11/6$ Hence; $a_2=a+d=5+(-11/6)=-19/6$ $a_3=a+2d=5+2(-11/6)=-8/6=-4/3$ Ans:Missing term: $-19/6;-4/3$ (iv) – 4,________, ________ ,_________,_______ ,6Answer Since $a_n=a+(n-1)d $ Here $a=-4;a_2=?,a_3=?;a_4=?;a_5=?;a_6=6$ $a_6=a+(6-1)d $ $6=(-4)+(5)d $ $d=10/5=2$ Hence, $a_2=a+d=(-4)+2=-2$ $a_3=a+2d=(-4)+2(2)=0$ $a_4=a+3d=(-4)+3(2)=2$ $a_5=a+4d=(-4)+4(2)=4$ Ans:Missing term:-2,0,2,4 (v) ________, 38, ________,________,________, – 22Answer Here $a=?;a_2=38,a_3=?;a_4=?;a_5=?;a_6=-22$ Since $a_n=a+(n-1)d $ $a_2=a+(2-1)d $ $38=a+(1)d .....................(1)$ $a_6=a+(6-1)d $ $-22=a+(5)d ....................(2)$ Solving equation (1)and (2) $38-(-22)=d-5d$ $60/{-4}=d$ i.e $d=-15$ putting value of $d$ in equation (1) Hence, $38=a+(1)d$;$38=a+(1)(-15)$ $a=38+15=53$ $a_3=a+(3-1)d =53+2(-15)=23$ $a_4=a+(4-1)d =53+3(-15)=8$ $a_5=a+(5-1)d =53+4(-15)=-7$ Ans:Missing term:53,23,8,-7 4. Which term of theAP : 3, 8, 13, 18, . . . ,is 78?Answer Since $a_n=a+(n-1)d $ Here $a=3$ and $d=(8-3)=5$ Let n term be 78 $78=3+(n-1)5 $ ${78-3}/5=(n-1)$ $(n-1)=15$ $n=16$ Ans:So, 16th term of the AP is 78. 5. Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205Answer Let $n$ be the number of terms . $a$=7; $a_2$=13; $a_n$=205; Therefore $d$=13-7=6 Since $a_n=a+(n-1)d $ $205=7+(n-1)6 $ $205-7=(n-1)6$ $198/6=(n-1)$ $n=33+1=34$ Ans:No of terms=34 (ii) 18, 15$1/2$, 13, . . . , – 47Answer Let $n$ be the number of terms . $a$=18; $a_2=15{1/2}$ $a_n=-47$ $d=a_2-a_1=31/2-18={31-36}/2=-5/2$ Since $a_n=a+(n-1)d $ $-47=18+(n-1)(-5/2)$ $5/2(n-1)=18+47=65$ $(n-1)=65(2/5)=26$ $n=27$ Ans:No of terms=27 6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .Answer Let $-150$ be nth term in a AP. Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d$. Here, $a_1=a=11$ $a_2=8$ $a_3=5$ Therefore $d=a_2-a_1=8-11=-3$ Let assume $-150$ be nth term in a AP. So, $n$ will be a whole number. Since $a_n=a+(n-1)d $ $-150 =11+(n-1)(-3) $ $-150-11 =(n-1)(-3) $ ${-161}/{-3} =(n-1) $ $(n-1)={-161}/{-3}$ $n=53.66$ Ans:Since $n$ is not a whole number , so $-150$ is not a term of the A.P. 7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.Answer Since $a_n=a+(n-1)d $ Therefore $a_11=a+(11-1)d$ $38=a+10d ................(1)$ $a_16=a+(16-1)d$ $73=a+15d ................(2)$ Solving equation (1) and (2) $35=5d$ i.e $d=7$ Putting the value of $d$ in equation(1) $38=a+10(7)$ $a=38-70$ $a=-32$ So the 31st term of an AP is $a_31=a+(31-1)d$ $a_31=-32+(30)7$ $a_31=-32+210$ $a_31=-32+210$ $a_31=178$ Ans: So, the 31st term of the AP is 178 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.Answer Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d$. Since $a_n=a+(n-1)d $ Therefore $a_3=a+(3-1)d$ $12=a+2d ..................(1)$ and $a_50=a+(50-1)d$ $106=a+49d..................(2)$ Solving equation (1) and (2) $47d=94$ i.e $d=2$ Putting value of $d$ in equation (1) $12=a+2(2)$ $a=12/4=3$ Therefore $a_29=a+(29-1)d$ =3+28(2) =59 Ans: So, the 29th term of the AP is 59 9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?Answer Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d$. Since $a_n=a+(n-1)d $ Therefore $a_3=a+(3-1)d$ $4=a+2d ..............(1)$ and $a_9=a+(9-1)d$ $-8=a+8d...............(2)$ Solving (1)and (2) $12=-6d$ $d=-2$ puting the value of $d$ in equation (1) $4=a+2(-2)$ $a=8$ Let $a_n$th term be zero. $a_n=a+(n-1)d$ $0=8+(n-1)(-2)$ $(n-1)=4$ i.e $n=5$ Ans: So, the 5th term of the AP is zero 10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.Answer Since $a_n=a+(n-1)d $ $a_17=a+(17-1)d$ and $a_10=a+(10-1)d$ Since $a_17=a_10+7$ $a+(17-1)d =a+(10-1)d +7 $ $16d=9d+7$ $7d=7$ i.e $d=1$ Ans: Common difference $d$ is 1 11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?Answer Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d$. Here $a=3$ $a_2=15$ $a_3=27$ Therefore $d=a_2-a=15-3=12$ Since $a_n=a+(n-1)d $ Therefore $a_54=a+(54-1)d$ $a_54=3+53(12)$ Let kth term be 132 more than its 54th term. Therefore $a_k=a+(k-1)d$ $a_k=3+(k-1)(12)$ Given, $a_k = a_54 + 132$ $3+(k-1)12=3+53(12)+132$ $(k-1)12=504$ $k=43$ Ans: 43th term of the AP is 132 more than its 54th term 12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?Answer Let the two AP's be $a_1,a_2,a_3....a_n$ and $b_1,b_2,b_3....b_n$ Since their common difference $d$ is same and their 100th terms are given by $a_100=a+(100-1)d$ $b_100=b+(100-1)d$ Given $a_100-b_100=100$ Therefore $[a+(100-1)d]-[b+(100-1)d]=100$ i.e $a-b=100$ Therefore the difference between their 1000th terms is $a_1000-b_1000$=$[a+(1000-1)d]-[b+(1000-1)d]$ =$a+999d-b-999d$ =$a-b$ =$100$ Ans: Therefore the difference between their 1000th term is 100 13. How many three-digit numbers are divisible by 7?Answer Let $n$ be the numbers of 3 digit numbers divisible by 7. The first 3 digit number divisible by 7 is 105. The last 3 digit number divisible by 7 is 994. Since the number shall be divisible by 7, so the common difference $d=7$ Since $a_n=a+(n-1)d$ Therefore $994=105+(n-1)7$ $994-105=(n-1)7$ $n-1=889/7=127$ $n=128$ Ans:There are 128 many three-digit number divisible by 7 14. How many multiples of 4 lie between 10 and 250?Answer Let $n$ be the many multiples of 4 between 10 and 250. The first number divisible by 4 is 12. The last number divisible by 4 is 248. Since the number shall be divisible by 4, so the common difference $d=4$ Since $a_n=a+(n-1)d$ Therefore $248=12+(n-1)4$ $248-12=(n-1)4$ $n-1=236/4=59$ $n=60$ Ans:There are 60 many multiples of 4 between 10 and 250 15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?Answer Since $a_n=a+(n-1)d$ For first AP Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d_1$. $a=63$,$a_2=65$,$a_3=67$ Therefore $d_1=a_2-a=65-63=2$ For second AP Let $b_1,b_2,b_3, ....$ be an AP whose first term $b_1$ is $b$ and the common difference $d_2$. $b=3$,$b_2=10$,$b_3=17$ Therefore $d_2=7$ Given: Since the nth term of the two AP's equal $a+(n-1)d_1=b+(n-1)d_2$ $63+(n-1)2=3+(n-1)7$ $(n-1)7-(n-1)2=63-3$ $7n-7-2n+2=60$ $5n-5=60$ $n=13$ Ans: 13th term of the both AP's are equal. 16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.Answer Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference is $d$. Since $a_n=a+(n-1)d$ Given:$a_3=16$;and $a_7=a_5+12$ Therefore $a_3=a+(3-1)d$ $16=a+2d................(1)$ and $a_5=a+(5-1)d=a+4d $ $a_7=a+(7-1)d=a+6d$ Since $a_7=a_5+12$ $a+6d=a+4d+12$ $2d=12$ $d=6$ Putting the value of $d$ in equation (1) $16=a+2(6)$ $a=16-12=4$ Ans:Therefore the AP series is 4,10,16,22,28,34,40 17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.Answer Let n be the total number of terms in the given AP:3, 8, 13, . . ., 253 Given $a=3,a_2=8,a_3=13,....,a_n=253$ Therefore common difference,$d=8-3=5$ $a_n=a+(n-1)d$ $253=3+(n-1)5$ $n-1=50$ $n=51$ Therefore the 20th term from the last term shall be the 31st term from the first term. $a_31=a+(31-1)d$ $a_31=3+30(5)$ $a_31=153$ Ans:Therefore the 20th term from the last term is 153 18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.Answer Let $a_1,a_2,a_3, ....$ be an AP whose first term $a_1$ is $a$ and the common difference $d$. Since nth term of an AP is given as $a_n=a+(n-1)d$ Therefore: $a_4=a+(4-1)d$ $a_8=a+(8-1)d$ $a_6=a+(6-1)d$ $a_10=a+(10-1)d$ Given: $a_4+a_8=24$ and $a_6+a_10=44$ Therefore: $a_4+a_8=24$ $a+3d+a+7d=24$ $2a+10d=24............(1)$ and $a_6+a_10=44$ $a+5d+a+9d=44$ $2a+14d=44 ............(2)$ Solving equation (1) and (2) $4d=20$ i.e $d=5$ Putting the value of $d$ in equation (1) $2a+10(5)=24$ $2a=24-50$ $a=-13$ First term=$a$=-13 Second term =$a+d$=-13+5=-8 Third term=$a+2d$=-13+10=-3 Ans: The first 3 terms of the AP are -13,-8,-3 19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?Answer Given: Annual Salary=₹ 5000 i.e$a=5000$ Increment =₹ 200 =$d=200$ Let $n$ be the no of year from 1995 when his income reached ₹ 7000 i.e $a_n=7000$ Since $a_n=a+(n-1)d$ $7000=5000+(n-1)200$ $2000=(n-1)200$ $n-1=10$ $n=11$ SO, 11 term from the year 1995, his salary will reached ₹ 7000 Ans: So, his salary will reached ₹ 7000 in the year 2005.Remember: Since 1st term of the AP starts at 2005 and 11 term of the AP is ₹ 7000, So while calculating the year,1 year shall be subracted from the no of year (i.e 1995+11-1=2005) Illustration is given below $a$ year=1995 ;$a_2$ year=1996;$a_3$ year=1997;$a_4$ year=1998 ;$a_5$ year=1999 $a_6$ year=2000 ;$a_7$ year=2001 ; $a_8$ year=2002 ;$a_9$ year=2003 ;$a_10$ year=2004 $a_11$ year=2005 20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n^{th}$ week, her weekly savings become ₹ 20.75, find $n$.Answer Given: $a=5$; $d=1.75$;$a_n=20.75$ Since $a_n=a+(n-1)d$ $20.75=5+(n-1)1.75$ $20.75-5=(n-1)1.75$ $15.75/1.75=(n-1)$ $n-1=9$ $n=10$ Ans: Her saving become ₹ 20.75 on 10th week.
