ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.3

 Exercise 9.3

Question 1

Find the sum of the following A.P.s :

(i) 2, 7, 12, … to 10 terms

(ii) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

Sol :

(i) 2, 7, 12, … to 10 terms

Here a = 2, d = 7 – 2 = 5 and n = 10

$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 2+(10-1) \times 5]$

$=5(4+45)=5 \times 49=245$

(ii) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

$a=\frac{1}{15}$

$d=\frac{1}{12}-\frac{1}{15}$

$=\frac{5-4}{60}=\frac{1}{60}$ or $\frac{1}{10}-\frac{1}{12}$

$=\frac{6-5}{60}=\frac{1}{60}$

$n=11$

$\therefore \mathrm{S}_{11}=\frac{n}{2} \times[2 a+(n-1) d]$

$=\frac{11}{2} \times\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right]$

$=\frac{11}{2} \times\left[\frac{2}{15}+\frac{1}{6}\right]=\frac{11}{2} \times\left[\frac{4+5}{30}\right]$

$=\frac{11}{2} \times \frac{9}{30}=\frac{33}{20}$ or

$=1 \frac{13}{20}$

Question 2

How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?

Sol :

A.P. = 27, 24, 21,…

a = 27

d = 24 – 27 = -3

$S_{n}=0$

Let n terms be there in A.P.

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 0=\frac{n}{2}[(2 \times 27)+(n-1)(-3)]$

$\Rightarrow 0=n[54-3 n+3]$

$\Rightarrow n[57-3 n]=0$

$\Rightarrow(57-3 n)=\frac{0}{n}=0$

$\Rightarrow 3 n=57$

$\therefore n=\frac{57}{3}=19$

Question 3

Find the sums given below :

(i) 34 + 32 + 30 + … + 10

(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)

Sol :

(i) 34 + 32 + 30 + … + 10

Here, a = 34, d = 32 – 34 = -2, l = 10

Tn = a + (n – 1)d

10 = 34 + (n – 1)(-2)

-24 = -2 (n – 1)

$=\frac{-24}{-2}=n-1 \Rightarrow n-1=12$

$\therefore n=12+1=13$

$\mathrm{~S}_{n}=\frac{n}{2}[a+l]$

$=\frac{13}{2}[34+10]=\frac{13}{2} \times 44=286$

(ii) -5+(-8)+(-11)+....+(-230)

Here, a=-5, d=-8-(-5)=-8+5=-3

l=-230

$\therefore l=a+(n-1) d$

$\Rightarrow-230=-5+(n-1)(-3)$

$-230+5=-3(n-1)$

$\Rightarrow-225=-3(n-1)$

$\frac{-225}{-3}=n-1$

$\Rightarrow n-1=75$

$\Rightarrow n=75+1=76$

$\therefore \mathrm{S}_{n}=\frac{n}{2}[a+l]$

$=\frac{76}{2}[-5+(-230)]$

$=38[-5-230]=38 \times(-235)=-8930$

Question 4

In an A.P. (with usual notations) :

(i) given a = 5, d = 3, a_n = 50, find n and S_n

(ii) given a = 7, a₁₃ = 35, find d and S₁₃

(iii) given d = 5, S₉ = 75, find a and a₉

(iv) given a = 8, a_n = 62, S_n = 210, find n and d

(v) given a = 3, n = 8, S = 192, find d.

Sol :

(i) a = 5, d = 3, an = 50

an = a + (n – 1 )d

50 = 5 + (n – 1) x 3

⇒ 50 – 5 = 3(n – 1)

$\Rightarrow 45=3(n-1)$

$\Rightarrow \frac{45}{3}=n-1$

$\Rightarrow n-1=15$

$\Rightarrow n=15+1=16$

$\therefore n=16$

and $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ 

$=\frac{16}{2}[2 \times 5+(16-1) \times 3]=8[10+45]$

$=8 \times 55=440$

(ii) $a=7, a_{13}=35$

$a_{n}=a+(n-1) d$

$35=7+(13-1) d$

$\Rightarrow 35-7=12 d$

$\Rightarrow 28=12 d$

$\Rightarrow d=\frac{28}{12}=\frac{7}{3}=2 \frac{1}{3}$

and $\mathrm{S}_{13}=\frac{n}{2} \cdot[2 a+(n-1) d]$

$=\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right]$

$=\frac{13}{2}[14+28]=\frac{13}{2} \times(42)$

$=13 \times 21=273$

(iii) $d=5, \mathrm{~S}_{9}=75$

$a_{n}=a+(n-1) d$

$a_{9}=a+(9-1) \times 5$

=a+40...(i)

$\mathrm{S}_{9}=\frac{n}{2}[2 a+(n-1) d]$

$75=\frac{9}{2}[2 a+8 \times 5]$

$\frac{150}{9}=2 a+40$

$2 a=\frac{150}{9}-40=\frac{50}{3}-40$

$2 a=\frac{-70}{3} \Rightarrow a=\frac{-70}{2 \times 3}$

$a=\frac{-35}{3}$

From (i)

$a_{9}=a+40=\frac{-35}{3}+40$

$=\frac{-35+120}{3}=\frac{85}{3}$

$\therefore a=\frac{-35}{3}, a_{9}=\frac{85}{3}$

(iv) $a=8, a_{n}=62, \mathrm{~S}_{n}=210$

$a_{n}=a+(n-1) d$

$62=8+(n-1) d$

$(n-1) d=62-8=54$...(i)

$\mathrm{~S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$210=\frac{n}{2}[2 \times 8+54]$ [From (i)]

$420=n(16+54) \Rightarrow 420=70 n$

$n=\frac{420}{70}=6$

$\therefore(6-1) d=54$

$\Rightarrow 5 d=54$

$\Rightarrow d=\frac{54}{5}$

Hence $d=\frac{54}{5}$ and $n=6$

(v) $a=3, n=8, \mathrm{~S}=192$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$192=\frac{8}{2}[2 \times 3+7 \times d]$

192=4[6+7d]

$\Rightarrow \frac{192}{4}=6+7 d$

$\Rightarrow 48=6+7 d$

$\Rightarrow 7 d=48-6=42$

$d=\frac{42}{7}=6$

$\therefore d=6$

Question 5

(i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

Sol :

(i) First term of an A.P. (a) = 5

Last term (l) = 45

Sum = 400

l = a + (n – 1 )d

45 = 5 + (n – 1)d

⇒ (n – 1)d = 45 – 5 = 40 …(i)

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$400=\frac{n}{2}[2 \times 5+40]$

$\Rightarrow 800=n(10+40)$

50n=800

$\Rightarrow n=\frac{800}{50}=16$

From (i)

(16-1) d=40 

$\Rightarrow 15 d=40$

$\Rightarrow d=\frac{40}{15}$

$\therefore d=\frac{8}{3}$ and $n=16$

(ii) Let a be the first term and d be the common difference.

Now, a=15

Sum of first n terms of an AP is given by,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[2 a+(15-1) d]$

$\Rightarrow 750=\frac{15}{2}(2 a+14 d)$

$\Rightarrow a+7 d=50$

$\Rightarrow 15+7 d=50$

$\Rightarrow 7 d=35$

$\Rightarrow d=5$

Now, 20 th term $=a_{20}=a+19 d$

$=15+19 \times 5=15+95=110$

Question 6

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Sol :

First term of an A.P. (a) = 17

and last term (l) = 350

d= 9

$1=T_{n}=a+(n-1) d$

$350=17+(n-1) \times 9$

$\Rightarrow 350-17=9(n-1)$

$\Rightarrow 333=9(n-1)$

$\Rightarrow n-1=\frac{333}{9}=37$

n=37+1=38

and $\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{38}{2}[2 \times 17+(38-1) \times 9]$

$=19[34+37 \times 9]=19[34+333]$

$=19 \times 367=6973$

Hence n=38 and $\mathrm{S}_{n}=6973$

Question 7

Solve for x : 1 + 4 + 7 + 10 + … + x = 287.

Sol :

1 + 4 + 7 + 10 + .. . + x = 287

Here, a = 1, d = 4 – 1 = 3, n = x

l = x = a = (n – 1)d = 1 + (n – 1) x 3

$\Rightarrow x-1=(n-1) d$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$287=\frac{n}{2}[2 \times 1+(n-1) 3]$

574=n(2-3n-3)

$\Rightarrow 3 n^{2}-n-574=0$

$\Rightarrow 3 n^{2}-42 n+41 n-574=0$

$\Rightarrow 3 n(n-14)+41(n-14)=0$

$\Rightarrow(n-14)(3 n+41)=0$

Either n-14=0, then n=14

or 3n+41=0, then 3n=-41 

$\Rightarrow n=\frac{-41}{3}$

which is not possible being negative.

$\therefore n=14$

Now, $x=a+(n-1) d$

$=1+(14-1) \times 3=1+13 \times 3$

$=1+39=40$

$\therefore x=40$

Question 8

(i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.

(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.

Sol :

(i) A.P. is 25, 22, 19, …

Sum = 116

Here, a = 25, d = 22 – 25 = -3

Let number of terms be n, then

$116=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 232=n[2 \times 25+(n-1)(-3)]$

$\Rightarrow 232=n[50-3 n+3] \Rightarrow 232=n(53-3 n)$

$\Rightarrow 232=53 n-3 n^{2}$

$\Rightarrow 3 n^{2}-53 n+232=0$

$\left\{\begin{array}{l}\because 232 \times 3=696 \\ \therefore 696=-24 \times(-29) \\ -53=-24-29\end{array}\right\}$

$\Rightarrow 3 n^{2}-24 n-29 n+232=0$

$\Rightarrow 3 n(n-8)-29(n-8)=0$

$\Rightarrow(n-8)(3 n-29)=0$

Either n-8=0, then n=8

or 3n-29=0, then 3n=29 

$\Rightarrow n=\frac{29}{3}$

which is not possible because of fractioin

$\therefore n=8$

Now, $\mathrm{T}=a+(n-1) d$

$=25+7 \times(-3)=25-21=4$

(ii) A.P. is 24,21,18, ...

Sum =78

Here, a=24, d=21-24=-3

$\mathrm{~S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 78=\frac{n}{2}[2 \times 24+(n-1)(-3)]$

$\Rightarrow 156=n(48-3 n+3)$

$\Rightarrow 156=51 n-3 n^{2}$

$\Rightarrow 3 n^{2}-51 n+156=0$

$\Rightarrow 3 n^{2}-12 n-39 n+156=0$

$\left\{\begin{array}{c}\because 156 \times 3=468 \\ \therefore 468=-12 \times-39 \\ -51=-12-39\end{array}\right\}$

$\Rightarrow 3 n(n-4)-39(n-4)=0$

$\Rightarrow(n-4)(3 n-39)=0$

Either n-4=0, then n=4

or 3n-39=0, then 3n=39 

$\Rightarrow n=13$

$\therefore n=4$ and 13

$n_{4}=a+(n-1) d=24+3(-3)$

=24-9=15

$n_{13}=24+12(-3)=24-36=-12$

$\therefore$ Sum of 5 th term to 13 term $=0$

$(\because 12+9+6+3+0+(-3)+(-6)+(-9)+(-12)=0$

Question 9

Find the sum of first 22 terms, of an A.P. in which d = 7 and $a_{22}$ is 149.

Sol :

Sum of first 22 terms of an A.P. whose d = 7

$a_{22}=149$ and n=22

149=a+(n-1)d$=a+21 \times 7$

$149=a+147 \Rightarrow a=149-147=2$

$\therefore \mathrm{S}_{22}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{22}{2}[2 \times 2+(22-1)(7)]$

$=11[4+21 \times 7]=11 \times[4+147]$

$=11 \times 151=1661$

Question 10

(i) Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.

(ii) If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.

Sol :

Sum of first 51 terms of an A.P. in which

$T_{2}=14, T_{3}=18$

$\therefore d=T_{3}-T_{2}=18-14=4$

and $a=T_{1}=14-4=10, n=51$

Now, $\mathrm{S}_{51}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{51}{2}[2 \times 10+(51-1) \times 4]$

$=\frac{51}{2}[20+50 \times 4]$

$=\frac{51}{2}[20+200]$

$=\frac{51}{2} \times 220=5610$

(ii) $\mathrm{T}_{3}=1, \mathrm{~T}_{6}=-11, n=32$

$a+2 d=1$..(i)

$a+5 d=-11$..(ii)

Subtracting (i) and (ii),

-3d=12 

$\Rightarrow d=\frac{12}{-3}=-4$

Substitute the value of d in eq. (i)

$a+2(-4)=1 \Rightarrow a-8=1$

a=1+8=9

$\therefore a=9, d=-4$

$\mathrm{S}_{32}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{32}{2}[2 \times 9+(32-1) \times(-4)]$

$=16[18+31 \times(-4)]$

$=16[18-124]=16 \times(-106)]=-1696$

Question 11

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Sol :

$S_{6}=36$

$S_{16}=256$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{6}=\frac{6}{2}[2 a+(6-1) d]=36$

$\Rightarrow 3[2 a+5 d]=36$

$\Rightarrow 2 a+5 d=12$...(i)

and $S_{16}=\frac{16}{2}[2 a+(16-1) d]=256$

$8[2 a+15 d]=256$

$2 a+15 d=32$...(ii)

Subtracting (i) from (ii)

-10d=-20

$\Rightarrow d=\frac{-20}{-10}=2$

Substitute the value of d in eq. (i)

2a+5d=12 

$\Rightarrow 2 a+5 \times 2=12$

$\Rightarrow 2 a+10=12$

$\Rightarrow 2 a=12-10=2$

Now, $S_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 1+(10-1) \times 2]$

$=5[2+9 \times 2]$

=5[2+18]

$=5 \times 20=100$

Question 12

Show that $a_{1}, a_{2}, a_{3}, \ldots$ form an A.P. where $\mathbf{a}_{\mathbf{n}}$ is defined as $a_{n}=3+4 n$ Also find the sum of first 15 terms.

Sol :

$a_{n}=3+4 n$

$a_{1}=3+4 \times 1=3+4=7$

$a_{2}=3+4 \times 2=3+8=11$

$a_{3}=3+4 \times 3=3+12=15$

$a_{4}=3+4 \times 4=3+16=19$

and so on Here, a = 1 and d = 11 – 7 = 4

$\mathrm{S}_{15}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[2 \times 7+(15-1) \times 4]$

$=\frac{15}{2}[14+14 \times 4]$

$=\frac{15}{2}[14+56]$

$=\frac{15}{2} \times 70=525$

Question 13

(i)If $a_{n}=3-4 n$show that $a_{1}, a_{2}, a_{3}, \ldots$ form an A.P. Also find $S_{20}$.

(ii) Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.

Sol :

(i) $a_n$ = 3 – 4n

$a_1$ = 3 – 4 x 1 = 3 – 4 = -1

$a_2$ = 3 – 4 x 2 = 3 – 8 = -5

$a_3$ = 3 – 4 x 3 = 3 – 12 = -9

$a_4$ = 3 – 4 x 4 = 3 – 16 = -13 and so on

Here, a = -1, d = -5 – ( -1) = -5 + 1 = -4

$\mathrm{Now}, \mathrm{S}_{20}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{20}{2}[2 \times(-1)+(20-1) \times(-4)]$

$=10[-2+19 \times(-4)]$

$=10[-2-76]=10 \times(-78)=-780$

(ii) Let a and d be the first term and common difference of A.P. respectively. Given, a=5

difference of A.P. respectively. 

Given, a=5

$a_{1}+a_{2}+a_{3}+a_{4}=\frac{1}{2}\left(a_{5}+a_{6}+a_{7}+a_{8}\right)$

$\therefore a+(a+d)+(a+2 d)+(a+3 d)=\frac{1}{2}[(a+4 d)+(a+5 d)+(a+6 d)+(a+7 d)]$

$\Rightarrow 2(4 a+6 d)=(4 a+22 d)$

$\Rightarrow 2(20+6 d)=(20+22 d)$ $(\because a=5)$

$\Rightarrow 40+12 d=20+22 d$

$\Rightarrow 10 d=20$

$\Rightarrow d=2$

Thus, the common difference of A.P. is 2 .

Question 14

The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is – 30 and the common difference is 8. Find n.

Sol :

In an A.P.

$S_{n}=S_{2 n}$

For the first A.P. a = 8, d = 20

and for second A.P. a = -30, d = 8

$\mathrm{Now}, \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2}[2 \times 8+(n-1) \times 20]$

$=\frac{n}{2}[16+20 n-20]=\frac{n}{2}[20 n-4]$

$=10 n^{2}-2 n$...(i)

Similarly,

$\mathrm{S}_{2 n}=2 \frac{n}{2}[2 a+(2 n-1) d]$

$=n[2 \times(-30)+(2 n-1) \times 8]$

=n[-60+16 n-8]

=n(16 n-68)

$=16 n^{2}-68 n$

$\because \mathrm{S}_{n}=2_{2 n}$

$\therefore 10 n^{2}-2 n=16 n^{2}-68 n$

$\Rightarrow 16 n^{2}-10 n^{2}-68 n+2 n=0$

$\Rightarrow 6 n^{2}-66 n=0$

$\Rightarrow n^{2}-11 n^{\doteq}=0$

n(n-11)=0

Either n=0 which is not possible

or n-11=0, then n=11

$\therefore n=11$

Question 15

The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is $\frac{1}{3}$. Calculate the first and the thirteenth term.

Sol :

$\mathrm{T}_{10}: \mathrm{T}_{30}=1: 3, \mathrm{~S}_{6}=42$

Let a be the first term and d be a common difference, then

$\frac{a+9 d}{a+29 d}=\frac{1}{3} \Rightarrow 3 a+27 d=a+29 d$

$\Rightarrow 3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d \Rightarrow a=d$

Now, $S_{6}=42=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 42=\frac{6}{2}[2 a+(6-1) d]$

$\Rightarrow 42=3[2 a+5 d]$

$\Rightarrow 14=2 a+5 d$

$\Rightarrow 14=2 a+5 a$ $(\because d=a)$

$\Rightarrow 7 a=14$

$\Rightarrow a=\frac{14}{7}=2$

$\therefore a=d=2$

Now, $T_{13}=a+(n-1) d$

$=2+(13-1) \times 2=2+12 \times 2$

=2+24=26

$\therefore$ 1st term is 2 and thirteenth term is 26

Question 16

In an A.P., the sum of its first n terms is 6n – n². Find is 25th term.

Sol :

$S_{n}=6 n-n^{2}$

$T_{25}=?$

$\mathrm{S}_{(n-1)}=6(n-1)-(n-1)^{2}$

$=6 n-6-\left(n^{2}-2 n+1\right)$

$=6 n-6-n^{2}+2 n-1=8 n-n^{2}-7$

$a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$

$=6 n-n^{2}-8 n+n^{2}+7$

=-2 n+7

$a_{25}=-2(25)+7=-50+7=-43$

Question 17

If the sum of first n terms of an A.P. is 4n – n², what is the first term (i. e. $S_1$)? What is the sum of the first two terms? What is the second term? Also, find the 3rd term, the 10th term, and the nth terms?

Sol :

$S_{n}=4 n-n^{2}$

$S_{n}-1=4(n-1)-(n-1)^{2}$

$=4 n-4-\left(n^{2}-2 n+1\right)$

$=4 n-4-n^{2}+2 n-1=6 n-n^{2}-5$

$\therefore a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}=4 n-n^{2}-\left(6 n-n^{2}-5\right)$

$=4 n-n^{2}-6 n+n^{2}+5$

=-2 n+5

$a_{1}=-2 \times 1+5=-2+5=3$

$a_{2}=-2 \times 2+5=-4+5=1$

$a_{3}=-2 \times 3+5=-6+5=-1$

$a_{4}=-2 \times 4+5=-8+5=-3$

$a_{10}=-2 \times 10+5=-20+5=-15$

$\mathrm{S}_{2}=4 n-n^{2}=4 \times 2-(2)^{2}$

=8-4=4

Hence, $\mathrm{S}_{2}=4, a_{1}=1, a_{3}=-1, a_{10} \mid=-15$

$a_{n}=-2 n+5$

Question 18

If $S_{n}$ denotes the sum of first n terms of an A.P., prove that $S_{30} = 3(S_{20} – S_{10})$.

Sol :

Sn denotes the sum of first n terms of an A.P.

To prove: $S_{30} = 3(S_{20} – S_{10})$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{10}=\frac{10}{2}[2 a+(10-1) d]=5(2 a+9 d)$

=10 a+45d

$\mathrm{S}_{20}=\frac{20}{2}[2 a+(20-1) d]=10(2 a+19 d)$

=20a+190d

$\mathrm{S}_{30}=\frac{30}{2}[2 a+(30-1) d=15(2 a+29 d)$

=30a+435d

Now, $\mathrm{R.H.S.}=3\left(\mathrm{~S}_{20}-\mathrm{S}_{10}\right)$

$=3[20 a+190 d-10 a-45 d]$

$=3[10 a+145 d]$

$=30 a+435 d$

$=\mathrm{S}_{30}=\mathrm{L} . \mathrm{H.S}$

Question 19

(i) Find the sum of first 1000 positive integers.

(ii) Find the sum of first 15 multiples of 8.

Sol :

(i) Sum of first 1000 positive integers

i. e., 1 + 2 + 3+ 4 + … + 1000

Here, a = 1, d = 1, n = 1000

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{1000}{2}[2 \times 1+(1000-1) 1]$

=500[2+999]

$=500 \times 1001$

=500500

(ii) Sum of first 15 multiples of 8

8+16+24+32+...120

Here, a=8, d=8, n=15

$\therefore \mathrm{S}_{15}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[2 \times 8+(15-1) \times 8]$

$=\frac{15}{2}[16+14 \times 8]$

$=\frac{15}{2}[16+112]$

$=\frac{15}{2} \times 128=15 \times 64=960$

Question 20

(i) Find the sum of all two digit natural numbers which are divisible by 4.

(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

(iii) Find the sum of all multiples of 9 lying between 300 and 700.

(iv) Find the sum of all natural numbers less than 100 which are divisible by 6.

Sol :

(i) Sum of two digit natural numbers which are divisible by 4

which are 12, 16, 20, 24, …, 96

Here, a = 12, d = 16 – 12 = 4, l = 96

$\therefore l=96=a+(n-1) d=12+(n-1) \times 4$

96=12+4n-4=8+4n

$\Rightarrow 4 n=96-8=88$

$\Rightarrow n=\frac{88}{4}=22$

$\therefore \mathrm{S}_{22}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{22}{2}[2 \times 12+(22-1) \times 4]$

$=11[24+21 \times 4]=11[24+84]$

$=11 \times 108=1188$

(ii) Sum of all natural numbers between 100 and 200 which are divisible by 4 which are

104,108,112,116, ...196

Here, $a=104, d=108-104=4, l=196$

$l=a_{n}=196=a+(n-1) d$

$\Rightarrow 196=104+(n-1) \times 4$

196-104=(n-1) 4

$\Rightarrow 92=(n-1) 4$

$n-1=\frac{92}{4}=23$

$\therefore n=23+1=24$

$\mathrm{Now}, \mathrm{S}_{24}=\frac{n}{2}[2 \dot{a}+(n-1) d]$

$=\frac{24}{2}[2 \times 104+(24-1) \times 4]$

$=12[208+23 \times 4]$

$=12 \times[208+92]$

$=12 \times 300=3600$

(iii) Sum of all natural numbers multiple of 9 lying between 300 and 700 which are

306,315,324,333, .... 693

Here, a=306, d=9, l=693

$l=a_{n}=693=a+(n-1) d$

$=306+(n-1) \times 9$

693-306=9(n-1)

387=9(n-1)

$\Rightarrow n-1=\frac{387}{9}=43$

$\therefore n=43+1=44$

$\mathrm{S}_{44}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{44}{2}[2 \times 306+(44-1) \times 9]$

$=22[6.12+43 \times 9]$

$=22[612+387]=22 \times 999=21978$

$\begin{array}{r}999 \\\times 22 \\\hline 1998 \\1998 \times \\\hline 21978 \\\hline\end{array}$

(iii) Sum of all natural numbers less then 100 which are divisible by 6 which are

6,12,18,24, ..., 96

Here, a=6, d=6, l=96

$a_{n}$=l=96=a+(n-1) d

$\Rightarrow 96=6+(n-1) \times 6$

$96-6=6(n-1)$

$\frac{90}{6}=n-1$

$\Rightarrow n-1=15$

n=15+1=16

$\therefore \mathrm{S}_{16}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{16}{2}[2 \times 6+(16-1) \times 6]$

$=8[12+15 \times 6]=8[12+90]$

$=8 \times 102=816$

Question 21

(i) Find the sum of all two digit odd positive numbers.

(ii) Find the sum of all 3-digit natural numbers which are divisible by 7.

(iii) Find the sum of all two digit numbers which when divided by 7 yield 1 as the

Sol :

(i) Sum of all two-digit odd positive numbers which are 11, 13, 15, …, 99

Here, a = 11, d = 2, l = 99

$a_{n}=l=a+(n-1) d$

$99=11+(n-1) \times 2$

$\Rightarrow 99-11=2(n-1)$

$\Rightarrow 88=2(n-1)$

$\Rightarrow n-1=\frac{88}{2}=44$

$\therefore n=44+1=45$

$\mathrm{S}_{45}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{45}{2}[2 \times 11+(45-1) \times 2]$

$=\frac{45}{2}[22+44 \times 2]$

$=\frac{45}{2}[22+88]$

$=\frac{45}{2} \times 110=2475$

(ii) Sum of all 3-digit natural numbers which are divisible by 7 which are 105,112,119 ... 994

Here, a=105, d=112-105=7, l=994

$\therefore l=T_{n}=994=a+(n-1) d$

$\Rightarrow 994=105+(n-1) \times 7$

$\Rightarrow 994-105=(n-1) 7$

$\Rightarrow 889=7(n-1)$

$\frac{889}{7}=n-1$

$\Rightarrow n-1=127$

$\Rightarrow n=127+1=128$

$\therefore \mathrm{S}_{128}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{128}{2}[2 \times 105+(128-1) \times 7]$

$=64[210+889]=64 \times 1099=70336$

(iii) The 2 -digit number which when divided by 7 gives remainder 1 are :

15,22,29, ..., 99

Here,a=15 and d=22-15=7

We have, $a_{n}=99$

nth term of an $\mathrm{AP}$ is $a_{n}=a+(n-1) d$

$\Rightarrow 99=15+(n-1) 7$

$\Rightarrow 99=15+7 n-7$

$\Rightarrow 99=8+7 n$

$\Rightarrow 7 n=99-8$

$\Rightarrow n=\frac{91}{7}=13$

$\therefore \quad n=13$

Now, we know

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{13}=\frac{13}{2}[2 \times 15+(13-1) \times 7]$

$=\frac{13}{2}[30+12 \times 7]$

$=\frac{13}{2}[30+84]=\frac{13}{2} \times 114$

$=13 \times 57=741$

Question 22

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days ?

Sol :

Penalty for

First day = Rs 200

Second day = Rs 250

Third day = Rs 300

Here, a=₹ 200, d=₹ 250-200=₹ 50

n=30 days

$\therefore \mathrm{S}_{30}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{30}{2}[2 \times 200+(30-1) \times 50]$

$=15[400+29 \times 50]=15[400+1450]$

$=15 \times 1850=₹ 27750$

Question 23

Kanika was given her pocket money on 1st Jan, 2016. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued on doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and was found that at the end of the month she still has Rs 100 with her. How much money was her pocket money for the month ?

Sol :

Pocket money for Jan. 2016

Out of her pocket money, Kanika puts

Rs 1 on the first day i.e., 1 Jan.

Rs 2 on second Jan

Rs 3 on third Jan

Rs 31 on 31st Jan

Here, a=₹ 1 and d=₹ 1, n=31

$\therefore$ Amount for 31 days =1+2+3+...+31

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{31}{2}[2 \times 1+(31-1) \times 1]$

$=\frac{31}{2}[2+30]=\frac{31}{2} \times 32=₹ 496$

Amount spent during the period =₹ 204

Saving at the end of month =₹ 100

$\therefore$ Total savings =₹ 496+₹ 204+₹ 100=₹ 800

Question 24

Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

Sol :

Savings for the first month = Rs 32

For the second month = Rs 36

For the third month = Rs 40

Total savings for the period = Rs 2000

Here, a=₹ 32, d=36-32=₹ 4,$\mathrm{~S}_{n}=₹ 2000$

$\mathrm{S}_{n}=₹ 2000=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 4000=n[2 \times 32+(n-1) \times 4]$

$\Rightarrow 4000=n[64+4 n-4]=n[60+4 n]$

$\therefore 60 n+4 n^{2}-4000=0$

$\Rightarrow n^{2}+15 n-1000=0$ (Dividing by 4)

$\Rightarrow n^{2}+40 n-25 n-1000=0$

$\Rightarrow n(n+40)-25(n-40)=0$

$\Rightarrow(n+40)(n-25)=0$

Either n+40=0, then n=-40 which is not possible being negative

or n-25=0, then n=25

$\therefore$ Required period =25 months

Question 25

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

Sol :

Total number of flags = 27

To fixed after every = 2 m

The flag is stored at the middlemost flag

i.e. $\frac{27+1}{2}$ th flag i.e. at 14 th flag

$\mathrm{S}_{13}$=2[4+8+12+16+....-13 terms ]

$=2\left[\frac{n}{2}(2 a+(n-1) d]$

$=2\left[\frac{13}{2}[2 \times 4+(13-1) \times 4]\right]\right.$

$=2\left[\frac{13}{2} \times(8+48)\right] \mathrm{m}$

$=2\left[\frac{13}{2} \times 56\right]=2 \times 13 \times 28 \mathrm{~m}$

$=2 \times 364=728 \mathrm{~m}$

Maximum distance traveled in carrying a flag $=2 \times 13=26 \mathrm{~m}$