Solution

(i) $2,7,12, \ldots$, to 10 terms

For this A.P.,

$a=2$

$d=a_2-a_1=7-2=5$

$n=10$

We know that,

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ S _{10} & =\dfrac{10}{2}[2(2)+(10-1) 5] \\ & =5[4+(9) \times(5)] \\ & =5 \times 49=245 \end{aligned} $

(ii) - $37,-33,-29, \ldots$, to 12 terms

For this A.P.,

$a=-37$

$d=a_2-a_1=(-33)-(-37)$

$=-33+37=4$

$n=12$

We know that,

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ S _{12} & =\dfrac{12}{2}[2(-37)+(12-1) 4] \\ & =6[-74+11 \times 4] \\ & =6[-74+44] \\ & =6(-30)=-180 \end{aligned} $

(iii) $0.6,1.7,2.8, \ldots$, to 100 terms

For this A.P.,

$a=0.6$ $d=a_2-a_1=1.7-0.6=1.1$

$n=100$

We know that,

$ \begin{aligned} S_n= & \dfrac{n}{2}[2 a+(n-1) d] \\ S _{100} & =\dfrac{100}{2}[2(0.6)+(100-1) 1.1] \\ & =50[1.2+(99) \times(1.1)] \\ & =50[1.2+108.9] \\ & =50[110.1] \\ & =5505 \end{aligned} $

(iv) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}$, to 11 terms

For this A.P.,

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

$ n=11 $

$ \begin{aligned} d=a_2-a_1 & =\dfrac{1}{12}-\dfrac{1}{15} \\ & =\dfrac{5-4}{60}=\dfrac{1}{60} \end{aligned} $

We know that,

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ S _{11} & =\dfrac{11}{2}[2(\dfrac{1}{15})+(11-1) \dfrac{1}{60}] \\ & =\dfrac{11}{2}[\dfrac{2}{15}+\dfrac{10}{60}] \\ & =\dfrac{11}{2}[\dfrac{2}{15}+\dfrac{1}{6}]=\dfrac{11}{2}[\dfrac{4+5}{30}] \\ & =(\dfrac{11}{2})(\dfrac{9}{30})=\dfrac{33}{20} \end{aligned} $

Solution

(i) $7+10 \dfrac{1}{2}+14+$ $+84$

For this A.P.,

$a=7$

$I=84$

$d=a_2-a_1=10 \dfrac{1}{2}-7=\dfrac{21}{2}-7=\dfrac{7}{2}$

Let 84 be the $n^{\text{th }}$ term of this A.P.

$I=a+(n-1) d$

$84=7+(n-1) \dfrac{7}{2}$

$77=(n-1) \dfrac{7}{2}$

$22=n-1$

$n=23$

We know that,

$ \begin{aligned} S_n & =\dfrac{n}{2}(a+l) \\ S_n & =\dfrac{23}{2}[7+84] \\ & =\dfrac{23 \times 91}{2}=\dfrac{2093}{2} \\ & =1046 \dfrac{1}{2} \end{aligned} $

(ii) $34+32+30+$ $+10$

For this A.P.,

$a=34$

$d=a_2-a_1=32-34=-2$

$I=10$

Let 10 be the $n^{\text{th }}$ term of this A.P.

$I=a+(n-1) d$

$10=34+(n-1)(-2)$

$-24=(n-1)(-2)$

$12=n-1$

$n=13$

$S_n=\dfrac{n}{2}(a+l)$

$=\dfrac{13}{2}(34+10)$

$=\dfrac{13 \times 44}{2}=13 \times 22$ $=286$

(iii) $(-5)+(-8)+(-11)+$ $+(-230)$

For this A.P.,

$a=-5$

$I=-230$

$d=a_2-a_1=(-8)-(-5)$

$=-8+5=-3$

Let -230 be the $n^{\text{th }}$ term of this A.P.

$I=a+(n-1) d$

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

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

$(n-1)=75$

$n=76$

And, $S_n=\dfrac{n}{2}(a+l)$

$ \begin{aligned} & =\dfrac{76}{2}[(-5)+(-230)] \\ & =38(-235) \\ & =-8930 \end{aligned} $

Solution

(i) Given that, $a=5, d=3, a_n=50$

As $a_n=a+(n-1) d$,

$\therefore 50=5+(n-1) 3$

$45=(n-1) 3$

$15=n-1$

$n=16$

$ \begin{aligned} S_n & =\dfrac{n}{2}[a+a_n] \\ S _{16} & =\dfrac{16}{2}[5+50] \\ & =8 \times 55 \\ & =440 \end{aligned} $

(ii) Given that, $a=7, a _{13}=35$

As $a_n=a+(n-1) d$,

$\therefore a _{13}=a+(13-1) d$

$35=7+12 d$

$35-7=12 d$

$28=12 d$

$d=\dfrac{7}{3}$

$S_n=\dfrac{n}{2}[a+a_n]$

$S _{13}=\dfrac{n}{2}[a+a _{13}]$

$=\dfrac{13}{2}[7+35]$

$=\dfrac{13 \times 42}{2}=13 \times 21$

$=273$

(iii)Given that, $a _{12}=37, d=3$

As $a_n=a+(n-1) d$,

$a _{12}=a+(12-1) 3$

$37=a+33$

$a=4$

$S_n=\dfrac{n}{2}[a+a_n]$

$S_n=\dfrac{12}{2}[4+37]$

$S_n=6(41)$

$S_n=246$

(iv) Given that, $a_3=15, S _{10}=125$

As $a_n=a+(n-1) d$,

$a_3=a+(3-1) d$

$15=a+2 d($ i) $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

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

$125=5(2 a+9 d)$

$25=2 a+9 d$

(ii)

On multiplying equation (1) by 2 , we obtain

$30=2 a+4 d$ (iii)

On subtracting equation (iii) from (ii), we obtain

$-5=5 d$

$d=-1$

From equation (i),

$15=a+2(-1)$

$15=a-2$

$a=17$

$a _{10}=a+(10-1) d$

$a _{10}=17+(9)(-1)$

$a _{10}=17-9=8$

(v)Given that, $d=5, S_9=75$

$ _{\text{As }} S_n=\dfrac{n}{2}[2 a+(n-1) d]$,

$S_9=\dfrac{9}{2}[2 a+(9-1) 5]$

$75=\dfrac{9}{2}(2 a+40)$

$25=3(a+20)$

$25=3 a+60$

$3 a=25-60$

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

$a_n=a+(n-1) d$

$a_9=a+(9-1)(5)$ $=\dfrac{-35}{3}+8(5)$

$=\dfrac{-35}{3}+40$

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

(vi) Given that, $a=2, d=8, S_n=90$

As $S_n=\dfrac{n}{2}[2 a+(n-1) d]$,

$90=\dfrac{n}{2}[4+(n-1) 8]$

$90=n[2+(n-1) 4]$

$90=n[2+4 n-4]$

$90=n(4 n-2)=4 n^{2}-2 n$

$4 n^{2}-2 n-90=0$

$4 n^{2}-20 n+18 n-90=0$

$4 n(n-5)+18(n-5)=0$

$(n-5)(4 n+18)=0$

Either $n-5=0$ or $4 n+18=0$

$n=5$ or $n=-\dfrac{18}{4}=\dfrac{-9}{2}$

However, $n$ can neither be negative nor fractional.

Therefore, $n=5$

Solution

Let there be $n$ terms of this A.P.

For this A.P., $a=9$

$d=a_2-a_1=17-9=8$ $S_n=\dfrac{n}{2}[2 a+(n-1) d]$

$636=\dfrac{n}{2}[2 \times a+(n-1) 8]$

$636=\dfrac{n}{2}[18+(n-1) 8]$

$636=n[9+4 n-4]$

$636=n(4 n+5)$

$4 n^{2}+5 n-636=0$

$4 n^{2}+53 n-48 n-636=0$

$n(4 n+53)-12(4 n+53)=0$

$(4 n+53)(n-12)=0$

Either $4 n+53=0$ or $n-12=0$

$n=\dfrac{-53}{4}$ or $n=12$

$n$ cannot be $\quad-\dfrac{53}{4}$. As the number of terms can neither be negative nor fractional, therefore, $n=12$ only.

Solution

Given that,

$a=5$

$I=45$

$S_n=400$

$S_n=\dfrac{n}{2}(a+l)$

$400=\dfrac{n}{2}(5+45)$

$400=\dfrac{n}{2}(50)$ $n=16$

$I=a+(n-1) d$

$45=5+(16-1) d$

$40=15 d$

$d=\dfrac{40}{15}=\dfrac{8}{3}$

Solution

Given that,

$a=17$

$I=350$

$d=9$

Let there be $n$ terms in the A.P.

$I=a+(n-1) d$

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

$333=(n-1) 9$

$(n-1)=37$

$n=38$

$S_n=\dfrac{n}{2}(a+l)$

$\Rightarrow S_n=\dfrac{38}{2}(17+350)=19(367)=6973$

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Solution

$d=7$

$a _{22}=149$

$S _{22}=?$

$a_n=a+(n-1) d$

$a _{22}=a+(22-1) d$

$149=a+21 \times 7$

$149=a+147$

$a=2$

$ \begin{aligned} S_n & =\dfrac{n}{2}(a+a_n) \\ & =\dfrac{22}{2}(2+149) \\ & =11(151)=1661 \end{aligned} $

Solution

Given that,

$a_2=14$

$a_3=18$

$d=a_3-a_2=18-14=4$

$a_2=a+d$

$14=a+4$

$a=10$

$ \begin{aligned} & S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ & S _{51}=\dfrac{51}{2}[2 \times 10+(51-1) 4] \\ &=\dfrac{51}{2}[20+(50)(4)] \\ &=\dfrac{51(220)}{2}=51(110) \\ &=5610 \end{aligned} $

Solution

Given that,

$S_7=49$

$S _{17}=289$

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

$S_7=\dfrac{7}{2}[2 a+(7-1) d]$

$49=\dfrac{7}{2}(2 a+6 d)$

$7=(a+3 d)$

$a+3 d=7(i)$

Similarly, $S _{17}=\dfrac{17}{2}[2 a+(17-1) d]$

$289=\dfrac{17}{2}[2 a+16 d]$

$17=(a+8 d)$

$a+8 d=17$ (ii)

Subtracting equation (i) from equation (ii),

$5 d=10$ $d=2$

From equation (i),

$a+3(2)=7$

$a+6=7$

$a=1$

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

$=\dfrac{n}{2}[2(1)+(n-1)(2)]$

$=\dfrac{n}{2}(2+2 n-2)$

$=\dfrac{n}{2}(2 n)$

$=n^{2}$

Solution

(i) $a_n=3+4 n$

$a_1=3+4(1)=7$

$a_2=3+4(2)=3+8=11$

$a_3=3+4(3)=3+12=15$

$a_4=3+4(4)=3+16=19$

It can be observed that

$a_2-a_1=11-7=4$

$a_3-a_2=15-11=4$

$a_4-a_3=19-15=4$

i.e., $a _{k+1}-a_k$ is same every time. Therefore, this is an AP with common difference as 4 and first term as 7 .

$ \begin{aligned} & S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ & S _{15}=\dfrac{15}{2}[2(7)+(15-1) 4] \\ & =\dfrac{15}{2}[(14)+56] \\ & =\dfrac{15}{2}(70) \\ & =15 \times 35 \\ & =525 \end{aligned} $

It can be observed that

$a_2-a_1=-1-4=-5$

$a_3-a_2=-6-(-1)=-5$

$a_4-a_3=-11-(-6)=-5$

i.e., $a _{k+1}-a_k$ is same every time. Therefore, this is an A.P. with common difference as - 5 and first term as 4.

$ \begin{aligned} & S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ & S _{15}=\dfrac{15}{2}[2(4)+(15-1)(-5)] \\ &=\dfrac{15}{2}[8+14(-5)] \\ &=\dfrac{15}{2}(8-70) \\ &=\dfrac{15}{2}(-62)=15(-31) \\ &=-465 \end{aligned} $

Solution

Given that,

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

First term, $a=S_1=4(1)-(1)^{2}=4-1=3$

Sum of first two terms $=S_2$

$=4(2)-(2)^{2}=8-4=4$

Second term, $a_2=S_2-S_1=4-3=1$

$d=a_2-a=1-3=-2$

$a_n=a+(n-1) d$

$=3+(n-1)(-2)$

$=3-2 n+2$

$=5-2 n$

Therefore, $a_3=5-2(3)=5-6=-1$

$a _{10}=5-2(10)=5-20=-15$

Hence, the sum of first two terms is 4 . The second term is $1.3^{\text{rd }}, 10^{\text{th }}$, and $n^{\text{th }}$ terms are $-1,-15$, and 5 -

$2 n$ respectively.

Solution

The positive integers that are divisible by 6 are

$6,12,18,24 \ldots$

It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.

$a=6$

$d=6$

$S _{40}=?$

$ \begin{aligned} & S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ & S _{40}=\dfrac{40}{2}[2(6)+(40-1) 6] \\ & =20[12+(39)(6)] \\ & =20(12+234) \\ & =20 \times 246 \\ & =4920 \end{aligned} $

Solution

The multiples of 8 are

$8,16,24,32 \ldots$

These are in an A.P., having first term as 8 and common difference as 8.

Therefore, $a=8$

$d=8$

$S _{15}=?$

$ \begin{aligned} & S_n=\dfrac{n}{2}[2 a+(n-1) d] \\ &=\dfrac{15}{2}[2(8)+(15-1) 8] \\ &=\dfrac{15}{2}[16+14(8)] \\ &=\dfrac{15}{2}(16+112) \\ &=\dfrac{15(128)}{2}=15 \times 64 \\ &=960 \end{aligned} $

Solution

The odd numbers between 0 and 50 are

$1,3,5,7,9 \ldots 49$

Therefore, it can be observed that these odd numbers are in an A.P.

$a=1$

$d=2$

$I=49$

$I=a+(n-1) d$

$49=1+(n-1) 2$

$48=2(n-1)$

$n-1=24$

$n=25$

$ S_n=\dfrac{n}{2}(a+l) $

$S _{25}=\dfrac{25}{2}(1+49)$

$=\dfrac{25(50)}{2}=(25)(25)$

$=625$

Solution

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.

$a=200$

$d=50$

Penalty that has to be paid if he has delayed the work by 30 days $=S _{30}$ $=\dfrac{30}{2}[2(200)+(30-1) 50]$

$=15[400+1450]$

$=15(1850)$

$=27750$

Therefore, the contractor has to pay Rs 27750 as penalty.

Solution

Let the cost of $1^{\text{st }}$ prize be $P$.

Cost of $2^{\text{nd }}$ prize $=P-20$

And cost of $3^{\text{rd }}$ prize $=P-40$

It can be observed that the cost of these prizes are in an A.P. having common difference as - 20 and first term as $P$.

$a=P$

$d=-20$

Given that, $S_7=700$

$\dfrac{7}{2}[2 a+(7-1) d]=700$

$\dfrac{[2 a+(6)(-20)]}{2}=100$

$a+3(-20)=100$

a $-60=100$

$a=160$

Therefore, the value of each of the prizes was Rs 160 , Rs 140 , Rs 120 , Rs 100, Rs 80, Rs 60, and Rs 40.

Solution

It can be observed that the number of trees planted by the students is in an AP.

$1,2,3,4,5$…. 12

First term, $a=1$

Common difference, $d=2-1=1$

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

$S _{12}=\dfrac{12}{2}[2(1)+(12-1)(1)]$

$=6(2+11)$

$=6(13)$

$=78$

Therefore, number of trees planted by 1 section of the classes $=78$

Number of trees planted by 3 sections of the classes $=3 \times 78=234$

Therefore, 234 trees will be planted by the students.

Solution

Semi-perimeter of circle $=\pi r$

$I_1=\pi(0.5)=\dfrac{\pi}{2} cm$

$I_2=\pi(1)=\pi cm$

$l_3=\pi(1.5)=\dfrac{3 \pi}{2} cm$

Therefore, $l_1, l_2, l_3$, i.e. the lengths of the semi-circles are in an A.P.,

$\dfrac{\pi}{2}, \pi, \dfrac{3 \pi}{2}, 2 \pi, \ldots \ldots \ldots \ldots$

$a=\dfrac{\pi}{2}$

$d=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}$

$S _{13}=?$

We know that the sum of $n$ terms of an a A.P. is given by

$ \begin{aligned} S_n & =\dfrac{n}{2}[2 a+(n-1) d] \\ & =\dfrac{13}{2}[2(\dfrac{\pi}{2})+(13-1)(\dfrac{\pi}{2})] \\ & =\dfrac{13}{2}[\pi+\dfrac{12 \pi}{2}] \\ & =(\dfrac{13}{2})(7 \pi) \\ & =\dfrac{91 \pi}{2} \\ & =\dfrac{91 \times 22}{2 \times 7}=13 \times 11 \end{aligned} $

$=143$

Therefore, the length of such spiral of thirteen consecutive semi-circles will be $143 cm$.

Fig. 5.4

[Hint : Length of successive semicircles is $l_1, l_2, l_3, l_4, \ldots$ with centres at A, B, A, B, .., respectively.]

Solution

It can be observed that the numbers of logs in rows are in an A.P.

$20,19,18 \ldots$

For this A.P.,

$a=20$

$d=a_2-a_1=19-20=-1$

Let a total of 200 logs be placed in $n$ rows.

$S_n=200$

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

$200=\dfrac{n}{2}[2(20)+(n-1)(-1)]$

$400=n(40-n+1)$

$400=n(41-n)$

$400=41 n-n^{2}$

$n^{2}-41 n+400=0$

$n^{2}-16 n-25 n+400=0$

$n(n-16)-25(n-16)=0$

$(n-16)(n-25)=0$

Either $(n-16)=0$ or $n-25=0$

$n=16$ or $n=25$

$a_n=a+(n-1) d$

$a _{16}=20+(16-1)(-1)$

$a _{16}=20-15$

$a _{16}=5$

Similarly,

$a _{25}=20+(25-1)(-1)$

$a _{25}=20-24$

$=-4$

Clearly, the number of logs in $16^{\text{th }}$ row is 5 . However, the number of logs in $25^{\text{th }}$ row is negative, which is not possible.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the $16^{\text{th }}$ row is 5 .

Solution

The distances of potatoes are as follows.

$5,8,11,14 \ldots$

It can be observed that these distances are in A.P.

$a=5$

$d=8-5=3$

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

$S _{10}=\dfrac{10}{2}[2(5)+(10-1) 3]$ $=5[10+9 \times 3]$

$=5(10+27)=5(37)$

$=185$

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run $=2 \times 185$

$=370 m$

Alternatively,

The distances of potatoes from the bucket are $5,8,11,14 \ldots$

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are

$10,16,22,28,34, \ldots \ldots \ldots$

$a=10$

$d=16-10=6$

$S _{10}=$ ?

$S _{10}=\dfrac{10}{2}[2 \times 10+(10-1) 6]$

$=5[20+54]$

$=5(74)$

$=370$

Therefore, the competitor will run a total distance of $370 m$.