NCERT Solutions for Class 10 Maths Chapter 5 Exercise 5.3 - Arithmetic Progressions

Q1 (i) Find the sum of the following APs: $2,7,12,...,$ to $10$ terms.

Answer:

Given series $2,7,12,...,$ to $10$ terms

Here, $a=2andn=10$ and $d=a_2-a_1=7-2=5$

Now, we know that

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

$\Rightarrow S=\frac{10}{2}\{2\times 2+(10-1)5\}$

$\Rightarrow S=5\{4+45\}$

$\Rightarrow S=5\left\{49\right\}$

$\Rightarrow S=245$

Therefore, the sum of AP $2,7,12,...,$ to $10$ terms is 245

Q1 (ii) Find the sum of the following APs: $-37,-33,-29,...,$ to $12$ terms.

Answer:

Given series $-37,-33,-29,...,$ to $12$ terms.

Here, $a=-37andn=12$ and $d=a_2-a_1=-33-(-37)=4$

Now, we know that

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

$\Rightarrow S=\frac{12}{2}\{2\times (-37)+(12-1)4\}$

$\Rightarrow S=6\{-74+44\}$

$\Rightarrow S=5\{-30\}$

$\Rightarrow S=-180$

Therefore, the sum of AP $-37,-33,-29,...,$ to $12$ terms. is -180

Q1 (iii) Find the sum of the following APs: $0.6,1.7,2.8,...,$ to $100$ terms.

Answer:

Given series $0.6,1.7,2.8,...,$ to $100$ terms..

Here, $a=0.6andn=100$ and $d=a_2-a_1=1.7-0.6=1.1$

Now, we know that

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

$\Rightarrow S=\frac{100}{2}\{2\times (0.6)+(100-1)(1.1)\}$

$\Rightarrow S=50\{1.2+108.9\}$

$\Rightarrow S=50\left\{110.1\right\}$

$\Rightarrow S=5505$

Therefore, the sum of AP $0.6,1.7,2.8,...,$ to $100$ terms. is 5505

Q1 (iv) Find the sum of the following APs: $\frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $11$ terms.

Answer:

Given series $\frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $11$ terms.

Here, $a=\frac{1}{15}andn=11$ and $d=a_2-a_1=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

Now, we know that

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

$\Rightarrow S=\frac{11}{2}\{2\times \frac{1}{15}+(11-1)(\frac{1}{60})\}$

$\Rightarrow S=\frac{11}{2}\{\frac{2}{15}+\frac{1}{6}\}$

$\Rightarrow S=\frac{11}{2}\left\{\frac{9}{30}\right\}$

$\Rightarrow S=\frac{99}{60}=\frac{33}{20}$

Therefore, the sum of AP $\frac{1}{15},\frac{1}{12},\frac{1}{10},...,$ to $11$ terms. is $\frac{33}{20}$

Q2 (i) Find the sums given below: $7+10\frac{1}{2}+14+...+84$

Answer:

Given series $7+10\frac{1}{2}+14+...+84$

First we need to find the number of terms

Here, $a=7and{a}_n=84$ and $d=a_2-a_1=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$

Let suppose there are n terms in the AP

Now, we know that

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

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

$\Rightarrow \frac{7n}{2}=77+\frac{7}{2}$

$\Rightarrow n=23$

Now, we know that

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

$\Rightarrow S=\frac{23}{2}\{2\times 7+(23-1)(\frac{7}{2})\}$

$\Rightarrow S=\frac{23}{2}\{14+77\}$

$\Rightarrow S=\frac{23}{2}\left\{91\right\}$

$\Rightarrow S=\frac{2093}{2}=1046\frac{1}{2}$

Therefore, the sum of AP $7+10\frac{1}{2}+14+...+84$ is $1046\frac{1}{2}$

Q2 (ii) Find the sums given below: $34+32+30+...+10$

Answer:

Given series $34+32+30+...+10$

First we need to find the number of terms

Here, $a=34and{a}_n=10$ and $d=a_2-a_1=32-34=-2$

Let suppose there are n terms in the AP

Now, we know that

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

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

$\Rightarrow -26=-2n$

$\Rightarrow n=13$

Now, we know that

$S=\frac{n}{2}\{a+a_n\}$

$\Rightarrow S=\frac{13}{2}\left\{44\right\}$

$\Rightarrow S=13\times 22=286$

Therefore, the sum of AP $34+32+30+...+10$ is 286

Q2 (iii) Find the sums given below: $-5+(-8)+(-11)+...+(-230)$

Answer:

Given series $-5+(-8)+(-11)+...+(-230)$

First we need to find the number of terms

Here, $a=-5and{a}_n=-230$ and $d=a_2-a_1=-8-(-5)=-3$

Let suppose there are n terms in the AP

Now, we know that

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

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

$\Rightarrow -228=-3n$

$\Rightarrow n=76$

Now, we know that

$S=\frac{n}{2}\{a+a_n\}$

$\Rightarrow S=\frac{76}{2}\{(-5-230)\}$

$\Rightarrow S=38\{-235\}$

$\Rightarrow S=-8930$

Therefore, the sum of AP $-5+(-8)+(-11)+...+(-230)$ is -8930

Q3 (i) In an AP: given $a=5$ , $d=3$ , $a_n=50$ , find $n$ and $S_n$ .

Answer:

Given $a=5,d=3and{a}_n=50$

Let suppose there are n terms in the AP

Now, we know that

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

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

$\Rightarrow 48=3n$

$\Rightarrow n=16$

Now, we know that

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

$\Rightarrow S=\frac{16}{2}\{2\times (5)+(16-1)(3)\}$

$\Rightarrow S=8\{10+45\}$

$\Rightarrow S=8\left\{55\right\}$

$\Rightarrow S=440$

Therefore, the sum of the given AP is 440

Q3 (ii) In an AP: given $a=7$ , $\text{Double subscripts: use braces to clarify}$ , find $d$ and $\text{Double subscripts: use braces to clarify}$ .

Answer:

Given $a=7and{a}_{13}=35$

$a_{13}=a+12d=35$

$=12d=35-7=28$

$d=\frac{28}{12}=\frac{7}{3}$

Now, we know that

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

$\Rightarrow S_{13}=\frac{13}{2}\{2\times (7)+(13-1)(\frac{7}{3})\}$

$\Rightarrow S_{13}=\frac{13}{2}\{14+28\}$

$\Rightarrow S_{13}=\frac{13}{2}\left\{42\right\}$

$\Rightarrow S_{13}=13\times 21=273$

Therefore, the sum of given AP is 273

Q3 (iii) In an AP: given $\text{Double subscripts: use braces to clarify}$ find $a$ and $\text{Double subscripts: use braces to clarify}$ .

Answer:

Given $d=3and{a}_{12}=37$

$a_{12}=a+11d=37$

$=a=37-11\times 3=37-33=4$

Now, we know that

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

$\Rightarrow S_{12}=\frac{12}{2}\{2\times (4)+(12-1)3\}$

$\Rightarrow S_{12}=6\{8+33\}$

$\Rightarrow S_{12}=6\left\{41\right\}$

$\Rightarrow S_{12}=246$

Therefore, the sum of given AP is 246

Q3 (iv) In an AP: given $\text{Double subscripts: use braces to clarify}$ find $d$ and $\text{Double subscripts: use braces to clarify}$

Answer:

Given $\text{Double subscripts: use braces to clarify}$

$a_3=a+2d=15-(i)$

Now, we know that

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

$\Rightarrow S_{10}=\frac{10}{2}\{2\times (a)+(10-1)d\}$

$\Rightarrow 125=5\{2a+9d\}$

$\Rightarrow 2a+9d=25-(ii)$

On solving equation (i) and (ii) we will get

$a=17andd=-1$

Now, $a_{10}=a+9d=17+9(-1)=17-9=8$

Therefore, the value of d and 10th terms is -1 and 8 respectively

Q3 (v) In an AP: given $d=5,S_9=75$ , find $a$ and $a_9$ .

Answer:

Given $d=5,S_9=75$

Now, we know that

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

$\Rightarrow S_9=\frac{9}{2}\{2\times (a)+(9-1)5\}$

$\Rightarrow 75=\frac{9}{2}\{2a+40\}$

$\Rightarrow 150=18a+360$

$\Rightarrow a=-\frac{210}{18}=-\frac{35}{3}$

Now, $a_9=a+8d=-\frac{35}{3}+8(5)=-\frac{35}{3}+40=\frac{-35+120}{3}=\frac{85}{3}$

Q3 (vi) In an AP: given $a=2,d=8,S_n=90,$ find $n$ and $a_n$ .

Answer:

Given $a=2,d=8,S_n=90,$

Now, we know that

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

$\Rightarrow 90=\frac{n}{2}\{2\times (2)+(n-1)8\}$

$\Rightarrow 180=n\{4+8n-8\}$

$\Rightarrow 8n^2-4n-180=0$

$\Rightarrow 4(2n^2-n-45)=0$

$\Rightarrow 2n^2-n-45=0$

$\Rightarrow 2n^2-10n+9n-45=0$

$\Rightarrow (n-5)(2n+9)=0$

$\Rightarrow n=5andn=-\frac{9}{2}$

We know that 'n' can not be negative so the only the value of n is 5

Now,

$a_5=a+4d=2+4\times 8=2+32=34$

Therefore, value of n and nth term is 5 and 34 respectively

Q3 (vii) In an AP: given $a=8,a_n=62,S_n=210,$ find $n$ and $d$.

Answer:

Given $a=8,a_n=62,S_n=210,$

Now, we know that

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

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

$(n-1)d=54-(i)$

Now, we know that

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

$\Rightarrow 210=\frac{n}{2}\{2\times (8)+(n-1)d\}$

$\Rightarrow 420=n\{16+54\}(using(i))$

$\Rightarrow 420=n\left\{70\right\}$

$\Rightarrow n=6$

Now, put this value in (i) we will get

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

Therefore, value of n and d are $6and\frac{54}{5}$ respectively

Q3 (viii) In an AP: given $a_n=4,d=2,S_n=-14,$ find $n$ and $a$ .

Answer:

Given $a_n=4,d=2,S_n=-14,$

Now, we know that

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

$4=a+(n-1)2$

$a+2n=6\Rightarrow a=6-2n-(i)$

Now, we know that

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

$\Rightarrow -14=\frac{n}{2}\{2\times (a)+(n-1)2\}$

$\Rightarrow -28=n\{2(6-2n)+2n-2\}(using(i))$

$\Rightarrow -28=n\{10-2n\}$

$\Rightarrow 2n^2-10n-28=0$

$\Rightarrow 2(n^2-5n-14)=0$

$\Rightarrow n^2-7n+2n-14=0$

$\Rightarrow (n+2)(n-7)=0$

$\Rightarrow n=-2andn=7$

Value of n cannot be negative so the only the value of n is 7

Now, put this value in (i) we will get

a = -8

Therefore, the value of n and a are 7 and -8 respectively

Q3 (ix) In an AP: given $a=3,n=8,S=192,$ find $d$ .

Answer:

Given $a=3,n=8,S=192,$

Now, we know that

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

$\Rightarrow 192=\frac{8}{2}\{2\times (3)+(8-1)d\}$

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

$\Rightarrow 7d=48-6$

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

Therefore, the value of d is 6

Q3 (x) In an AP: given $l=28,S=144andn=9$ and there are total $9$ terms. Find $a$ .

Answer:

Given $l=28,S=144andn=9$

Now, we know that

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

$28=a_n=a+(n-1)d-(i)$

Now, we know that

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

$\Rightarrow 144=\frac{9}{2}\{a+a+(n-1)d\}$

$\Rightarrow 288=9\{a+28\}-(using(i))$

$\Rightarrow a+28=32$

$\Rightarrow a=4$

Therefore, the value of a is 4

Q4 How many terms of the AP: $9,17,25,...$ must be taken to give a sum of $636$ ?

Answer:

Given series $9,17,25,...$

Here, $a=9andd=8$

And $S_n=636$

Now, we know that

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

After putting values we get:

$\Rightarrow 636=\frac{n}{2}\{18+(n-1)8\}$

$\Rightarrow 1272=n\{10+8n\}$

$\Rightarrow 8n^2+10n-1272=0$

$\Rightarrow 2(4n^2+5n-636)=0$

$\Rightarrow 4n^2+53n-48n-636=0$

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

$\Rightarrow n=12andn=-\frac{53}{4}$

The value of n can not be negative, so the only value of n is 12

Therefore, the sum of 12 terms of AP $small9,17,25,...$ must be taken to give a sum of $636$.

Q5 The first term of an AP is $5$ , the last term is $45$ and the sum is $400$ . Find the number of terms and the common difference.

Answer:

Given $a=5,a_n=45,S_n=400,$

Now, we know that

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

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

$(n-1)d=40-(i)$

Now, we know that

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

$\Rightarrow 400=\frac{n}{2}\{2\times (5)+(n-1)d\}$

$\Rightarrow 800=n\{10+40\}(using(i))$

$\Rightarrow 800=n\left\{50\right\}$

$\Rightarrow n=16$

Now, put this value in (i), we will get

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

Therefore, value of n and d are $16and\frac{8}{3}$ respectively

Q6 The first and the last terms of an AP are $17$ and $350$ respectively. If the common difference is $9$ , how many terms are there and what is their sum?

Answer:

Given $a=17,l=350,d=9,$

Now, we know that

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

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

$(n-1)9=333$

$(n-1)=37$

$n=38$

Now, we know that

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

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

$\Rightarrow S_{38}=19\{34+333\}$

$\Rightarrow S_{38}=19\left\{367\right\}$

$\Rightarrow S_{38}=6973$

Therefore, there are 38 terms and their sum is 6973.

Q7 Find the sum of first $22$ terms of an AP in which $d=7$ and $22$ nd term is $149$ .

Answer:

Given $a_{22}=149,d=7,n=22$

Now, we know that

$a_{22}=a+21d$

$149=a+21\times 7$

$a=149-147=2$

Now, we know that

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

$\Rightarrow S_{22}=\frac{22}{2}\{2\times (2)+(22-1)7\}$

$\Rightarrow S_{22}=11\{4+147\}$

$\Rightarrow S_{22}=11\left\{151\right\}$

$\Rightarrow S_{22}=1661$

Therefore, there are 22 terms and their sum is 1661.

Q8 Find the sum of first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively.

Answer:

Given $a_2=14,a_3=18,n=51$ and $d=a_3-a_2=18-14=4$

Now,

$a_2=a+d$

$a=14-4=10$

Now, we know that

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

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

$\Rightarrow S_{51}=\frac{51}{2}\{20+200\}$

$\Rightarrow S_{51}=\frac{51}{2}\left\{220\right\}$

$\Rightarrow S_{51}=51\times 110$

$\Rightarrow S_{51}=5304$

Therefore, there are 51 terms and their sum is 5610.

Q9 If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$ , find the sum of first $n$ terms.

Answer:

Given $S_7=49and{S}_{17}=289$

Now, we know that

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

After putting values we get:

$\Rightarrow S_7=\frac{7}{2}\{2\times (a)+(7-1)d\}$

$\Rightarrow 98=7\{2a+6d\}$

$\Rightarrow a+3d=7-(i)$

Similarly,

$\Rightarrow S_{17}=\frac{17}{2}\{2\times (a)+(17-1)d\}$

$\Rightarrow 578=17\{2a+16d\}$

$\Rightarrow a+8d=17-(ii)$

On solving equations (i) and (ii), we will get

a = 1 and d = 2

Now, the sum of the first n terms is

$S_n=\frac{n}{2}\{2\times 1+(n-1)2\}$

$S_n=\frac{n}{2}\{2+2n-2\}$

$S_n=n^2$

Therefore, the sum of n terms is $n^2$.

Q10 (i) Show that $a_1,a_2,...,a_n,...$ form an AP where an is defined as below: $a_n=3+4n$ Also find the sum of the first $15$ terms.

Answer:

Given $a_n=3+4n$

We will check the values of $a_n$ for different values of n

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

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

$a_3=3+4(3)=3+12=15$ and so on.

From the above, we can see that this is an AP with the first term(a) equal to 7 and a common difference (d) equal to 4

Now, we know that

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

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

$\Rightarrow S_{15}=\frac{15}{2}\{14+56\}$

$\Rightarrow S_{15}=\frac{15}{2}\left\{70\right\}$

$\Rightarrow S_{15}=15\times 35$

$\Rightarrow S_{15}=525$

Therefore, the sum of 15 terms is 525.

Q10 (ii) Show that $a_1,a_2,...,a_n,...$ form an AP where an is defined as below: $a_n=9-5n$ . Also find the sum of the first $small15$ terms in each case .

Answer:

Given $a_n=9-5n$

We will check the values of $a_n$ for different values of n

$a_1=9-5(1)=9-5=4$

$a_2=9-5(2)=9-10=-1$

$a_3=9-5(3)=9-15=-6$ and so on.

From the above, we can see that this is an AP with the first term (a) equal to 4 and common difference (d) equal to -5

Now, we know that

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

$\Rightarrow S_{15}=\frac{15}{2}\{2\times (4)+(15-1)(-5)\}$

$\Rightarrow S_{15}=\frac{15}{2}\{8-70\}$

$\Rightarrow S_{15}=\frac{15}{2}\{-62\}$

$\Rightarrow S_{15}=15\times (-31)$

$\Rightarrow S_{15}=-465$

Therefore, the sum of 15 terms is -465.

Q11 If the sum of the first $n$ terms of an AP is $4n-n^2$ , what is the first term (that is $S_1$ )? What is the sum of first two terms? What is the second term? Similarly, find the $3$ rd, the $10$ th and the $n$ th terms