## 200 logs are stacked in the following manner :

[question] Question. 200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row? [/question] [solution] 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 $\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=19-20=-1$ Let a total of 200 logs be placed in n rows. $S_{n}=200$ $S_{n}=\...

## A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance.

[question] Question. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes. [/question] [solution] Solution: Let the Ist prize be of Rs. a. Then the next prize will be of Rs. (a – 20) Then the next prize will be of Rs. {(a – 20) – 20}, i.e., Rs. (a – 40) Thus, the seven prizes are of Rs. a, Rs. $(a-20)$, Rs. $(a-40), \ldots($ an AP) Then $...

## Find the sum of the odd numbers between 0 and 50.

[question] Question. Find the sum of the odd numbers between 0 and 50. [/question] [solution] Solution: 1, 3, 5, 7 ..., 49 a = 1, d = 2 $\ell=\mathrm{t}_{\mathrm{n}}=49$ $\Rightarrow a+(n-1) d=49$ $\Rightarrow 1+(n-1)(2)=49$ $\Rightarrow 1+2 n-2=49$ $\Rightarrow 2 n=50$ or $n=25$ The sum $=\frac{25}{2}\{a+\ell\}=\frac{25}{2}\{1+49)$ $=\frac{25}{2} \times 50=625$ [/solution]...

## Find the sum of the first 15 multiples of 8.

[question] Question. Find the sum of the first 15 multiples of 8. [/question] [solution] Solution: The multiples of 8 are 8, 16, 24, 3... These are in an A.P., having first term as 8 and common difference as 8. Therefore, a = 8 d = 8 $S_{15}=?$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $=\frac{15}{2}[2(8)+(15-1) 8]$ $=\frac{15}{2}[16+14(8)]$ $=\frac{15}{2}(16+112)$ $=\frac{15(128)}{2}=15 \times 64$ $=960$ [/solution]...

## Find the sum of the first 40 positive integers divisible by 6.

[question] Question. Find the sum of the first 40 positive integers divisible by 6. [/question] [solution] Solution: The positive integers that are divisible by 6 are 6, 12, 18, 24 .... 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 $\mathrm{S}_{40}=?$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\mathrm{S}_{40}=\frac{40}{2}[2(6)+(40-1) 6]$ = 20[12 + (39) (6)] = 20(12 + 234) = 20 × 246 = 4920 [/solution]...

## If the sum of the first $\mathrm{n}$ terms of an $\mathrm{AP}$ is $4 \mathrm{n}-\mathrm{n}^{2}$,

[question] Question. If the sum of the first $n$ terms of an $A P$ is $4 n-n^{2}$, 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 $3 \mathrm{rd}$, the 10 th and the nth terms. [/question] [solution] Solution: $\mathrm{S}_{\mathrm{n}}=4 \mathrm{n}-\mathrm{n}^{2}$ Putting $n=1$, we get $S_{1}=4-1=3$ i.e., $\mathrm{t}_{1}=3$ $\mathrm{S}_{2}=4(2)-(2)^{2}=8-4=4$, i.e., $\mathrm{S}_{2}=4$ $\Rightarrow \mathrm{t}_{1}+\mathrm{...

## Show that $a_{1}, a_{2}, \ldots a_{n}, \ldots$ form an $A P$ where $a_{n}$ is defined as below :

[question] Question. Show that $a_{1}, a_{2}, \ldots a_{n}, \ldots$ form an $A P$ where $a_{n}$ is defined as below : (i) $a_{n}=3+4 n$ (ii) $a_{n}=9-5 n$ Also find the sum of the first 15 terms in each case. [/question] [solution] Solution: (i) $a_{n}=3+4 n$ Putting $\mathrm{n}=1,2,3,4, \ldots$ in $(1)$, we get $a_{1}=3+4=7, a_{2}=3+8=11$ $a_{3}=3+12=15, a_{4}=3+16=19, \ldots$ Thus, the sequence (list of numbers) is 7, 11, 15, 19, ..... Here, $\quad a_{2}-a_{1}=11-7=4$ $a_{3}-a_{2}=15-11=4$ $a_...

## If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,

[question] Question. If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of n terms. [/question] [solution] Solution: $\mathrm{S}_{7}=49$ $\Rightarrow \frac{7}{2}\{2 \mathrm{a}+6 \mathrm{~d}\}=49 \Rightarrow \mathrm{a}+3 \mathrm{~d}=7 \ldots(1)$ $\mathrm{S}_{17}=289$ $\Rightarrow \frac{17}{2}\{2 a+16 d\}=289 \Rightarrow a+8 d=17 \ldots(2)$ Subtracting (1) from (2), we get $5 \mathrm{~d}=17-7=10$ $\Rightarrow d=2$ From (1), a + 3 × 2 = 7 $\Rightarrow \mathrm{a}=1$ $\mat...

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

[question] Question. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively [/question] [solution] Solution: $\mathrm{t}_{2}=14, \mathrm{t}_{3}=18$ $\mathrm{d}=\mathrm{t}_{3}-\mathrm{t}_{2}=18-14=4$, i.e., $\mathrm{d}=4$ Now $\quad \mathrm{t}_{2}=14 \quad \Rightarrow \mathrm{a}+\mathrm{d}=14$ $\Rightarrow a+4=14 \quad \Rightarrow a=10$ $\mathrm{S}_{51}=\frac{51}{2}\{2 \mathrm{a}+50 \mathrm{~d}\}=\frac{51}{2}\{2 \times 10+50 \times 4\}$ $=\frac{51}{2} \tim...

## Find the sum of first 22 terms of an AP in which d = 7

[question] Question. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. [/question] [solution] Solution: $d=7$ $a_{22}=149$ \mathrm{S}_{22}=? $a_{22}=a+(22-1) d$ 149 = a + 21 × 7 149 = a + 147 a = 2 $S_{n}=\frac{n}{2}\left(a+a_{n}\right)=\frac{22}{2}(2+149)=11(151)=1661$ [/solution]...

## The first and the last term of an AP are 17 and 350 respectively

[question] Question. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? [/question] [solution] Solution: Given that, $a=17$ $\ell=350$ d = 9 Let there be n terms in the A.P. $\ell=a+(n-1) d$ 350 = 17 + (n – 1)9 333 = (n – 1)9 (n – 1) = 37 n = 38 $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\ell)$ $\Rightarrow S_{n}=\frac{38}{2}(17+350)=19(367)=6973$ Thus, this A.P. contains 38 terms and the ...

## The first term of an AP is 5,

[question] Question. 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. [/question] [solution] Solution: $a=5$, last term $t_{n}=45$ and $S_{n}=400$ $\mathrm{S}_{\mathrm{n}}=400 \Rightarrow \frac{\mathrm{n}}{2}\left\{\mathrm{t}_{1}+\mathrm{t}_{\mathrm{n}}\right\}=400$ $\Rightarrow \frac{n}{2}\{5+45\}=400 \Rightarrow \frac{n}{2} \times 50=400$ $\Rightarrow \mathrm{n}=16$ Now, $\mathrm{t}_{\mathrm{n}}=45 \quad \Rightarrow \math...

## How many terms of the AP :

[question] Question. How many terms of the AP : 9, 17, 25,.... must be taken to give a sum of 636? [/question] [solution] Solution: $a=9, d=8$ Let $S_{n}=636$ $\Rightarrow \frac{n}{2}[2 a+(n-1) d]=636$ $\Rightarrow \frac{n}{2}\{2 \times 9+(n-1)(8)\}=636$ $\Rightarrow \frac{n}{2}\{18+8 n-8\}=636$ $\Rightarrow \frac{n}{2}\{8 n+10\}=636 \Rightarrow n(4 n+5)=636$ $\Rightarrow 4 n^{2}+5 n-636=0$ $\Rightarrow \mathrm{n}=\frac{-5 \pm \sqrt{25+10176}}{8}=\frac{-5 \pm \sqrt{10201}}{8}$ $=\frac{-5 \pm 101...

## In an AP :

[question] Question. In an AP : (1) Given $a=5, d=3, a_{n}=50$, find $n$ and $S_{n}$. (ii) Given $\mathrm{a}=7, \mathrm{a}_{13}=35$, find $\mathrm{d}$ and $\mathrm{S}_{13}$. (iii) Given $\mathrm{a}_{12}=37, \mathrm{~d}=3$, find a and $\mathrm{S}_{12}$. (iv) Given $a_{3}=15, S_{10}=125$, find $d$ and $a_{10}$. (v) Given $\mathrm{d}=5, \mathrm{~S}_{9}=75$, find a and $\mathrm{a}_{9}$. (vi) Given $\mathrm{a}=2, \mathrm{~d}=8, \mathrm{~S}_{\mathrm{n}}=90$, find $\mathrm{n}$ and $\mathrm{a}_{\mathrm{...

## Find the sums given below :

[question] Question. Find the sums given below : (i) $7+10 \frac{1}{2}+14+\ldots+84$ (ii) 34 + 32 + 30 + ... + 10 (iii) – 5 + (– 8) + (– 11) +...+ (– 230). [/question] [solution] Solution: (i) $\mathrm{a}=7, \mathrm{~d}=10 \frac{1}{2}-7=3 \frac{1}{2}=\frac{7}{2}$ $\ell=t_{n}=84 \Rightarrow a+(n-1) d=84$ $\Rightarrow 7+(n-1) \times \frac{7}{2}=84$ $\Rightarrow(\mathrm{n}-1) \times \quad \frac{7}{2}=77$ $\Rightarrow \mathrm{n}-1=77 \times \frac{2}{7}=22$ $\Rightarrow \mathrm{n}=23$ The sum $=\frac...

## Find the sum of the following APs :

[question] Question. Find the sum of the following APs : (i) 2, 7, 12, .... to 10 terms. (ii) – 37, – 33, – 29, ... to 12 terms. (iii) 0.6, 1.7, 2.8, ... to 100 terms (iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms [/question] [solution] Solution: (i) $a=2, d=5$ $S_{10}=\frac{10}{2}\{2 a+9 d\}$ $\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}\}\right)$ $=5 \times\{2 \times 2+9 \times 5)=5 \times 49=245$ (ii) $\mathrm{a}=-37,...

## Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week,

[question] Question. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n. [/question] [solution] Solution: $\mathrm{t}_{1}=$ Rs. 5 (savings in the lst week) $t_{2}=R s .5+R s .1 .75=R s .6 .75$ (savings in the 2 nd week) $t_{3}=$ Rs. $6.75+$ Rs. $1.75=$ Rs. $8.50$ (savings in the $3 \mathrm{rd}$ week) ...................... $\mathrm{t}_{\mathrm{n}}=$ Rs. $20.75$ $\Rightarrow a+(n-1) d=2...

## Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year.

[question] Question. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000 ? [/question] [solution] Solution: It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs. 200. Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, ..... Here, a = 5000 d = 200 Let after $\mathrm{n}^{\text {th }}$ year, his sal...

## The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44.

[question] Question. 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. [/question] [solution] Solution: $\mathrm{t}_{4}+\mathrm{t}_{8}=24 ; \mathrm{t}_{6}+\mathrm{t}_{10}=44$ $\Rightarrow(a+3 d)+(a+7 d)=24$ $(a+5 d)+(a+9 d)=44$ $\Rightarrow 2 \mathrm{a}+10 \mathrm{~d}=24 ; 2 \mathrm{a}+14 \mathrm{~d}=44$ We have $a+5 d=12$ ...(1) and $\mathrm{a}+7 \mathrm{~d}=22$ ...(2) Subtracting (1) from (2), we get $2 d=10 \Righ...

## Find the 20th term from the last term of the AP

[question] Question. Find the 20th term from the last term of the AP 3, 8, 13, ....., 253. [/question] [solution] Solution: The AP is 3, 8, 13, ..., 253 Its first term = 3 and the common difference = 5. Now, the AP in the reverse order will have the first term = 253 and the common difference = – 5. The 20th term from the end of the AP (1) = The 20 term of the AP in the reverse order $=a+19 d$ $=253+19 \times(-5)=253-95=158$ [/solution]...

## Determine the AP whose third term is 16

[question] Question. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. [/question] [solution] Solution: $a_{3}=16$ $a+(3-1) d=16$ $a+2 d=16$ ...(1) $a_{7}-a_{5}=12$ $[a+(7-1) d]-[a+(5-1) d]=12$ $(a+6 d)-(a+4 d)=12$ $2 d=12$ d = 6 From equation (1), we obtain a + 2 (6) = 16 a + 12 = 16 a = 4 Therefore, A.P. will be $4,10,16,22, \ldots$ [/solution]...

## For what value of n, are the nth terms of two APs 63, 65, 67, ...

[question] Question. For what value of n, are the nth terms of two APs 63, 65, 67, ... and 3, 10, 17, .... equal? [/question] [solution] Solution: l. Two APs are 63, 65, 67, ..., 3, 10, 17, ... From (1), First term = 63 and common difference = 2 Its $n$th term $=63+(n-1) \times 2=2 n+61$ From $(2)$, First term $=3$ and common difference $=7$ Its nth term $=3+(n-1) \times 7=7 n-4$ Putting $7 \mathrm{n}-4=2 \mathrm{n}+61$ $\Rightarrow 7 n-2 n=61+4 \Rightarrow 5 n=65 \Rightarrow n=13$ [/solution]...

## How many multiples of 4 lie between 10 and 250?

[question] Question. How many multiples of 4 lie between 10 and 250? [/question] [solution] Solution: The multiples of 4 between 10 and 250 are 12, 16, 20, 24...., 248. Let these numbers be n. $a=12, d=4$ $\mathrm{t}_{\mathrm{n}}=248$ $\Rightarrow a+(n-1) d=248$ $\Rightarrow 12+(n-1) \times 4=248$ $\Rightarrow 4 n+8=248 \Rightarrow n=60$ [/solution]...

## How many three-digit numbers are divisible by 7?

[question] Question. How many three-digit numbers are divisible by 7? [/question] [solution] Solution: First three-digit number that is divisible by 7 = 105 Next number = 105 + 7 = 112 Therefore, 105, 112, 119,\ldots All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7. The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 - 5 = 994 is the maximum p...

## Two APs have the same common difference.

[question] Question. Two APs have the same common difference. The difference between their 100 th terms is 100, what is the difference between their 1000th terms? [/question] [solution] Solution: Let the two APs with same common difference d be $a, a+d, a+2 d, \ldots$ $b, b+d, b+2 d, \ldots .(ab)$ we are given that $\{100$ th term of the first $A P\}$ $-\{100$ th term of the second $A P\}=100$ $\Rightarrow\{a+99 d\}-\{b+99 d\}=100$ $\Rightarrow a-b=100$ ...(1) Now, $\{1000$ th term of the first ...

## Which term of the AP :

[question] Question. Which term of the AP : 3, 15, 27, 39, .... will be 132 more than its 54th term? [/question] [solution] Solution: $a=3, d=12$ Let us suppose $t_{n}=t_{54}+132$ $\Rightarrow a+(n-1) d=a+53 d+132$ $\Rightarrow(\mathrm{n}-1) \mathrm{d}-53 \mathrm{~d}=132$ $\Rightarrow\{n-1-53) d=132$ $\Rightarrow(n-54) \times 12=132$ $\Rightarrow \mathrm{n}-54=11$ $\Rightarrow \mathrm{n}=65$ Hence, $\mathrm{t}_{65}$ is 132 more than $\mathrm{t}_{54}$. [/solution]...

## The 17th term of an AP exceeds its 10th term by 7.

[question] Question. The 17th term of an AP exceeds its 10th term by 7. Find the common difference [/question] [solution] Solution: $a_{17}-a_{10}=7$ $(a+16 d)-(a+9 d)=7$ $7 d=7$ $d=1$ Therefore, the common difference is 1 . [/solution]...

## If the 3rd and 9th terms of an AP are 4 and – 8 respectively,

[question] Question. If the 3rd and 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? [/question] [solution] Solution: Given that, $a_{2}=4$ $a_{9}=-8$ We know that, $a_{n}=a+(n-1) d$ $a_{2}=a+(3-1) d$ $4=a+2 d$ ...(i) $a_{9}=a+(9-1) d$ $-8=a+8 d$ ...(ii) On subtracting equation (I) from (II), we obtain $-12=6 d$ $d=-2$ From equation (I), we obtain $4=a+2(-2)$ $4=a+2(-2)$ $4=a-4$ $a=8$ Let $\mathrm{n}^{\text {th }}$ term of this A.P. be zero. $a_{n}=a+(n-1) d$ 0 = 8 +...

## An AP consists of 50 terms of which 3rd term is 12 and the last term is 106.

[question] Question. An AP consists of 50 terms of which $3 \mathrm{rd}$ term is 12 and the last term is 106 . Find the 29 th term. [/question] [solution] Solution: $\mathrm{t}_{3}=12, \mathrm{t}_{50}$ (last term) $=106$ $\Rightarrow a+2 d=12$ ..(1) and $a+49 d=106$ ...(2) Subtracting (1) from (2), we get $47 \mathrm{~d}=106-12=94 \Rightarrow \mathrm{d}=2$ From (1), a + 2 × 2 =12 $\quad \Rightarrow a=8$ $\mathrm{t}_{29}=\mathrm{a}+28 \mathrm{~d}=8+28 \times 2=64$ [/solution]...

## Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

[question] Question. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. [/question] [solution] Solution: Given that, $a_{11}=38$ $a_{16}=73$ We know that, $a_{n}=a+(n-1) d$ $a_{11}=a+(11-1) d$ $38=a+10 d$ ..(1) Similarly, $a_{16}=a+(16-1) d$ $73=a+15 d$ $73=a+15 d$ ...(2) On subtracting (1) from (2), we obtain $35=5 \mathrm{~d}$ $d=7$ From equation (1), $38=a+10 \times(7)$ $38-70=a$ $a=-32$ $a_{31}=a+(31-1) d$ $=-32+30(7)$ $=-32+210$ $=178$ Hence, $31^{\text {st }}$ term ...

## Check whether – 150 is a term of the AP

[question] Question. Check whether – 150 is a term of the AP : 11, 8, 5, 2, .... . [/question] [solution] Solution: $a=11, d=-3$ Let if possible $\mathrm{t}_{\mathrm{n}}=-150$ $\Rightarrow a+(n-1) d=-150$ $\Rightarrow 11+(n-1) \times(-3)=-150$ $\Rightarrow 11-3 n+3=-150$ $\Rightarrow 14-3 n=-150$ $\Rightarrow 3 n=14+150=164$ $\Rightarrow \mathrm{n}=\frac{164}{3}=54 \frac{2}{3}$ It is not possible because n is to be natural number. Hence, – 150 cannot be a term of the AP. [/solution]...

## Find the number of terms in each of the following AP's :

[question] Question. Find the number of terms in each of the following AP's : (i) $7,13,19 \ldots \ldots, 205$ (ii) $18,15 \frac{1}{2}, 13, \ldots,-47$ [/question] [solution] Solution: (i) $\mathrm{a}=7, \mathrm{~d}=6$, $t_{n}=205$ $\Rightarrow a+(n-1) d=205$ $\Rightarrow 7+(n-1) \times 6=205 \quad \Rightarrow 6 n+1=205$ $\Rightarrow 6 n=204 \quad \Rightarrow n=34$ Hence, 34 terms (ii) $a=18$ $\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=15 \frac{1}{2}-18$ $\mathrm{d}=\frac{31-36}{2}=-\frac{5}{2}$ L...

## Which term of the AP :

[question] Question. Which term of the AP : 3, 8, 13, 18, ... is 78? [/question] [solution] Solution: $a=3, d=5$ Let $\mathrm{t}_{\mathrm{n}}=78$ $\Rightarrow a+(n-1) d=78$ $\Rightarrow 3+(n-1) \times 5=78 \Rightarrow 5 n-2=78$ $\Rightarrow 5 \mathrm{n}=80 \quad \Rightarrow \mathrm{n}=16$ Hence, $\mathrm{t}_{16}=78$ [/solution]...

## In the following APs,

[question] Question. In the following APs, find the missing terms in the boxes : (i) $2, \square, 26$ (ii) $\square, 13, \square, 3$ (iii) $5, \square, \square, 9 \frac{1}{2}$ (iv) $-4, \square, \square, \square, \square, 6$ (v) $\square, 38, \square, \square, \square,-22$ [/question] [solution] Solution: (i) $\mathrm{a}=2, \mathrm{a}+2 \mathrm{~d}=26 \quad \Rightarrow 2+2 \mathrm{~d}=26$ $\Rightarrow 2 \mathrm{~d}=26-2=24 \quad \Rightarrow \mathrm{d}=12$ Then the missing term $t_{2}=a+d=2+12=14...

## Choose the correct choice in the following and justify

[question] Question. Choose the correct choice in the following and justify (i) 30th term of the AP: $10,7,4, \ldots$ is (A) 97 (B) 77 (C) $-77$ (D) $-87$ (ii) 11th term of the AP: $-3,-\frac{1}{2}, 2, \ldots$ is (A) 28 (B) 22 (C) $-38$ (D) $-48 \frac{1}{2}$ [/question] [solution] Solution: (i) $a=10, d=-3$ $\mathrm{t}_{30}=\mathrm{a}+29 \mathrm{~d}=10+29 \times(-3)$ $=10-87=-77$ Hence, the correct option is (C) (ii) $\mathrm{a}=-3, \mathrm{~d}=5 / 2$ $\mathrm{t}_{11}=\mathrm{a}+10 \mathrm{~d}=-...

## Fill in the blanks in the following table,

[question] Question. Fill in the blanks in the following table, given that a is the first term, $d$ the common difference and $a_{n}$, the $n^{\text {th }}$ term of the AP. [/question] [solution] Solution: (i) $\mathrm{a}=7, \mathrm{~d}=3, \mathrm{n}=8$ $a_{8}=a+7 d=7+7 \times 3=28$ Hence, $a_{8}=28$. (ii) $\mathrm{a}=-18, \mathrm{n}=10, \mathrm{a}_{\mathrm{n}}=0, \mathrm{~d}=?$ $a_{n}=a+(n-1) d$ $0=-18+(10-1) d$ $18=9 d \quad \Rightarrow d=\frac{18}{9}=2$ Hence, $d=2$ (iii) $d=-3, n=18, a_{n}=-...

## Which of the following are APs ?

[question] Question. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) $2,4,8,16, \ldots$ (ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$ (iii) $-1.2,-3.2,-5.2,-7.2, \ldots$ (iv) $-10,-6,-2,2, \ldots \ldots$ (v) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$ (vi) $0.2,0.22,0.222,0.2222, \ldots .$ (vii) $\quad 0,-4,-8,-12, \ldots .$ (viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$ (ix) $1,3,9,27, \ldots \ldots$ ...

## For the following APs,

[question] Question. For the following APs, write the first term and the common difference : (i) $3,1,-1,-3, \ldots$ (ii) $-5,-1,3,7, \ldots \ldots$ (iii) $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots$ (iv) $0.6,1.7,2.8,3.9, \ldots$ [/question] [solution] Solution: (i) $\mathrm{a}=3, \mathrm{~d}=\mathrm{t}_{2}-\mathrm{t}_{1}=1-3=-2$, i.e., $d=-2$ (ii) $a=-5, d=4$ (iii) $a=\frac{1}{3}$ $\mathrm{d}=\mathrm{t}_{2}-\mathrm{t}_{1}=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$ (iv) $0.6,1.7,2.8,...

## Write first four terms of the AP,

[question] Question. Write first four terms of the AP, when the first term a and the common difference d are given as follows: (i) $a=10, d=10$ (ii) $a=-2 \quad d=0$ (iii) $\mathrm{a}=4, \mathrm{~d}=-3$ (iv) $\mathrm{a}=-1, \mathrm{~d}=1 / 2$ (v) $\mathrm{a}=-1.25, \quad \mathrm{~d}=-0.25$ [/question] [solution] Solution: (i) $\mathrm{t}_{1}=\mathrm{a}=10$, $t_{2}=10+d=10+10=20$ $t_{3}=20+d=20+10=30$ $\mathrm{t}_{4}=30+\mathrm{d}=30+10=40, \ldots .$ Thus, the AP is $10,20,30,40, \ldots$ (ii) Giv...

## In which of the following situations,

[question] Question. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each s...