Question 12 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = 10, d = 5$
Answer$a = 10, d = 5$
Let $a_1 = a = 10$
Since, the common difference $d = 5$
Using formula $a_{n + 1} = a_n + d$
Thus, $a_2 = a_1 + d = 10 + 5 = 15$
$a_{3 =} a_2 + d = 15 + 5 = 20$
$a_4 = a_3 + d = 20 + 5 = 25$
Hence, An A.P with common difference $5$ is $10, 15, 20, 25,….$
View full question & answer→Question 22 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$127, 132, 137, . . .$
Answer$127, 132, 137, . . .$
Here, the first term, $a_1 = 127$
Second term, $a_2 = 132$
$a_3 = 137$
Now, common difference = $a_2 – a_1 = 132 – 127 = 5$
Also, $a_3 – a_2 = 137 – 132 = 5$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d = 5.$
View full question & answer→Question 32 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$
Answer$3,3+\sqrt{ } 2,3+2 \sqrt{ } 2,3+3 \sqrt{ } 2, \ldots$
Here, the first term, $a_1=3$
Second term, $a_2=3+\sqrt{ } 2$
$a_3=3+2 \sqrt{ } 2$
Now, common difference $=a_2-a_1=3+\sqrt{2}-3=\sqrt{ } 2$
Also, $a_3-a_2=3+2 \sqrt{ } 2-(3+\sqrt{ } 2)=3+2 \sqrt{ } 2-3-\sqrt{ } 2=\sqrt{ } 2$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d=\sqrt{ } 2$
View full question & answer→Question 42 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$-\frac{1}{5},-\frac{1}{5},-\frac{1}{5}, \ldots$
Answer$-\frac{1}{5},-\frac{1}{5},-\frac{1}{5}, \ldots$
Here, the first term, $a _1=-\frac{1}{5}$
Second term, $a _2=-\frac{1}{5}$
$a_3=-\frac{1}{5}$
Now, common difference $=a_2-a_1=-\frac{1}{5}-\left(-\frac{1}{5}\right)=-\frac{1}{5}+\frac{1}{5}=0$
$\text { Also, }=a_3-a_2=-\frac{1}{5}-\left(-\frac{1}{5}\right)=-\frac{1}{5}+\frac{1}{5}=0$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d =0$.
View full question & answer→Question 52 Marks
The given sequence is AP or not ? If they are A.P. find the common difference.
$0, – 4, – 8, – 12, . . .$
Answer$0, – 4, – 8, – 12, . . .$
Here, the first term, $a_1 = 0$
Second term, $a_2 = – 4$
$a_3 = – 8$
Now, common difference =$ a_2 – a_1 = – 4 – 0 = – 4$
Also, $a_3 – a_2 = – 8 – ( – 4) = – 8 + 4 = – 4$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d = – 4.$
View full question & answer→Question 62 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$0.3, 0.33, .0333, . . .$
Answer$0.3, 0.33, 0.333,…..$
Here, the first term, $a_1 = 0.3$
Second term, $a_2 = 0.33$
$a_3 = 0.333$
Now, common difference $= a_2 – a_1 = 0.33 – 0.3 = 0.03$
Also, $a_3 – a_2 = 0.333 – 0.33 = 0.003$
Since, the common difference is not same.
Hence the terms are not in Arithmetic progression
View full question & answer→Question 72 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$– 10, – 6, – 2, 2, . . .$
Answer$– 10, – 6, – 2,2, . . .$
Here, the first term, $a_1 = – 10$
Second term, $a_2 = – 6$
$a_3 = – 2$
Now, common difference $= a_2 – a_1 = – 6 – ( – 10) = – 6 + 10 = 4$
Also, $a_3 – a_2 = – 2 – ( – 6) = – 2 + 6 = 4$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d = 4.$
View full question & answer→Question 82 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$2, \frac{5}{2}, 3, \frac{7}{3}, \ldots \ldots$
Answer$2, \frac{5}{2}, 3, \frac{7}{3}, \ldots \ldots$
Here, the first term, $a_1=2$
Second term, $a _2=\frac{5}{2}$
Third Term, $a_3=3$
Now, common difference $=a_2-a_1=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}$
Also, $a_3-a_2=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d =\frac{1}{2}$.
View full question & answer→Question 92 Marks
In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
AnswerGiven: $t _{10}=46$
$t _5+ t _7=52$
Now, By using $n ^{\text {th }}$ term of an A.P. formula
$t_n=a+(n-1) d$
where $n = no$. of terms
$a =$ first term
$d =$ common difference
$t _{ n }= n ^{\text {th }} \text { terms }$
Hence, by given condition we get,
$\Rightarrow t_{10}=46$
$\Rightarrow a+(10-1) d=46$
$\Rightarrow a+9 d=46 \ldots \ldots(1)$
$\Rightarrow t_5+t_7=52$
$\Rightarrow[a+(5-1) d]+[a+(7-1) d]=52$
$\Rightarrow[a+4 d]+[a+6 d]=52$
$\Rightarrow 2 a+10 d=52 \ldots \ldots .(2)$
Multiply eq. $(2)$ by $2$ we get,
$\Rightarrow 2 a+18 d=92$
Subtract eq. $(2)$ by eq. $(3)$
$\Rightarrow[2 a+18 d]-[2 a+10 d]=92-52$
$\Rightarrow 8 d=40$
$\Rightarrow d = \frac{40}{8} = 5$
Substitute "d" in (1)
$\Rightarrow a+9 \times 5=46$
$\Rightarrow a+45=46$
$\Rightarrow a=t_1=46-45=1$
we know that, $t_{n+1}=t_n+d$
$\Rightarrow t_2=t_1+d=1+5=6$
$\Rightarrow t_3=t_2+d=6+5=11$
Hence, an A.P. is $1,6,11, \ldots$
View full question & answer→Question 102 Marks
Find the first term and common difference for each of the A.P.
$\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots$
Answer$\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots$
First term $a _1=\frac{1}{4}$
Second term $a _2=\frac{3}{4}$
Third term $a _3=\frac{5}{4}$
We know that $d=a_{n+1}-a_n$
Thus, $d = a _2- a _1=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Hence, the common difference $d =\frac{1}{2}$ and first term is $\frac{1}{4}$
View full question & answer→Question 112 Marks
Find the first term and common difference for each of the A.P.
$127, 135, 143, 151, . . .$
Answer$127, 135, 143, 151, . . .$
First term $a_1 = 127$
Second term $a_2 = 135$
Third term $a_3 = 143$
We know that $d = a_{n + 1} – a_n$_
Thus, $d = a_2 – a_1 = 135 – 127 = 8$
Hence, the common difference $d = 8$ and first term is $127$
View full question & answer→Question 122 Marks
Find the first term and common difference for each of the A.P.
$0.6, 0.9, 1.2, 1.5, . . .$
Answer$0.6, 0.9, 1.2, 1.5, . . .$
First term $a_1 = 0.6$
Second term $a_2 = 0.9$
Third term $a_3 = 1.2$
We know that $d = a_{n + 1} – a_n$_
Thus, $d = a_2 – a_1 = 0.9 – 0.6 = 0.3$
Hence, the common difference $d = 0.3$ and first term is $0.6$
View full question & answer→Question 132 Marks
Find the first term and common difference for each of the A.P.
$5, 1, – 3, – 7, . . .$
Answer$5, 1, – 3, – 7, . . .$
First term $a_1 = 5$
Second term $a_2 = 1$
Third term $a_3 = – 3$
We know that $d = a_{n + 1} – a_n$_
Thus, $d = a_2 – a_1 = 1 – 5 = – 4$
Hence, the common difference $d = – 4$ and first term is $5$
View full question & answer→Question 142 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = – 19, d = – 4$
Answer$a = – 19, d = – 4$
Let $a_1 = a = – 19$
Since, the common difference d = – 4
Using formula $a_{n + 1} = a_n + d$
Thus, $a_2 = a_1 + d = – 19 + ( – 4) = – 19 – 4 = – 23$
$a_{3 =} a_2 + d = – 23 + ( – 4) = – 23 – 4 = – 27$
$a_4 = a_3 + d = – 27 + ( – 4) = – 27 – 4 = – 31$
Hence, An A.P with common difference $– 4$ is $– 19, – 23, – 27, – 31,….$
View full question & answer→Question 152 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = 6, d = – 3$
Answer$a = 6, d = – 3$
Let $a_1 = a = 6$
Since, the common difference $d = – 3$
Using formula $a_{n + 1} = a_n + d$
Thus, $a_2 = a_1 + d = 6 + ( – 3) = 6 – 3 = 3$
$a_{3 =} a_2 + d = 3 + ( – 3) = 3 – 3 = 0$
$a_4 = a_3 + d = 0 + ( – 3) = – 3$
Hence, An A.P with common difference $– 3$ is $6, 3, 0, – 3…$
View full question & answer→Question 162 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = – 1.25, d = 3$
Answer$a = – 1.25, d = 3$
Let $a_1 = a = – 1.25$
Since, the common difference $d = 3$
Using formula $a_{n + 1} = a_n + d$
Thus, $a_2 = a_1 + d = – 1.25 + 3 = 1.75$
$a_{3 =} a_2 + d = 1.75 + 3 = 4.75$
$a_4 = a_3 + d = 4.75 + 3 = 7.7$5
Hence, An A.P with common difference $3$ is $– 1.25, 1.75, 4.75, 7.75$
View full question & answer→Question 172 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a=-7, d=\frac{1}{2}$
Answer$a=-7, d=\frac{1}{2}$
Let $a_1=a=-7$
Since, the common difference $d =\frac{1}{2}$
Using formula $a_{n+1}=a_n+d$
Thus, $a _2= a _1+ d =-7+\frac{1}{2}=\frac{-14+1}{2}=-\frac{13}{2}$
$a_3=a_2+d=-\frac{13}{2}+\frac{1}{2}=\frac{-13+1}{2}=-\frac{12}{2}=-6 $
$a_4=a_3+d=-6+\frac{1}{2}=\frac{-12+1}{2}=-\frac{11}{2}$
Hence, An A.P with common difference $\frac{1}{2}$ is $-7,-\frac{13}{2},-6,-\frac{11}{2}, \ldots .$.
View full question & answer→Question 182 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = – 3, d = 0$
Answer$a = – 3, d = 0$
Let $a_1 = a = – 3$
Since, the common difference $d = 0$
Using formula $a_{n + 1} = a_n + d$
Thus, $a_2 = a_1 + d = – 3 + 0 = – 3$
$a_{3 =} a_2 + d = – 3 + 0 = – 3$
$a_4 = a_3 + d = – 3 + 0 = – 3$
Hence, An A.P with common difference $0$ is $– 3, – 3, – 3, – 3,….$
View full question & answer→Question 192 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$2, 4, 6, 8, . . .$
Answer$2, 4, 6, 8, . . .$
Here, the first term, $a_1 = 2$
Second term, $a_2 = 4$
$a_3 = 6$
Now, common difference $= a_2 – a_1 = 4 – 2 = 2$
Also, $a_3 – a_2 = 6 – 4 = 2$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d = 2.$
View full question & answer→Question 202 Marks
Find the fourth term from the end in an A.P. $– 11, – 8, – 5, . . ., 49.$
AnswerFirst term from end $a = 49$
$t_n = – 11$
$t_{n – 1} = – 8$
Common difference $d = t_n – t_{n – 1} = – 11 – ( – 8) = – 11 + 8 = – 3$
Now, By using $n^{th}$ term of an A.P. formula
$t_n = a + (n – 1)d$
where n = no. of terms
$a$ = first term
$d$ = common difference
$t_n$ = $n^{th}$ terms
no. of terms $n = 4$
$\Rightarrow t_4 = 49 + (4 – 1) \times ( – 3)$
$\Rightarrow t_4 = 49 + 3 \times ( – 3)$
$\Rightarrow t_4 = 49 – 9 = 40$
View full question & answer→Question 212 Marks
In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020?
Question 222 Marks
Find the sum of first n natural numbers.
Question 232 Marks
Which term of the following A.P. is $560 ?$
$2,11,20,29, \ldots$
AnswerGiven A.P. $2, 11, 20, 29,. .$
Here $a=2, d=11-2=9$
$n^{\text {th }}$ term of this A.P. is $560 .$
$ t_{\mathrm{n}}=a+(n-1) d$
$\therefore 560=2+(n-1) \times 9$
$\quad=2+9 n-9$
$\therefore 9 n=567$
$\therefore \quad n=\frac{567}{9}=63$
$\therefore 63^{\text {rd }} \text { term of given A.P. is } 560 . $
View full question & answer→Question 242 Marks
Find $t_{\mathrm{n}}$ for following A.P. and then find $30^{\text {th }}$ term of A.P.
$3,8,13,18, \ldots$
AnswerGiven A.P. $3,8,13,18$,.
Here $t_1=3, t_2=8, t_3=13, t_4=18, \ldots$
$d=t_2-t_1=8-3=5$
We know that $t_{\mathrm{n}}=a+(n-1) d$
$ \therefore t_{\mathrm{n}}=3+(n-1) \times 5 \because a=3, d=5$
$\therefore t_{\mathrm{n}}=3+5 n-5$
$\therefore t_{\mathrm{n}}=5 n-2$
$\therefore 30^{\text {th }} \text { term } \quad =t_{30}=5 \times 30-2$
$ =150-2 \quad=148$
View full question & answer→Question 252 Marks
The first term a and common difference d are given. Find first four terms of A.P :
$a=8, d=-5$
Answer$ a=8, d=-5$
$a=t_1=8$
$t_2=t_1+d=8+(-5)=3$
$t_3=t_2+d=3+(-5)=-2$
$t_4=t_3+d=-2+(-5)=-7$
$8,3,-2,-7, \ldots$
$\therefore \text { A.P. is }=8,3,-2,-7, \ldots $
View full question & answer→Question 262 Marks
The first term a and common difference d are given. Find first four terms of A.P :
$a=-1, d=-\frac{1}{2}$
Answer$ a=-1, d=-\frac{1}{2}$
$a=t_1=-1$
$t_2=t_1+d=-1+\left(-\frac{1}{2}\right)=-\frac{3}{2}$
$t_3=t_2+d=-\frac{3}{2}+\left(-\frac{1}{2}\right)=-\frac{4}{2}=-2$
$t_4=t_3+d=-2+\left(-\frac{1}{2}\right)$
$=-2-\frac{1}{2}=-\frac{5}{2}$
$\text { P. is }=-1,-\frac{3}{2},-2,-\frac{5}{2}, \ldots $
$\therefore$ A.P. is $=-1,-\frac{3}{2},-2,-\frac{5}{2}, \ldots$.
View full question & answer→Question 272 Marks
The first term a and common difference d are given. Find first four terms of A.P : $a=200, d=7$
AnswerGiven $a=200, d=7$
$
\begin{array}{c}
a=t_1=200 \\
t_2=t_1+d=200+7=207 \\
t_3=t_2+d=207+7=214 \\
t_4=t_3+d=214+7=221 \\
\therefore \text { A.P. is }=200,207,214,221, \ldots
\end{array}
$
View full question & answer→Question 282 Marks
The first term a and common difference d are given. Find first four terms of A.P :
$a=-3, d=4$
AnswerGiven $a=-3, d=4$
$ t_1=-3$
$t_2=t_1+d=-3+4=1$
$t_3=t_2+d=1+4=5$
$t_4=t_3+d=5+4=9$
$\therefore \text { A.P. is }=-3,1,5,9, \ldots $
View full question & answer→Question 292 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $\frac{3}{2}, \frac{1}{2},-\frac{1}{2}, \ldots$
View full question & answer→Question 302 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $1,1,2,2,3,3, \ldots$
View full question & answer→Question 312 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $2,-2,-6,-10, \ldots$
View full question & answer→Question 322 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $5,12,19,26, \ldots$
View full question & answer→Question 332 Marks
Find $t_{11}$ from the following A.P. 4, 9, 14,…
View full question & answer→Question 342 Marks
Find $n$, if the $n^{\text {th }}$ term of the following sequence is 68 .
$5,8,11,14, \ldots$
View full question & answer→Question 352 Marks
How many terms are there in the A.P. 187, 194, 201, ..., 439 ?
Question 362 Marks
Write one example of finite and infinite A.P. each.
AnswerFinite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ………………. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ……………..
View full question & answer→Question 372 Marks
In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.
AnswerGiven: t10 = 46
t5 + t7 = 52
Now, By using nth term of an A.P. formula
tn = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
tn = nth terms
Hence, by given condition we get,
⇒ t10 = 46
⇒ a + (10 – 1)d = 46
⇒ a + 9d = 46 ……(1)
⇒ t5 + t7 = 52
⇒ [a + (5 – 1)d] + [a + (7 – 1)d] = 52
⇒ [a + 4d] + [a + 6d] = 52
⇒ 2a + 10d = 52 ……(2)
Multiply eq. (2) by 2 we get,
⇒ 2a + 18d = 92 ……(3)
Subtract eq. (2) by eq. (3)
⇒ [2a + 18d] – [ 2a + 10d] = 92 – 52
⇒ 8d = 40
$\Rightarrow d=\frac{40}{8}=5$
Substitute “d” in (1)
⇒ a + 9 × 5 = 46
⇒ a + 45 = 46
⇒ a = t1 = 46 – 45 = 1
we know that, tn + 1 = tn + d
⇒ t2 = t1 + d = 1 + 5 = 6
⇒ t3 = t2 + d = 6 + 5 = 11
Hence, an A.P. is 1, 6, 11, . . .
View full question & answer→Question 382 Marks
Find the fourth term from the end in an A.P. – 11, – 8, – 5, . . ., 49.
AnswerFirst term from end a = 49tn = – 11
tn – 1 = – 8
Common difference d = tn – tn – 1 = – 11 – ( – 8) = – 11 + 8 = – 3
Now, By using nth term of an A.P. formula
tn = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
tn = nth terms
no. of terms n = 4
⇒ t4 = 49 + (4 – 1) × ( – 3)
⇒ t4 = 49 + 3 × ( – 3)
⇒ t4 = 49 – 9 = 40
View full question & answer→Question 392 Marks
To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
$1^3, 2^3, 3^3, 4^3$
Answer$1^3, 2^3, 3^3, \ldots$.
$(\text { next term })^3=(\text { previous term }+1)^3$
View full question & answer→Question 402 Marks
Find the first term and common difference for each of the A.P.
$\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots$
Answer$\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \ldots$
First term $a _1=\frac{1}{4}$
Second term ${a_2}=\frac{3}{4}$
Third term $a_3=\frac{5}{4}$
We know that $d=a_{n+1}-a_n$
Thus, $d = a _2- a _1=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Hence, the common difference $d =\frac{1}{2}$ and first term is $\frac{1}{4}$
View full question & answer→Question 412 Marks
Find the first term and common difference for each of the A.P.
127, 135, 143, 151, . . .
Answer127, 135, 143, 151, . . .
First term a1 = 127
Second term a2 = 135
Third term a3 = 143
We know that d = an + 1 – an
Thus, d = a2 – a1 = 135 – 127 = 8
Hence, the common difference d = 8 and first term is 127
View full question & answer→Question 422 Marks
Find the first term and common difference for each of the A.P.
0.6, 0.9, 1.2, 1.5, . . .
Answer0.6, 0.9, 1.2, 1.5, . . .
First term a1 = 0.6
Second term a2 = 0.9
Third term a3 = 1.2
We know that d = an + 1 – an
Thus, d = a2 – a1 = 0.9 – 0.6 = 0.3
Hence, the common difference d = 0.3 and first term is 0.6
View full question & answer→Question 432 Marks
Find the first term and common difference for each of the A.P.
5, 1, – 3, – 7, . . .
Answer5, 1, – 3, – 7, . . .
First term a1 = 5
Second term a2 = 1
Third term a3 = – 3
We know that d = an + 1 – an
Thus, d = a2 – a1 = 1 – 5 = – 4
Hence, the common difference d = – 4 and first term is 5
View full question & answer→Question 442 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
a = – 19, d = – 4
Answera = – 19, d = – 4
Let a1 = a = – 19
Since, the common difference d = – 4
Using formula an + 1 = an + d
Thus, a2 = a1 + d = – 19 + ( – 4) = – 19 – 4 = – 23
a3 = a2 + d = – 23 + ( – 4) = – 23 – 4 = – 27
a4 = a3 + d = – 27 + ( – 4) = – 27 – 4 = – 31
Hence, An A.P with common difference – 4 is – 19, – 23, – 27, – 31,….
View full question & answer→Question 452 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
a = 6, d = – 3
AnswerLet a1 = a = 6
Since, the common difference d = – 3
Using formula an + 1 = an + d
Thus, a2 = a1 + d = 6 + ( – 3) = 6 – 3 = 3
a3 = a2 + d = 3 + ( – 3) = 3 – 3 = 0
a4 = a3 + d = 0 + ( – 3) = – 3
Hence, An A.P with common difference – 3 is 6, 3, 0, – 3…
View full question & answer→Question 462 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
a = – 1.25, d = 3
Answera = – 1.25, d = 3
Let a1 = a = – 1.25
Since, the common difference d = 3
Using formula an + 1 = an + d
Thus, a2 = a1 + d = – 1.25 + 3 = 1.75
a3 = a2 + d = 1.75 + 3 = 4.75
a4 = a3 + d = 4.75 + 3 = 7.75
Hence, An A.P with common difference 3 is – 1.25, 1.75, 4.75, 7.75
View full question & answer→Question 472 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
$a =-7, d =\frac{1}{2}$
Answer$a =-7, d =\frac{1}{2}$
Let $a_1=a=-7$
Since, the common difference $d =\frac{1}{2}$
Using formula $a_{n+1}=a_n+d$
Thus, ${ }^2= a _1+ d =-7+\frac{1}{2}=\frac{-14+1}{2}=-\frac{13}{2}$
$\begin{array}{l}
a_3=a_2+d=-\frac{13}{2}+\frac{1}{2}=\frac{-13+1}{2}=-\frac{12}{2}=-6 \\
a_4=a_3+d=-6+\frac{1}{2}=\frac{-12+1}{2}=-\frac{11}{2}
\end{array}$
Hence, An A.P with common difference $\frac{1}{2}$ is $-7,-\frac{13}{2},-6,-\frac{11}{2}, \ldots .$.
View full question & answer→Question 482 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
a = – 3, d = 0
Answera = – 3, d = 0
Let a1 = a = – 3
Since, the common difference d = 0
Using formula an + 1 = an + d
Thus, a2 = a1 + d = – 3 + 0 = – 3
a3 = a2 + d = – 3 + 0 = – 3
a4 = a3 + d = – 3 + 0 = – 3
Hence, An A.P with common difference 0 is – 3, – 3, – 3, – 3,….
View full question & answer→Question 492 Marks
Write an A.P. whose first term is a and common difference is d in each of the following.
a = 10, d = 5
Answera = 10, d = 5
Let a1 = a = 10
Since, the common difference d = 5
Using formula an + 1 = an + d
Thus, a2 = a1 + d = 10 + 5 = 15
a3 = a2 + d = 15 + 5 = 20
a4 = a3 + d = 20 + 5 = 25
Hence, An A.P with common difference 5 is 10, 15, 20, 25,….
View full question & answer→Question 502 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
127, 132, 137, . . .
Answer127, 132, 137, . . .
Here, the first term, a1 = 127
Second term, a2 = 132
a3 = 137
Now, common difference = a2 – a1 = 132 – 127 = 5
Also, a3 – a2 = 137 – 132 = 5
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 5.
View full question & answer→Question 512 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$
Answer3, 3 + √2, 3 + 2√2, 3 + 3√2, ….
Here, the first term, a1 = 3
Second term, a2 = 3 + √2
a3 = 3 + 2√2
Now, common difference = a2 – a1 = 3 + √2 – 3 = √2
Also, a3 – a2 = 3 + 2√2 –(3 + √2) = 3 + 2√2 – 3 – √2 = √2
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = √2 .
View full question & answer→Question 522 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$-\frac{1}{5},-\frac{1}{5},-\frac{1}{5}, \ldots$
Answer$-\frac{1}{5},-\frac{1}{5},-\frac{1}{5}, \ldots$
Here, the first term, ${ }^{a_1}=-\frac{1}{5}$
Second term, $a_2=-\frac{1}{5}$
$a_3=-\frac{1}{5}$
Now, common difference $=a_2-a_1=-\frac{1}{5}-\left(-\frac{1}{5}\right)=-\frac{1}{5}+\frac{1}{5}=0$
Also, $=a_3-a_2=-\frac{1}{5}-\left(-\frac{1}{5}\right)=-\frac{1}{5}+\frac{1}{5}=0$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d =0$.
View full question & answer→Question 532 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
0, – 4, – 8, – 12, . . .
Answer0, – 4, – 8, – 12, . . .
Here, the first term, a1 = 0
Second term, a2 = – 4
a3 = – 8
Now, common difference = a2 – a1 = – 4 – 0 = – 4
Also, a3 – a2 = – 8 – ( – 4) = – 8 + 4 = – 4
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = – 4.
View full question & answer→Question 542 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
0.3, 0.33, .0333, . . .
Answer0.3, 0.33, 0.333,…..
Here, the first term, a1 = 0.3
Second term, a2 = 0.33
a3 = 0.333
Now, common difference = a2 – a1 = 0.33 – 0.3 = 0.03
Also, a3 – a2 = 0.333 – 0.33 = 0.003
Since, the common difference is not same.
Hence the terms are not in Arithmetic progression
View full question & answer→Question 552 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
– 10, – 6, – 2, 2, . . .
Answer– 10, – 6, – 2,2, . . .
Here, the first term, a1 = – 10
Second term, a2 = – 6
a3 = – 2
Now, common difference = a2 – a1 = – 6 – ( – 10) = – 6 + 10 = 4
Also, a3 – a2 = – 2 – ( – 6) = – 2 + 6 = 4
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 4.
View full question & answer→Question 562 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$2, \frac{5}{2}, 3, \frac{7}{3}, \ldots$
Answer$2, \frac{5}{2}, 3, \frac{7}{3}, \ldots .$
Here, the first term, $a_1=2$
Second term, ${a_2}=\frac{5}{2}$
Third Term, $a_3=3$
Now, common difference $=a_2-a_1=\frac{5}{2}-2=\frac{5-4}{2}=\frac{1}{2}$
Also, ${ }^{a_3}- a _2=3-\frac{5}{2}=\frac{6-5}{2}=\frac{1}{2}$
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, $d =\frac{1}{2}$.
View full question & answer→Question 572 Marks
Which of the following sequences are A.P. ? If they are A.P. find the common difference.
2, 4, 6, 8, . . .
Answer2, 4, 6, 8, . . .
Here, the first term, a1 = 2
Second term, a2 = 4
a3 = 6
Now, common difference = a2 – a1 = 4 – 2 = 2
Also, a3 – a2 = 6 – 4 = 2
Since, the common difference is same.
Hence the terms are in Arithmetic progression with common difference, d = 2.
View full question & answer→Question 582 Marks
In the year 2010 in the village there were 4000 people who were literate. Every year the number of literate people increases by 400. How many people will be literate in the year 2020?
Answer| Year | 2010 | 2011 | 2012 | ..... | 2020 |
| Literate People | 4000 | 4400 | 4800 | ..... | $\square$ |
$a=4000, \quad d=400$
$n=11 t_n=a+(n-1) d$
$=4000+(11-1) 400$
$=4000+4000$
$=8000$
In year 2020,8000 people will be literate. View full question & answer→Question 592 Marks
Find the sum of first n natural numbers.
AnswerFirst $n$ natural numbers are $1,2,3, \ldots, n$.
Here $a=1, d=1, n^{\text {th }}$ term $=n$
$\begin{aligned}
\therefore & S _{ n }=1+2+3+\ldots+n \\
S _{ n } & =\frac{n}{2}[\text { First term + last term] } \ldots \ldots \text { (by the formula) } \\
& =\frac{n}{2}[1+n] \\
& =\frac{n(n+1)}{2}
\end{aligned}$
$\therefore$ Sum of first $n$ natural number is $\frac{n(n+1)}{2}$.
View full question & answer→Question 602 Marks
Which term of the following A.P. is 560 ?
$
2,11,20,29, \ldots
$
AnswerGiven A.P. 2, 11, 20, 29,. .
Here $a=2, d=11-2=9$
$n^{\text {th }}$ term of this A.P. is 560 .
\[
\begin{array}{l}
t_{\mathrm{n}}=a+(n-1) d \\
\therefore 560=2+(n-1) \times 9 \\
\quad=2+9 n-9 \\
\therefore 9 n=567 \\
\therefore \quad n=\frac{567}{9}=63 \\
\therefore 63^{\text {rd }} \text { term of given A.P. is } 560 .
\end{array}
\]
View full question & answer→Question 612 Marks
Find $t_{\mathrm{n}}$ for following A.P. and then find $30^{\text {th }}$ term of A.P.
$
3,8,13,18, \ldots
$
AnswerGiven A.P. $3,8,13,18$,.
Here $t_1=3, t_2=8, t_3=13, t_4=18, \ldots$
$
d=t_2-t_1=8-3=5
$
We know that $t_{\mathrm{n}}=a+(n-1) d$
$
\begin{array}{l}
\therefore t_{\mathrm{n}}=3+(n-1) \times 5 \because a=3, d=5 \\
\therefore t_{\mathrm{n}}=3+5 n-5 \\
\therefore t_{\mathrm{n}}=5 n-2 \\
\therefore 30^{\text {th }} \text { term } \quad \begin{aligned}
& =t_{30}=5 \times 30-2 \\
& =150-2 \quad=148
\end{aligned}
\end{array}
$
View full question & answer→Question 622 Marks
The first term a and common difference d are given. Find first four terms of A.P : $a=8, d=-5$
Answer
$\begin{array}{l} a=8, d=-5 \\
a=t_1=8 \\
t_2=t_1+d=8+(-5)=3 \\
t_3=t_2+d=3+(-5)=-2 \\
t_4=t_3+d=-2+(-5)=-7 \\
8,3,-2,-7, \ldots \\
\therefore \text { A.P. is }=8,3,-2,-7, \ldots
\end{array}
$
View full question & answer→Question 632 Marks
The first term a and common difference d are given. Find first four terms of A.P : $a=-3, d=4$
AnswerGiven $a=-3, d=4$
$
\begin{array}{l}
t_1=-3 \\
t_2=t_1+d=-3+4=1 \\
t_3=t_2+d=1+4=5 \\
t_4=t_3+d=5+4=9 \\
\therefore \text { A.P. is }=-3,1,5,9, \ldots
\end{array}
$
View full question & answer→Question 642 Marks
The first term a and common difference d are given. Find first four terms of A.P : $a=200, d=7$
AnswerGiven $a=200, d=7$
$
\begin{array}{c}
a=t_1=200 \\
t_2=t_1+d=200+7=207 \\
t_3=t_2+d=207+7=214 \\
t_4=t_3+d=214+7=221 \\
\therefore \text { A.P. is }=200,207,214,221, \ldots
\end{array}
$
View full question & answer→Question 652 Marks
The first term a and common difference d are given. Find first four terms of A.P : $a=-1, d=-\frac{1}{2}$
Answer
$\begin{array}{l} a=-1, d=-\frac{1}{2} \
a=t_1=-1 \
t_2=t_1+d=-1+\left(-\frac{1}{2}\right)=-\frac{3}{2} \
t_3=t_2+d=-\frac{3}{2}+\left(-\frac{1}{2}\right)=-\frac{4}{2}=-2 \
t_4=t_3+d=-2+\left(-\frac{1}{2}\right) \
=-2-\frac{1}{2}=-\frac{5}{2} \
\text { P. is }=-1,-\frac{3}{2},-2,-\frac{5}{2}, \ldots
\end{array}
$
$\therefore$ A.P. is $=-1,-\frac{3}{2},-2,-\frac{5}{2}, \ldots$. View full question & answer→Question 662 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $\frac{3}{2}, \frac{1}{2},-\frac{1}{2}, \ldots$
AnswerIn the sequence $\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}, \ldots$,
$\begin{array}{l}
t_1=\frac{3}{2}, \quad t_2=\frac{1}{2}, t_3=-\frac{1}{2}, \quad t_4=-\frac{3}{2} \ldots \\
t_2-t_1=\frac{1}{2}-\frac{3}{2}=-\frac{2}{2}=-1 \\
t_3-t_2=-\frac{1}{2}-\frac{1}{2}=-\frac{2}{2}=-1 \\
t_4-t_3=-\frac{3}{2}-\left(-\frac{1}{2}\right)=-\frac{3}{2}+\frac{1}{2}=-\frac{2}{2}=-1
\end{array}$
Here the common difference $d=-1$ which is constant.
$\therefore$ Given sequence is an A.P. Let's find next two terms of this A.P.
$-\frac{3}{2}-1=-\frac{5}{2}, \quad \frac{5}{2}-1=-\frac{7}{2}$
$\therefore$ Next two terms are $-\frac{5}{2}$ and $-\frac{7}{2}$
View full question & answer→Question 672 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $5,12,19,26, \ldots$
AnswerIn this sequence $5,12,19,26, \ldots$,
First term $=t_1=5, \quad t_2=12, \quad t_3=19, \ldots$
$\begin{array}{l}
t_2-t_1=12-5=7 \\
t_3-t_2=19-12=7
\end{array}$
Here first term is 5 and common difference which is constant is $d=7$
$\therefore$ This sequence is an A.P.
Next two terms in this A.P. are $26+7=33$ and $33+7=40$.
Next two terms in given A.P. are 33 and 40
View full question & answer→Question 682 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $2,-2,-6,-10, \ldots$
AnswerIn the sequence $2,-2,-6,-10, \ldots$,
$\begin{array}{l}
t_1=2, \quad t_2=-2, t_3=-6, \quad t_4=-10 \ldots \\
t_2-t_1=-2-2=-4 \\
t_3-t_2=-6-(-2)=-6+2=-4 \\
t_4-t_3=-10-(-6)=-10+6=-4
\end{array}$
From this difference between two consecutive terms that is $t_n-t_{n-1}=-4$
$\therefore d=-4$, which is constant. $\quad \therefore$ It is an A.P.
Next two terms in this A.P. are $(-10)+(-4)=-14$ and $(-14)+(-4)=-18$
View full question & answer→Question 692 Marks
Which of the following sequences are A.P ? If it is an A.P, find next two terms : $1,1,2,2,3,3, \ldots$
AnswerIn the sequence $1,1,2,2,3,3, \ldots$,
$\begin{array}{l}
t_1=1, t_2=1, \quad t_3=2, \quad t_4=2, \quad t_5=3, \quad t_6=3 \ldots \\
t_2-t_1=1-1=0 \quad t_3-t_2=2-1=1 \\
t_4-t_3=2-2=0 \quad t_3-t_2 \neq t_2-t_1
\end{array}$
In this sequence difference between two consecutive terms is not constant.
$\therefore$ This sequence is not an A.P.
View full question & answer→Question 702 Marks
How many terms are there in the A.P. 187, 194, 201, ..., 439 ?
Question 712 Marks
Find $t_{11}$ from the following A.P. 4, 9, 14,…
View full question & answer→Question 722 Marks
Find $n$, if the $n^{\text {th }}$ term of the following sequence is 68 .
$5,8,11,14, \ldots$
View full question & answer→