Question 11 Mark
In an AP: $l = 28, S = 144,$ and there are total $9$ terms. Find $'a'.$
AnswerHere, $l = 28$
$S = 144$
$n = 9$
We know that
$S = \frac{n}{2}(a + l)$
$ \Rightarrow 144 = \frac{9}{2}(a + 28)$
$ \Rightarrow \frac{{(144)(2)}}{9} = a + 28$
$ \Rightarrow $ $32 = a + 28$
$ \Rightarrow $ $a = 32 - 28$
$ \Rightarrow $ $a = 4$
View full question & answer→Question 21 Mark
Find the sum of the APs: $0.6, 1.7, 2.8,….,$ to $100$ terms.
AnswerHere, $a = 0.6$
$d = 1.7 -0.6 = 1.1$
$n = 100$
We know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{100}} = \frac{{100}}{2}\left[ {2(0.6) + (100 - 1)(1.1)} \right]$
$ \Rightarrow {S_{100}} = 50[1.2 + 108.9]$
$ \Rightarrow {S_{100}} = 50[110.1]$
So, the sum of the first $100$ terms of the given $AP$ is $5505$.
View full question & answer→Question 31 Mark
Find the sum of the APs: $–37, –33, –29, …,$ to $12$ terms.
Answer$Here, a = -37$
$d = -33 - (-37) = -33 + 37 = 4$
$n = 12$
We know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{12}} = \frac{{12}}{2}[2( - 37) + (12 - 1)4]$
$ \Rightarrow {S_{12}} = 6[ - 74 + 44]$
$ \Rightarrow {S_{12}} = 6[ - 30]$
$ \Rightarrow {S_{12}} = - 180$
So, the sum of the first $12$ terms of the given AP is $-180$.
View full question & answer→Question 41 Mark
Find the sum of an AP given as: $2, 7, 12,...$ upto $10$ terms.
AnswerHere, $a = 2, d = 5$ and $n = 10$
Sum of $n$ terms can be given as follows:
$S_n=\frac{n}{2}[2 a+(n-1) d]$
$\mathrm{S}_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]$
$= 5(4 + 45)$
$ = 5 \times 49$
$= 245$
Thus, the sum of the $10$ terms of given AP, $(S_{10}) = 25$
View full question & answer→Question 51 Mark
In the AP $2, ? , 26,$ find the missing terms?
AnswerLet the common difference of the given $AP$ be $d$.
Then,
Third term $= 2 + d + d = 2 + 2d$
According to the question,
$2 + 2d = 26$
$ \Rightarrow $ $2d = 26 - 2$
$ \Rightarrow $ $2d = 24$
$ \Rightarrow d = \frac{{24}}{2} = 12$
So, second term $= 2 + d = 2 + 12 = 14$
Hence, the missing termed is $14$
View full question & answer→Question 61 Mark
Determine the $AP$ whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12$.
Answer$a_3= 16$
$ \Rightarrow a + 2d = 16 ..... (i)$
$a_7= a_5+ 12$
$ \Rightarrow a + 6d = a + 4d + 12$
$ \Rightarrow 2d = 12 $
$ \Rightarrow d = 6$
Put the value of d in eq. $(i)$
$a + 2 \times 6 = 16$
$ \Rightarrow a = 16 - 12$
$ \Rightarrow a = 4$
$4, 10, 16....$
View full question & answer→Question 71 Mark
How many three-digit numbers are divisible by $7?$
AnswerAll 3-digit numbers divisible by $7$ are
$105,112,119, ...,994.$
Clearly, these numbers form an AP with $a = 105, d = (112-105) = 7$ and $l = 994.$
Let it contain n terms. Then,
$T_n= 994$ $ \Rightarrow a + (n-1)d = 994$
$\Rightarrow 105 + (n -1) 7 = 994$
$\Rightarrow 98 + 7n = 994$
$\implies7n = 896$
$ \Rightarrow n = 128.$
Hence, there are $128$ three-digit numbers divisible by $7.$
View full question & answer→Question 81 Mark
Are the given numbers $1, 3, 9, 27, ........$ forms an $AP?$ If It forms an AP, then find the common difference d and write the next three terms.
AnswerFrom the given numbers, we can have
$ a_2-a_1=3-1=2$
$ a_3-a_2=9-3=6$
$ a_4-a_3=27-9=18$
$\text { since } a_{k+1}-a_k \text { i.e. the common difference is not same for all values of } k$
Hence, it is not an AP. So, we can not find next three terms.
View full question & answer→Question 91 Mark
Is the given series: $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$ forms an AP? If It forms an AP, then find the common difference d and write the next three terms.
AnswerHere, it is given that all terms are same, so the common difference $(d) = 0$
since $a_{k+1}-a_k$ is the same for all values of k
Hence, it forms an AP.
The next three terms will be the same, i.e. $ - \frac{1}{2}$,$ - \frac{1}{2}$,$ - \frac{1}{2}$
View full question & answer→Question 101 Mark
Is the sequence $0.2, 0.22, 0.222, 0.2222, ….$ forms an AP$?$ If it forms an AP, find the common difference $d$ and write three more terms.
Answer$0.2, 0.22, 0.222, 0.2222, 000$
$a_2- a_1= 0.22 - 0.2 = 0.02$
$a_3- a_2= 0.222 - 0.22 = 0.002$
As $a_2-a_1 \ne a_3- a_2$, the given list of numbers does not form an AP.
View full question & answer→Question 111 Mark
Is the given sequence: $1^2, 3^2, 5^2, 7^2, .....$ forms an AP? If it forms an AP, then find the common difference d and write the next three terms.
AnswerThe given terms can be written as follows:
$1, 9, 25, 49,....$
$ \text { Here, } a_2-a_1=9-1=8 $
$ a_3-a_2=25-9=16 $
$a_4-a_3=49-25=24$
$\text { Since } a_{k+1}-a_k \text { is not same for all values of } k$
Hence, it is not an AP. So we can not find next three terms.
View full question & answer→Question 121 Mark
Is the given sequence: $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$ form an AP? If it forms an AP, then find the common difference d and write the next three terms.
AnswerFrom the given information, we can have
${a_{2}-a_{1}=\sqrt{6}-\sqrt{3}}$
${a_{3}-a_{2}=\sqrt{9}-\sqrt{6}=3-\sqrt{6}}$
$a_{4}-a_{3}=\sqrt{12}-\sqrt{9}=2 \sqrt{3}-3$ $(since\sqrt{12}=\sqrt{2} \times 2 \times 3=2 \sqrt{3}) $
since $a_{k+1}-a_{k}$ is not the same for all values of $k$.
Hence, it is not an AP.
View full question & answer→Question 131 Mark
For the AP $0.6, 1.7, 2.8, 3.9 ........,$ write the first term and the common difference.
Answer$0.6, 1.7, 2.8, 3.9...$
First term $= a= 0.6$
Common difference $(d) =$ Second term -First term
= Third term - Second term and so on
Therefore, Common difference $(d) = 1.7−0.6=1.1$
View full question & answer→Question 141 Mark
For the AP $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},.....$, write the first term and the common difference.
Answer$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{{13}}{3},.....$
First term $(a)$ $ = \frac{1}{3}$
Common difference $(d)$ $ = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$
View full question & answer→Question 151 Mark
For the AP $–5, –1, 3, 7,...$ write the first term and the common difference.
AnswerGiven A.P is: $-5, - 1, 3, 7, ...$
First term $(a) = -5$
Common difference $(d) = -1 - (-5) = -1 + 5 = 4$
View full question & answer→Question 161 Mark
For the AP $3, 1, -1, -3 ......,$ write the first term and the common difference.
Answer$3, 1, -1, -3...$
First term $= a= 3,$
Common difference $(d) =$ Second term - first term = Third term - second term and so on
Therefore, Common difference $(d) = 1 - 3 = -2$
View full question & answer→Question 171 Mark
Write first four terms of the AP, when the first term $a = -1.25$ and the common difference, $d = –0.25$
AnswerHere, $a_1= a = -1.25, d = -0.25$
First term, $a = -1.25$
Second term, $a_2= -1.25 + d = -1.25 + (-0.25) = -1.50$
Third term $= -1.50 + d = -1.50 + (-0.25) = -1.75$
Fourth term $= -1.75 + d = -1.75 + (-0.25) = -2.00$
Hence, first four terms of the given AP are $-1.25, -1.75, -2.00$
View full question & answer→Question 181 Mark
Write the first four terms of the AP, when the first term $a = -1$ and the common difference d = $\frac{1}{2}$
Answer$a = -1,$ $d = \frac{1}{2}$
First term $= a = -1$
Second term $= -1 + d$ $ = - 1 + \frac{1}{2} = - \frac{1}{2}$
Third term $ = - \frac{1}{2} + d = - \frac{1}{2} + \frac{1}{2} = 0$
Fourth term $= 0 + d$ $ = 0 + \frac{1}{2} = \frac{1}{2}$
Hence, the first four terms of the given AP are $-1,$ $ - \frac{1}{2},0,\frac{1}{2}$.
View full question & answer→Question 191 Mark
Write first four terms of the AP, when the first term $a = 4$ and the common difference $d = -3$
Answer$a = 4, d = -3$
First term $= a = 4$
Second term $= 4 + d = 4 + (-3) = 1$
Third term $= 1 + d = 1 + (-3) = -2$
Fourth term = $- 2 + d = - 2 + ( - 3 ) = - 5$
Hence, four first terms of the AP are $4, 1, -2, -5.$
View full question & answer→Question 201 Mark
Write the first four terms of the AP, when the first term $a = -2$ and the common difference $d = 0$
Answer$a = -2, d = 0$
First term $= a = -2$
Second term $= -2 + d = -2 + 0 = -2$
Third term $= -2 + d = -2 + 0 = -2$
Fourth term $= -2 + d = -2 + 0 = -2$
Hence, first four terms of the given AP are $-2, -2, -2, -2$ respectively.
View full question & answer→Question 211 Mark
Write first four terms of the AP, when the first term $a = 10$ and the common difference $d = 10.$
Answer$a = 10, d = 10$
First term $a = 10$
Second term $= 10 + d = 10 + 10 = 20$
Third term $= 20 + d = 20 + 10 = 30$
Fourth term $= 30 + d = 30 + 10 = 40$
Hence, first four terms of the given AP are $10, 20, 30, 40$.
View full question & answer→Question 221 Mark
Find the $11th$ term from the last term (towards the first term) of the AP : $10, 7, 4, ..., -62.$
Answer$a = – 62, d = 3$
$a_{11}= a + 10d$
$= – 62 + 10(3)$
$= – 32$
View full question & answer→Question 231 Mark
How many two-digit numbers are divisible by $3?$
AnswerThe two -digit numbers divisible by $3$ start from $12,15,18,21,...,99$
Here,
$a=12$
$d=3$
$a_n=a+(n-1)d$
$\Rightarrow$ $99=12+(n-1)(3)$
$\Rightarrow$$ 99=12+3n-3$
$\Rightarrow$ $90=3n$
$\Rightarrow$ $n=30$
Thus, $30$ two-digit numbers are divisible by $3$.
View full question & answer→Question 241 Mark
Find the $10^{th}$ term of the AP $2, 7, 12,...$
AnswerHere $a = 2$;
$d = 7 - 2 = 5$.
So, $a_{10}= a + 9d = 2 + 9$ $\times$ $5 = 47$
View full question & answer→Question 251 Mark
Find the sum of first 24 terms of the list of numbers whose nth term is given by $a_n= 3 + 2n.$
Answer$a_n= 3 + 2n$
Put $n = 1, 2, 3,...$
$a_1= 5, a_2= 7, a_3= 9....$
$a = 5, d = 7 - 5 = 2$
${S_{24}} = \frac{{24}}{2}\left[ {2 \times 5 + (24 - 1) \times 2} \right]= 12[10 + 46] = 672$
View full question & answer→Question 261 Mark
Find the sum of the first 1000 positive integers.
Answer$S_n= \frac { n } { 2 } [ a + l ]$
$\Rightarrow S_{1000}= \frac { 1000 } { 2 }[1 + 1000] = 500500$
View full question & answer→Question 271 Mark
Find the sum of first $22$ terms of the AP $8,3, - 2,...$
AnswerHere $a = 8,\ d = 3 - 8 = -5.$
So,$ S_{22}= \frac{22}{2}[2a + (22 - 1)d]$
$\Rightarrow$ $S_{22} = 11(16 - 105) = -979$
View full question & answer→Question 281 Mark
In a flower bed, there are $23$ rose plants in the first row, $21$ in the second, $19$ in the third, and so on. There are $5$ rose plants in the last row. How many rows are there in the flower bed?
AnswerThe number of rose plants in the $1^{st}, 2^{nd}, ......$ are $23, 21, 19,...5$
$a = 23, d = 21 - 23 = - 2 , a_n= 5$
$\therefore a_n= a + (n - 1 )d$
or, $5 = 23 + (n - 1)(-2)$
or, $5 = 23 -2n + 2$
or, $5 = 25 -2n$
or, $2n = 20$
or, $n =10$
So, there are $10$ rows.
View full question & answer→Question 291 Mark
For the AP: $\frac { 3 } { 2 } , \frac { 1 } { 2 } , -\frac {1 } { 2 } , -\frac {3} {2}$, ... write the first term a and the common difference $d$.
AnswerGiven AP = $\frac { 3 } { 2 } , \frac { 1 } { 2 } , \frac { - 1 } { 2 } , \frac { - 3 } { 2 }$... $1^{st}$ term = $\frac{3}{2}$ Common difference = $\frac { 1 } { 2 } - \frac { 3 } { 2 }$ = -1
View full question & answer→