Question 13 Marks
The houses of a row are numbered consecutively from $1$ to $49$. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of $x.$
[Hint: $S_{x-1} = S_{49} - S_x]$
AnswerAccording to the question,we have to find the value of x.
We are given an AP, namely $1,2,3,..., (x -1), x, (x +1),..., 49$
such that $1 + 2 + 3 +... + (x -1) = (x +1) + (x + 2) +... + 49.$
Thus, we have $S_{x-1}= S_{49}- S_x ... (i)$
Using the formula, $S_n$ =$\frac{n}{2}$ (a + l) in (i), we have,
$\frac { ( x - 1 ) } { 2 } \cdot \{ 1 + ( x - 1 ) \} = \frac { 49 } { 2 } \cdot ( 1 + 49 ) - \frac { x } { 2 } \cdot ( 1 + x )$
$\Rightarrow \frac { x ( x - 1 ) } { 2 } + \frac { x ( x + 1 ) } { 2 } = 1225$
$\Rightarrow 2x^2=2450$
$\Rightarrow x^2 =1225$
$\Rightarrow x= \sqrt {1225} =35$
Hence, $x = 35.$
View full question & answer→Question 23 Marks
A ladder has rungs $25$ cm apart. The rungs decrease uniformly in length from $45$ cm at the bottom to $25$ cm at the top. If the top and bottom rungs are $2\frac 12$m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs = $\frac{250}{25} + 1$]
AnswerIt is given that the gap between two consecutive rungs is 25 cm and the top and bottom rungs are 2.5 metre i.e., 250 cm apart.
$\therefore$ Number of rungs = $\frac { 250 } { 25 } $+ 1 = 10 + 1 = 11.
It is given that the rungs are decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top.
Therefore, lengths of the rungs form an A.P. with first term a = 45 cm and $11^{th}$ term l = 25 cm. n = 11
$\therefore$ Length of the wood required for rungs = Sum of 11 terms of an A.P. with first term 45 cm and last term is 25 cm
$= \frac { 11 } { 2 }$ ( 45 + 25 ) cm $\left[ \because S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$= \frac { 11 } { 2 }$(70) cm
$= 11 (35) cm$
$= 385 cm$
Length of the wood required for rungs = $\frac{385}{100}$ = 3.85 metres ($\because$100 cm = 1 m)
The length of the wood required for the rungs is 3.85 metres.
View full question & answer→Question 33 Marks
Which term of the AP: $121, 117, 113, ....$ is its first negative term?
[Hint: Find n for $a_n < 0]$
AnswerGiven: $121, 117, 113, .......$
Here $a = 121, d = 117 - 121 = 4$
Now, $a_n = a + (n - 1)d$
$= 121 + (n - 1) (-4) = 121 - 4n + 4 = 125 - 4n$
For the first negative term, $a_n < 0$
$ \Rightarrow 125 - 4n < 0 \Rightarrow 125 < 4n \Rightarrow \frac{{125}}{4} < n$
$ \Rightarrow 31\frac{1}{4} < n$
n is an integer and $n > 31\frac{1}{4}$.
Hence, the first negative term is $32^{nd}$ term
View full question & answer→Question 43 Marks
If the sum of the first $7$ terms of an $A.P$. is $49$ and that of the first $17$ terms is $289$, find the sum of its first n terms.
Answer$S_n$= $\frac n2$[2a + (n - 1)d]
$S_7$ = $\frac 72$(2a + 6d) = 49
or, $a + 3d = 7...........(i)$
$S_{17}$ = $\frac{17}{2}$(2a + 16d) = 289
or, $a + 8d = 17..........(ii)$
On subtracting (i) from (ii), we get
or, 5d = 10 or, d = 2
Put $d = 2$ in (i)
$a + 3d = 7$
$a + 2(3) = 7$
$a + 6 = 7$
and $a = 1$
$S _ { n } = \frac { n } { 2 } [ 2 \times 1 + ( n - 1 ) 2 ]$
$= \frac { n } { 2 } [ 2 + 2 n - 2 ] $
$= \frac { n } { 2 } [ 2 n ] $
$= n ^ { 2 }$
Hence, sum of n terms $= n^2$
View full question & answer→Question 53 Marks
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
AnswerHere, $d = 7$
$a_{22} = 149$
Let the first term of the AP be a.
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_{22} = a + (22 - 1)d$
$ \Rightarrow a_{22} = a + 21d$
$ \Rightarrow 149 = a + (21) (7)$
$ \Rightarrow 149 = a + 147$
$ \Rightarrow a = 2$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{22}} = \frac{{22}}{2}\left[ {2(2) + (22 - 1)7} \right]$
$ \Rightarrow S_{22} = (11) [4 + 147]$
$ \Rightarrow S_{22} = (11) (151) $
$ \Rightarrow S_{22} = 1661$
Hence, the sum of the first $22$ terms of the AP is $1661.$
View full question & answer→Question 63 Marks
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.
AnswerHere, a = 5
l = 45
S = 400
We know that
$S = \frac{n}{2}(a + l)$
$ \Rightarrow 400 = \frac{n}{2}(5 + 45)$
$ \Rightarrow 400 = \frac{n}{2}(50)$
$ \Rightarrow 400 - 25n$
$ \Rightarrow n = \frac{{400}}{{25}}$
$ \Rightarrow n = 16$
Hence, the number of terms is 16.
Again, we know that
l = a + (n - 1)d
$ \Rightarrow $ 45 = 5 + (16 - 1)d
$ \Rightarrow $ 45 = 5 + 15d
$ \Rightarrow $ 40 = 15 d
$ \Rightarrow d = \frac{{40}}{{15}} = \frac{8}{3}$
Hence, the common difference is $\frac{8}{3}$.
View full question & answer→Question 73 Marks
How many terms of the AP :$ 9, 17, 25, ….$ must be taken to give a sum of 636?
AnswerThe given AP is $9,17,25, \ldots$
Here, $a =9$
$d=17-9=8$
Let $n$ terms of the AP must be taken
Then, $S _{ n }=636$
$ \Rightarrow \frac{n}{2}\left[ {2a + (n - 1)d} \right] = 636$
$ \Rightarrow \frac{n}{2}\left[ {2(9) + (n - 1)8} \right] = 636$
$ \Rightarrow n[9 + (n - 1)4] = 636$
$ \Rightarrow n[9 + 4n - 4] = 636$
$ \Rightarrow n[(4n + 5)] = 636$
$ \Rightarrow 4n^2 + 5n - 636 = 0$
$ \Rightarrow 4n^2 + 53n - 48n - 636 = 0$
$ \Rightarrow n(4n + 53) - 12(4n + 53) = 0$
$ \Rightarrow (4n + 53) (n - 12) = 0$
$ \Rightarrow 4n + 53 = 0 or n - 12 = 0$
$ \Rightarrow n = - \frac{{53}}{4}$ or n = 12
$n = - \frac{{53}}{4}$ is in admissible as n, being the number of terms, is a natural number
$\therefore n = 12$
Hence, $12$ terms of the $AP$ must be taken.
View full question & answer→Question 83 Marks
In an AP: a = 3, n = 8, s = 192, find d.
AnswerHere, a = 3
n = 8
S = 192
We know that
$S = \frac{n}{2}[2a + (n - 1)d]$
$ \Rightarrow 192 = \frac{8}{2}\left[ {2(3) + (8 - 1)d} \right]$
$ \Rightarrow 192 = 4[6 + 7d]$
$ \Rightarrow \frac{{192}}{4} = 6 + 7d$
$ \Rightarrow $ 48 = 6 + 7d
$ \Rightarrow $ 48 - 6 = 7d
$ \Rightarrow $ 42 = 7d
$ \Rightarrow $ 7d = 42
$ \Rightarrow d = \frac{{42}}{7}$
$ \Rightarrow $ d = 6
View full question & answer→Question 93 Marks
In an $AP: a = 8, a_n = 62, S_n = 210$, find n and d.
AnswerHere, $a = 8$
$a_n= 62$
$S_n = 210$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow 62 = 8 + (n - 1)d$
$ \Rightarrow 62 - 8 = (n - 1)d$
$ \Rightarrow 54 = (n - 1)d$
$ \Rightarrow (n - 1)d = 54 ........ (1)$
Again we know that
${S_n} = \frac{n}{2}\left[ {2n + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {2(8) + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + 54} \right]$ ......... Using (1)
$ \Rightarrow 210 = \frac{n}{2}(70)$
$ \Rightarrow 210 = 35n$
$ \Rightarrow n = \frac{{210}}{{35}}$
$ \Rightarrow n = 6$
Putting n = 6 in equation ( 1), we get
$(6 - 1)d = 54$
$ \Rightarrow 5d = 54$
$ \Rightarrow d = \frac{{54}}{5}$
View full question & answer→Question 103 Marks
In an AP:$ a = 2, d = 8, S_n = 90$, find n and $a_n.$
AnswerHere, $a = 2$
$d = 8$
$S_n = 90$
We know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow 90 = \frac{n}{2}\left[ {2(2) + (n - 1)8} \right]$
$ \Rightarrow 90 = n[2 + (n - 1) 4]$
$ \Rightarrow 90 = n[2 + 4n - 4]$
$ \Rightarrow 90 = n[4n - 2]$
$ \Rightarrow 90 = 2n[2n - 1]$
$ \Rightarrow 45 = n[2n - 1]$
$ \Rightarrow 45 = 2n^2 - n$
$ \Rightarrow 2n^2 - n - 45 = 0$
$ \Rightarrow 2n^2 - 10n + 9n - 45 = 0 $
$ \Rightarrow 2n(n - 5) + 9(n - 5) = 0 $
$ \Rightarrow (n - 5) (2n + 9) = 0$
$ \Rightarrow n - 5 = $0 or 2n + 9 = 0
$ \Rightarrow n = 5$ or $n = - \frac{9}{2}$
$n = - \frac{9}{2}$ is in admissible as n, being the number of terms, is a natural number.
$\therefore $ n = 5
Again, we know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_n = 2 (5 - 1)8$
$ \Rightarrow a_n = 2 + (4)8$
$ \Rightarrow a_n = 2 + 32$
$ \Rightarrow a_n = 34$
View full question & answer→Question 113 Marks
In an AP: $d = 5, S_9 = 75$, find a and $a_9 $.
AnswerHere, $d = 5$
$S_9 = 75$
We know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + (9 - 1)d} \right]$
$ \Rightarrow {S_9} = \frac{9}{2}\left[ {2a + 8d} \right]$
$ \Rightarrow {S_9} = 9\left[ {a + 4d} \right]$
$ \Rightarrow {S_9} = 9\left[ {a + 4 \times 5} \right]$
$ \Rightarrow S_9 = 9[a + 20]$
$ \Rightarrow 75 = 9a + 180$
$ \Rightarrow 9a = 75 - 180$
$ \Rightarrow 9a = -105$
$ \Rightarrow a = - \frac{{105}}{9}$
$ \Rightarrow a = - \frac{{35}}{3}$
Again, we know that
$a_n = a + (n - 1)d$
$ \Rightarrow $ $a_9 = a + (9 - 1)d$
$ \Rightarrow $ $a_9 = a + 8d$
$ \Rightarrow {a_9} = - \frac{{35}}{3} + 8(5)$
$ \Rightarrow {a_9} = - \frac{{35}}{3} + 40$
$ \Rightarrow {a_9} = \frac{{ - 35 + 120}}{3}$
$ \Rightarrow {a_9} = \frac{{85}}{3}$
View full question & answer→Question 123 Marks
In an AP: $a _{12} = 37, d = 3$, find a and $S_{12}.$
AnswerHere, $a_{12} = 37$
$d = 3$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_{12} = a + (12 - 1)d$
$ \Rightarrow a_{12} = a + 11d$
$ \Rightarrow 37 = a + 33$
$ \Rightarrow a = 37 - 33 = 4$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{12}} = \frac{{12}}{2}\left[ {2a + (12 - 1)d} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {2a + 11d} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {2 \times 4 + 11 \times 3} \right]$
$ \Rightarrow {S_{12}} = 6\left[ {8 + 33} \right]$
$ \Rightarrow {S_{12}} = 6 \times 41$
$ \Rightarrow {S_{12}} = 246$
View full question & answer→Question 133 Marks
In an AP: $a = 7, a_{13} = 35$, find d and $S_{13}.$
AnswerHere, $a = 7$
$a_{13} = 35$
$a_n= a + (n - 1)d$
$ \Rightarrow a_{13} = a + (13 - 1)d$
$ \Rightarrow a_{13} = a + 12d$
$ \Rightarrow 35 = 7 + 12d$
$ \Rightarrow 12d = 35 - 7$
$ \Rightarrow 12d = 28$
$ \Rightarrow d = \frac{{28}}{{12}}$
$ \Rightarrow d = \frac{7}{3}$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + (13 - 1)d} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}\left[ {2a + 12d} \right]$
$={S_{13}} = \frac{{13}}{2}\left[ {2(7) + 12\left( {\frac{7}{3}} \right)} \right]$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(14 + 28)$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(42)$
$ \Rightarrow {S_{13}} = (13)(21)$
$ \Rightarrow {S_{13}} = 273$
View full question & answer→Question 143 Marks
In an AP: $a = 5, d = 3, a_n = 50$, find n and $S_n.$
AnswerHere, $a = 5, d = 3, a_n = 50$
We know that
$a_n = a + (n – 1)d$
$ \Rightarrow 50 = 5 + (n - 1)3$
$ \Rightarrow (n – 1)3 = 50 - 5$
$ \Rightarrow (n - 1)3 = 45$
$ \Rightarrow n - 1 = \frac{{45}}{3}$
$ \Rightarrow n - 1 = 15$
$ \Rightarrow n = 15 + 1$
$ \Rightarrow n = 16$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_n} = \frac{{16}}{2}\left[ {2(5) + (16 - 1)3} \right]$
$ \Rightarrow S_n = 8[10 +45]$
$ \Rightarrow S_n = 8(55)$
$ \Rightarrow S_n = 440$
View full question & answer→Question 153 Marks
Find the sum of $-5 + (-8) + (-11) + …. + (-230).$
AnswerLet $S_n = –5 + (–8) + (–11) + …. + (–230)$
Clearly, the terms of the sum form an A.P.
with, $a = –5$
$d = –8 –(–5) = –8 + 5 = –3$
$l = –230$
Let the number of terms of the AP be n
We know that
$l = a + (n - 1)d$
$ \Rightarrow -230 = -5 + (n - 1) (-3)$
$ \Rightarrow (n - 1) (-3) = -230 + 5$
$ \Rightarrow (n - 1) (-3) = -225$
$ \Rightarrow n - 1 = \frac{{ - 225}}{{ - 3}} = 75$
$ \Rightarrow n = 75 + 1$
$ \Rightarrow n = 76$
Again, we know that
${S_n} = \frac{n}{2}(a + l)$
$ \Rightarrow {S_{76}} = \frac{{76}}{2}\left[ {( - 5) + ( - 230)} \right]$
$ \Rightarrow S_{76} = 38(-235)$
$ \Rightarrow S_{76} = -8930$
Hence, the required sum is $-8930.$
View full question & answer→Question 163 Marks
Find the sum of 34 + 32 + 30 + … + 10.
AnswerHere, a = 34
d = 32 - 34 = -2
l = 10
Let the number of terms of the AP be n.
We know that,
l = a + (n - 1)d
$ \Rightarrow $ 10 = 34 + (n - 1) (-2)
$ \Rightarrow $ (n - 1) (-2) = - 24
$ \Rightarrow $ $n - 1 = \frac{{ - 24}}{{ - 2}} = 12$
$ \Rightarrow n = 13$
Again, we know that,
${S_n} = \frac{n}{2}(a + l)$
$ \Rightarrow {S_{13}} = \frac{{13}}{2}(34 + 10)$
$ \Rightarrow {S_{13}} = 286$
Hence, the required sum is 286.
View full question & answer→Question 173 Marks
Find the sum of $7 + 10\frac{1}{2} + 14..... + 84$.
AnswerClearly, the terms of the given sum form an A.P,
with, a = 7
$d = 10\frac{1}{2} - 7 = 3\frac{1}{2} = \frac{7}{2}$
Let the number of terms of the AP be n.
We know that
l = a + (n - 1)d
$ \Rightarrow 84 = 7 + (n - 1)\frac{7}{2}$
$ \Rightarrow (n - 1)\frac{7}{2} = 84 - 7$
$ \Rightarrow (n - 1)\frac{7}{2} = 77$
$ \Rightarrow (n - 1) = 22$
$ \Rightarrow n = 22 + 1$
$ \Rightarrow n = 23$
Again, we know that
${S_n} = \frac{n}{2}(a + l)$
$ \Rightarrow {S_{23}} = \frac{{23}}{2}(7 + 84)$
$ \Rightarrow {S_{23}} = \frac{{2093}}{2}$
$ \Rightarrow {S_{23}} = \frac{{2093}}{2}$
$ \Rightarrow {S_{23}} = 1046\frac{1}{2}$
Hence, the required sum is $1046\frac{1}{2}$
View full question & answer→Question 183 Marks
In a potato race, a bucket is placed at the starting point, which is $5$ m from the first potato, and other potatoes are placed $3$ m apart in a straight line. There are n potatoes in the line (See Fig.).
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 $ \times$ 5 + 2 $ \times$ (5 + 3)] Answer
Let $d_1$_ = Distance run by the competitor to pick up first potato = 2 $ \times$ 5 m
$d_2$_ = Distance run by the competitor to pick up second potato = 2 (5 + 3) m
$d_3 $= Distance run by the competitor to pick up third potato = 2 (5 + 2 $ \times$ 3) m
$d_4 $= Distance run by the competitor to pick up fourth potato = 2 (5 + 3 $ \times$ 3) m .....
$d_n$_ = Distance run by the competitor to pick up $n^{th}$^ potato = 2 {5 + (n -1) $ \times$ 3} m
Therefore, total distance run by the competitor to pick up n potatoes
$= d_1 + d_2 + d_3 + ... + d_n$_
= 2 $ \times$ 5 + 2(5 + 3) + 2 (5 + 2 $ \times$ 3) + 2 (5 + 3 $ \times$ 3) +.....+2{5 + (n - 1) $ \times$ 3} metres
= 2 [5 + {5 + 3} + {5 + (2 $ \times$ 3)} + {5 + (3 $ \times$ 3} +....+ {5 + (n - 1) $ \times$ 3}]
$= 2\left[ {(5 + \mathop {5 + \cdots + }\limits_{n - times} 5) + \{ 3 + (2 \times 3) + (3 \times 3) + \cdots + (n - 1) \times 3\} } \right]$
$ = 2 [ 5 n + 3 \{ 1 + 2 + 3 + \cdots + ( n - 1 ) \} ]$
$ = 2 \left[ 5 n + 3 \left( \frac { n - 1 } { 2 } \right) \{ 1 + ( n - 1 ) \} \right]$$ \left[ \mathrm { U } \operatorname { sing } : \mathrm { S } _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$ = 2 \left\{ 5 n + \frac { 3 n ( n - 1 ) } { 2 } \right\}$= [10n + 3n (n - 1)] = 3$n^2 + 7n = n(3n + 7)$ metres View full question & answer→Question 193 Marks
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant $1$ tree, a section of Class II will plant $2$ trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
AnswerSince each section of each class plants the same number of trees as the class number and there are three sections of each class.
Three sections of class I will plant = 1 $\times$ 3 = 3
Three sections of class II will plant = 2 $\times$ 3 = 6
Three sections of class III will plant = 3 $\times$ 3 = 9 and so on.
Three sections of class XII will plant = 12 $\times$ 3 = 36
So, we get an A .P. $3, 6, 9, ..., 36.$
Here $a = 3$ and$ d = 6 -3 = 3$
$a_n = 36$
We know that, $a_n = a +(n -1) d$
$3 + (n - 1) 3 = 36$
$(n - 1) 3 = 33$
$(n - 1) = 11$
$n = 11+ 1$
$n = 12$
Sn $= \frac { n } { 2 } (a_1 +a_n) = \frac { 12 } { 2 } (3+ 36)= 234$
View full question & answer→Question 203 Marks
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:
₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc; the penalty for each succeeding day being ₹50 more than for the preceding day. How much does a delay of 30 days cost the contractor?
AnswerSince the penalty for each succeeding day is ₹ 50 more than for the preceding day.
Therefore, amount of penalty for different days forms an A.P. with first term a = 200 and common difference d = 250 - 200= 50.
We have to find how much does a delay of 30 days cost the contractor.
In other words, we have to find the sum of 30 terms of the A.P.
n = 30, a = 200 and d= 50
∴ Required sum $= \frac { 30 } { 2 } \{ 2 \times 200 + ( 30 - 1 ) \times 50 \}$$\left[ \because S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] \right]$
⇒ Required sum = 15 (400 + 29 $\times$ 50)
⇒ Required sum = 15 (400 + 1450)
⇒ Required sum = 15 $\times$ 1850 = 27750
Thus, a delay of 30 days will cost the contractor of ₹ 27750.
View full question & answer→Question 213 Marks
Find the sum of the first $15$ multiples of$ 8.$
AnswerThe first 15 multiples of $8$ are $8, 16, 24, 32,....$
Here, $a_2 - a_4 = 16 - 8 = 8$
$a_3 - a_2 = 24 - 16 = 8$
$a_4 - a_3 = 32 - 24 = 8$
i.e.$ a_{k-1} - a_k$_ is the same everytime.
So, the above list of numbers forms an AP.
Here, $a = 8$
$d = 8$
$n = 15$
$\therefore $ Sum of first 15 multiples of $8 =S_{15}$_
$ = \frac{{15}}{2}\left[ {2a + (15 - 1)d} \right]$ ......... $\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15 (a + 7d)$
$ = (15)(8 + 7 \times 8)$
$= (15) (8 + 56)$
$= (15) (64)$
$= 960$
View full question & answer→Question 223 Marks
Find the sum of the first $40$ positive integers divisible by $6$.
AnswerThe first $40$ positive integers divisible by $6$ are $6, 12, 18, 24, .....$
Here, $a_2 - a_1 = 12 - 6 = 6$
$a_3 - a_2 = 18 - 12 = 6$
$a_4 - a_3 = 24 - 18 = 6$
i.e.$ a_{k+1} - a_k$ is the same every time.
So, the above list of numbers form an AP.
Here, $a = 6$
$d = 6$
$n = 40$
$\therefore $ Sum of the first 40 positive integers = $S_{40}$
$ = \frac{{40}}{2}\left[ {2a + (40 - 1)d} \right]$ .........${\{\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]}\}$
= 20[2a + 39d]
$ = (20)[2 \times 6 + 39 \times 6]$
$= (20) (246)$
$= 4920$
View full question & answer→Question 233 Marks
If the sum of the first n terms of an A.P. is $4n - n^2$, what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the nth term.
AnswerGiven that,
$S_n = 4n - n^2$^
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, $a_2 = S_2 - S_1 = 4 - 3 = 1$
$d = a_2 - a = 1 - 3 = -2$
$a_n= a + (n - 1)d$
$= 3 + (n - 1)(-2)$
$= 3 - 2n + 2$
$= 5 - 2n$
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 $1. 3^{rd}, 10^{th}$^ and$ n^{th}$^ terms are $-1, -15$, and $5 - 2n$ respectivey.
View full question & answer→Question 243 Marks
Show that $a_1, a_2, ........., a_n,$ form an AP where $a_n = 3 + 4n.$
AnswerPut $n = 1, 2, 3, 4, .....$ in succession, we get
$a_1 = 3 + 4(1) = 3 + 4 = 7$
$a_2 = 3 + (2) = 3 + 8 = 11$
$a_3 = 3 + 4(3) = 3 + 12 = 15$
$a_4 = 3 + 4(4) = 3 + 16 = 19$
$\therefore $$ a_2 - a1 = 11 - 7 = 4$
$a_3 - a_2 = 15 - 11 = 4$
$a_4 - a_3 = 19 - 15 = 4$
i.e.$ a_{k+1} - a_k$_ is the same every time.
So,$ a_1, a_2, .....,$ an, .... from an AP.
Here, $a = a_1 = 7$
$d = a_2 - a_1 = 4$
$\therefore $ Sum of the first 15 terms $= S_{15}$_
$ = \frac{{15}}{2}\left[ {2a + (15 - 1)d} \right]$
$\because {S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ = \frac{{15}}{2}\left[ {2a + 14d} \right]$
$= 15(a + 7d)$
$ = (15)(7 + 7 \times 4)$
$= (15) (7 + 28)$
$= (15) (35)$
$= 525$
View full question & answer→Question 253 Marks
If the $3^{rd}$^ and the $9^{th} $terms of an AP are $4$ and -$ 8$ respectively, then which term of the given AP will be zero?
AnswerIt is given that $a_3= 4$ and $a_9= -8$
$a_3= a + 2d = 4 .....(i)$
$a_9= a + 8d = -8 .....(ii)$
Subtracting $eq^n$^(i) from$ eq^n$^(ii) we get;
$a + 8d - a - 2d = -8 - 4 = -12$
$6d = -12$
$d = - 2$
Substituting the value of d in $eq^n$^(i), we get
$a + 2 (-2) = 4$
$a - 4 = 4$
$a = 8$
Now;$ 0 = a + (n - 1)d$
$0 = 8 + (n - 1)(-2)$
$(n - 1)(- 2) = - 8$
$n - 1 = 4$
$n = 5$
Thus, $5^{th}$ term of the given AP is zero.
View full question & answer→Question 263 Marks
An AP consists of 50 terms of which $3^{\text {rd }}$ term is 12 and the last term is 106 . Find the $29^{\text {th }}$ term.
AnswerAn AP consists of $50$ terms and the $50^{\text {th }}$ term is equal to 106 and $a_3=12$
Using formula $a_n=a+(n-1) d$, to find $n^{\text {th }}$ term of arithmetic progression,
$a_{50}=a+(50-1) d \text { And } a_3=a+(3-1) d$
$\Rightarrow 106=a+49 d \text { and } 12=a+2 d$
These are equations consisting of two variables.
Using equation $106=a+49 d$, we get $a=106-49 d$
Putting value of $a$ in the equation $12=a+2 d$,
$12=106-49 d+2 d$
$\Rightarrow 47 d=94$
$\Rightarrow d=2$
Putting value of $d$ in the equation, $a=106-49 d$,
$a=106-49(2)=106-98=8$
Therefore, First term $= a =8$ and Common difference $= d =2$
To find $29^{\text {th }}$ term, we use formula $a_n=a+(n-1) d$ which is used to find $n^{\text {th }}$ term of arithmetic progression,
$a_{29}=8+(29-1) 2=8+56=64$
Therefore, 29th term of AP is equal to 64 .
View full question & answer→Question 273 Marks
Find the $31^{\text {st }}$ term of an AP whose $11^{\text {th }}$ term is 38 and the $16^{\text {th }}$ term is 73.
AnswerLet the first term and the common difference of the AP be a and d respectively.
Then,
$11th$ term $= 38 ....... $ Given
$ \Rightarrow a + (11 - 1)d = 38$
$\because {a_n} = a + (n - 1)d$
$ \Rightarrow a + 10d = 38 ........ (1)$
and, 16th term $= 73$
$ \Rightarrow a + (16 - 1)d = 73$
$\because {a_n} = a + (n - 1)d$
$ \Rightarrow a + 15d = 73 .......... (2)$
Solving (1) and (2), we get
$a = -32$
$d = 7$
Therefore, $31^{st}$ term
$= a + (31 - 1)d$
$= a + 30d$
$= -32 + (30) (7)$
$= - 32 + 210 = 178$
Hence, the $31^{st}$ term of the AP is $178.$
View full question & answer→Question 283 Marks
Check whether $–150$ is a term of the $AP: 11, 8, 5, 2, …..$
AnswerThe given list of numbers is $11, 8, 5, 2,.....$
$a_2- a_1= 8 - 11 = -3$
$a_3 - a_2 = 5 - 8 = -3$
$a_4 - a_3 = 2 - 5 = - 3$
i.e. $a_{k+1} - a_k$_ is the same every time.
So, the given list of numbers forms an $AP$ with first term $a = 11$ and the common difference $d = -3.$
Let $-150$ be the nth term of the given $AP$
Then, $a_n = -150$
$ \Rightarrow a + (n - 1) d = -150$
$ \Rightarrow 11+ (n - 1)(-3) = -150$
$ \Rightarrow (-3) (n - 1) = -150 - 11$
$ \Rightarrow (-3) (n - 1) = -161$
$ \Rightarrow 3(n - 1) = 161$
$ \Rightarrow n - 1 = \frac{{161}}{3}$
$ \Rightarrow n = \frac{{161}}{3} + 1$
$ \Rightarrow n = \frac{{164}}{3}$
But n should be a positive integer. So, -150 is not a term of $11, 8, 5, 2,....$
View full question & answer→Question 293 Marks
Find the number of terms in AP. 18, $15\frac{1}{2}$, $13, ….., – 47.$
Answer$18,15\frac{1}{2},13,..., - 47$
Here, $a = 18$
$d = 15\frac{1}{2} - 18 = \frac{{31}}{2} - 18 = - \frac{5}{2}$
$a_n = -47$
Let the number of terms be n.
Then,
$a_n = -47$
$ \Rightarrow a + (n - 1)d = -47$
$ \Rightarrow 18 + (n - 1)\left( { - \frac{5}{2}} \right) = - 47$
$ \Rightarrow - \frac{5}{2}(n - 1) = - 47 - 18$
$ \Rightarrow - \frac{5}{2}(n - 1) = - 65$
$ \Rightarrow \frac{5}{2}(n - 1) = 65$
$ \Rightarrow n - 1 = \frac{{65 \times 2}}{5}$
$ \Rightarrow n - 1 = 26$
$ \Rightarrow n = 26 + 1$
$ \Rightarrow n = 27$
Hence, the number of terms of the given $AP$ is $27.$
View full question & answer→Question 303 Marks
Find the number of terms in AP: $7, 13, 19, …., 205$
Answer$7, 13, 19, ...., 205$
Here $a = 7$
$d = 13 - 7 = 6$
$a_n = 205$
Let the number of terms be n.
Then, $a_n = 205$
$ \Rightarrow a + (n - 1)d = 205$
$ \Rightarrow 7 + (n - 1)6 = 205$
$ \Rightarrow 6(n - 1) = 205 - 7$
$ \Rightarrow 6(n - 1) = 198$
$ \Rightarrow n - 1 = \frac{{198}}{6}$
$ \Rightarrow n - 1= 33$
$ \Rightarrow n = 33 + 1$
$ \Rightarrow n = 34$
Hence, the number of terms of the given AP is 34.
View full question & answer→Question 313 Marks
Is the given sequence:$1^2, 5^2, 7^2, 73, .....$ forms an AP? If it forms an AP, then find the common difference d and write the next three terms.
AnswerWe have given the numbers as follows: $1^2, 5^2, 7^2, 73.....$
now find
$a_2 - a_1 = 5^2 - 1 = 25 - 1 = 24$
$a_3 - a_2 = 7^2 - 5^2 = 49 - 25 = 24$
$a_4 - a_3 = 73 - 7^2 = 73 - 49 = 24$
As, the common difference is the same. The sequence is in A.P.
Next three terms are: $a_5= a_4+ d = 73 + 24 = 97$
$a_6= a_5+ d = 97 + 24 = 121$
$a_7= a_6+ d = 121 + 24 = 145$
View full question & answer→Question 323 Marks
Is this $\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,...$ an AP? If it forms an AP, find the common difference d and write three more terms.
Answer$\sqrt 2 ,\sqrt 8 ,\sqrt {18} ,\sqrt {32} ,...$
${a_2} - {a_1} = \sqrt 8 - \sqrt 2 = 2\sqrt 2 - \sqrt 2 = \sqrt 2 $
${a_3} - {a_2} = \sqrt {18} - \sqrt 8 = 3\sqrt 2 - 2\sqrt 2 = \sqrt 2 $
${a_4} - {a_3} = \sqrt {32} - \sqrt {18} = 4\sqrt 2 - 3\sqrt 2 = \sqrt 2 $
i.e. $a_{k+1} - a_k$ is the same every time.
So, the given list of numbers forms an AP with the common difference d = $\sqrt 2 .$
The next three terms are:
$\sqrt {32} + \sqrt 2 = 4\sqrt 2 + \sqrt 2 = 5\sqrt 2 = \sqrt {50} $
$5\sqrt 2 + \sqrt 2 = 6\sqrt 2 = \sqrt {72} $
and $6\sqrt 2 + \sqrt 2 = 7\sqrt 2 = \sqrt {98} $
View full question & answer→Question 333 Marks
Is the given sequence $a, 2a, 3a, 4a,...$ forms an$ AP$? If it forms an $AP$, then find the common difference d and write the next three terms.
Answerfrom the given sequence, we can have
$a_{2}-a_{1}=2 a-a=a$
$a_{3}-a_{2}=3 a-2 a=a$
$ a_{4}-a_{3}=4 a-3 a=a$
since $a_{k+1}-a_{k}$ i.e. the common difference is the same for all values of k
Hence, the given sequence forms an AP.
Now the next three terms are:
$a_5= a + 4d = a + 4a = 5a$
$a_6= a + 5d = a + 5a = 6a$
$a_7= a + 6d = a + 6a = 7a$
Next three terms are: 5a, 6a and 7a
View full question & answer→Question 343 Marks
Which term of the AP: $3, 8, 13, 18, ………..,$ is $78$
AnswerThe given AP is $3, 8, 13, 18, .....$
Here $a = 3$
$d = 8 - 3 = 5$
Let the nth term of the AP be 78.
then, $a_n = a + (n - 1) d$
$ \Rightarrow 78 = 3 + (n - 1) (5)$
$ \Rightarrow 5(n - 1) = 78 - 3$
$ \Rightarrow 5(n - 1) = 75$
$ \Rightarrow n - 1 = \frac{{75}}{5}$
$ \Rightarrow n - 1 = 15$
$ \Rightarrow n = 15 + 1$
$ \Rightarrow n = 16$
View full question & answer→Question 353 Marks
In the AP, 38, ? , ? , ? , –22, find the missing terms?
AnswerLet the first terms and the common difference of the given AP be a and d respectively.
$ \Rightarrow $ a + (2 - 1)d = 38 $\because {a_n} = a + (n - 1)d$
$ \Rightarrow $ a + d = 38 ....... (1)
Sixth term = -22
$ \Rightarrow $ a + (6 - 1) d = -22
$ \Rightarrow $ a + 5d = -22 ......... (2)
Solving (1) and (2), we get
a = 53
d = -15
Therefore,
Third term = 53 + (3 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= 53 - 30
= 23
Fourth term = 53 + (4 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= 8
Fifth = 53 + (5 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= -7
Hence, the missing terms in the boxes are 53, 23, 8, -7
View full question & answer→Question 363 Marks
In the AP, –4, ?, ?, ?, ?, 6 find the missing terms?
AnswerLet the common difference of the given AP be d.
a = -4
6th term = 6
$ \Rightarrow $ -4 + (6 - 1)d = 6 $\because {a_n} = a + \left| {a - 1} \right|d$
$ \Rightarrow $ -4 + 5d = 6
$ \Rightarrow $ 5d = 6 + 4
$ \Rightarrow $ 5d = 10
$ \Rightarrow d = \frac{{10}}{5}$
$ \Rightarrow $ d = 2
Therefore,
Second term = -4 + 2 = -2
Third term = -2 + 2 = 0
Fourth term = 0 + 2 = 2
Fifth term = 2 + 2 = 4
Hence, the missing terms are -2, 0, 2, 4
View full question & answer→Question 373 Marks
In the AP, ?, 13, ?, 3 find the missing terms?
AnswerLet the first term and the common difference of
the given AP be a and d respectively.
Second term = 13
$ \Rightarrow $ a + (2 - 1)d = 13
$ \Rightarrow $ a + d = 13 ....... (1)
Fourth term = 3
$ \Rightarrow $ a + (4 - 1) d = 3
$ \Rightarrow $ a + 3d = 3 .......... (2)
Solving (1) and (2), we get
a = 18
d = -5
Therefore,
Third term = a + (3 - 1) d
= a + 2d
= 18 + 2(-5)
= 18 - 10
= 8
Hence, the missing terms are 18 and 8.
View full question & answer→Question 383 Marks
Ramkali saves ₹$5$ in the first week of a year and then increased her weekly savings by ₹$1.75$. If in the nth week, her weekly savings becomes ₹$20.75$, then find n.
AnswerHere,$ a = ₹ 5$
$d = ₹ 1.75$
$a_n = ₹ 20.75$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow 20.75 = 5 + (n - 1)d$
$ \Rightarrow (n - 1) (1.75) = 20.75 - 5$
$ \Rightarrow (n - 1) (1.75) = 15.75$
$ \Rightarrow n - 1 = \frac{{15.75}}{{1.75}}$
$ \Rightarrow n - 1 = 9$
$ \Rightarrow n = 10$
Hence, the required value of n is 10.
View full question & answer→Question 393 Marks
Suba Rao started work in $1995$ at an annual salary of ₹$5000$ and received a ₹$200$ raise each year. In what year did his annual salary will reach ₹$7000$?
AnswerAnnual salary received by Suba Rao in $1995,1996,1997, \ldots$ is
$\text { ₹5000, ₹5200, ₹5400,.............. } 7000 .$
Clearly, it is an arithmetic progression with first term $a=5000$ and common difference $d=200$.
Suppose Suba Rao's annual salary reaches to ₹7000 in nth years. Then,
$n ^{\text {th }}$ term of the above A.P. $=₹ 7000$
$\Rightarrow a+(n-1) d=7000$
$\Rightarrow 5000+(n-1) \times 200=7000$
$\Rightarrow(n-1) \times 200=2000$
$\Rightarrow \quad n-1=\frac{2000}{200} \Rightarrow n-1=10 \Rightarrow n=11$
Thus, 11th annual salary received by Suba Rao will be ₹$7000$. This means that after $10$ years i.e., in the year $2005$ his annual salary will reach to $₹ 7000$.
View full question & answer→Question 403 Marks
For what value of n, are the nth terms of two APs: $63, 65, 67, …$. and $3, 10, 17, ….$ equal?
AnswerFirst APs
$63, 65, 67, ......$
Here, $a = 63$
$d = 65 - 63 = 2$
$\therefore $ nth term
$\because a_n = a + (n - 1)d$
Second APs
$3, 10, 17, .....$
Here, $a = 3$
$d = 10 - 3 = 7$
$\therefore $ nth term $= 3 + (n - 1)7$
$\because a_n = a+(n - 1)d$
If the n th terms of two APs are equal then
$63 + (n - 1)2 = 3 + (n - 1)7$
$ \Rightarrow (n - 1)2 - (n - 1)7 = 3 - 63$
$ \Rightarrow (n - 1) (2 - 7) = -60$
$ \Rightarrow (n - 1) (-5) = -60$
$ \Rightarrow n - 1 = \frac{{ - 60}}{{ - 5}}$
$ \Rightarrow n - 1 = 12$
$ \Rightarrow n = 12 + 1$
$ \Rightarrow n = 13$
Hence, for n = 13th terms of the two APs are equal
View full question & answer→Question 413 Marks
How many multiples of $4$ lie between $10$ and $250?$
AnswerThe multiples of 4 that lie between $10$ and $250$ are:
$12, 16, 20, 24, ...., 248$
$a_2 - a_1 = 16 - 12 = 4$
$a_3 - a_2 = 20 - 16 = 4$
$a_4 - a_3 = 24 - 20 = 4$
As $a_{k+1} - a_k$ is the same for $k = 1, 2, 3$, etc.
The above list of numbers forms an AP with the first term $a = 12$
and the common difference $d = 4$
Last term $(l) = 248$
Let there be n term s in this AP. Then, nth term = l
$ \Rightarrow a + (n - 1)d = 248$
$ \Rightarrow 12 + (n - 1)4 = 248$
$ \Rightarrow (n - 1)d = 248 - 12$
$ \Rightarrow (n - 1) = 236$
$ \Rightarrow n - 1 = \frac{{236}}{4}$
$ \Rightarrow n - 1 = 59$
$ \Rightarrow n = 59 + 1$
$ \Rightarrow n = 60$
Hence, 60 multiples of 4 lie between 10 and 250.
View full question & answer→Question 423 Marks
Is the given series: $0, -4, -8, -12, ....$ forms an $AP$? If it forms an $AP$, then find the common difference d and write three more terms.
AnswerHere: $a_2- a_1= -4 - 0 = -4$
$a_3- a_2= - 8 + 4 = -4$
$a_4- a_3= -12 + 8 = -4$ , since $a_{k+1}- a_k$_ is same for all values of k
Hence, this is an AP.
The next three terms can be calculated as follows:
$a_5= a + 4d = 0 + 4(- 4) = -16$
$a_6= a + 5d = 0 + 5(- 4) = - 20$
$a_7= a + 6d = 0 + 6(- 4) = - 24$
Thus, the next three terms are:$ -16, -20$ and$ -24$
View full question & answer→Question 433 Marks
Is this $3,3 + \sqrt 2 ,3 + 2\sqrt 2 ,3 + 3\sqrt 2$, ..... an AP? If it forms an AP, find the common difference d and write three more terms.
AnswerAccording to question we are given that a spiral is made up of successive semi-circles, with centres alternately at A and B, starting with centre at A, of radii $0.5 \ cm, 1.0\ cm, 1.5\ cm, 2.0\ cm$, ....as shown in Fig.
Let $l_1, l_2, l_3, l_4,..l_{13}$_ be the lengths (circumferences) of semi-circles of radii $r_1= 0.5\ cm, r_2 =1.0\ cm, r_3 = 1.5\ cm, r_4 = 2.0\ cm, r_5 = 2.5\ cm,..$. respectively.
Now, Semi-perimeter of circle = $\pi\cdot r $
Therefore,
$l _ { 1 } = \pi r _ { 1 } = \pi \times 0.5 = \frac { \pi } { 2 } \mathrm { cm }$
$l _ { 2 } = \pi r _ { 2 } = \pi \times 1 = 2 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 3 } = \pi r _ { 3 } = \pi \times \frac { 3 } { 2 } = 3 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
$l _ { 4 } = \pi r _ { 4 } = \pi \times 2 = 4 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
and
$l _ { 13 } = \pi r _ { 13 } = \pi \times \frac { 13 } { 2 } \mathrm { cm } = 13 \left( \frac { \pi } { 2 } \right) \mathrm { cm }$
Therefore total length of the spiral = l_1 + l_2 + l_3 +...+ l_{13}
$\bf= \left\{ \frac { \pi } { 2 } + 2 \left( \frac { \pi } { 2 } \right) + 3 \left( \frac { \pi } { 2 } \right) + \dots + 13 \left( \frac { \pi } { 2 } \right) \right\} $
$\bf= \frac { \pi } { 2 } ( 1 + 2 + 3 + \cdots + 13 ) $
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } ( 1 + 13 ) \quad \left[ \text { Using } S _ { n } = \frac { n } { 2 } ( a + l ) \right]$
$\bf= \frac { \pi } { 2 } \times \frac { 13 } { 2 } \times 14$ = $\bf\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times 13 \times 7$ = $\bf {143 cm}$
which is required length of the spiral made up of thirteen consecutive semi-circles. View full question & answer→Question 443 Marks
Is this $–10, –6, –2, 2, ....$ an $AP$? If it forms an $AP$, find the common difference d and write three more terms.
Answer$-10, -6, -2, 2, ....$
$a_2 - a_1 = -6 - (-10) = -6 + 10 = 4$
$a_3 - a_2 = -2 - (-6) = -2 + 6 = 4$
$a_4 - a_3 = 2 - (-2) = 2 + 2 = 4$
i.e. $a_{k+1} - a_k$_ is the same every time.
So, the given lists of numbers form an AP with the common difference $d = 4.$
The next three terms are:
$2 + 4 = 6, 6 + 4 = 10$ and $10 + 4 = 14$
View full question & answer→Question 453 Marks
Is this $–1.2, –3.2, –5.2, –7.2, …$. an AP? If it forms an $AP$, find the common difference d and write three more terms.
Answer$-1.2, -3.2, -5.2, -7.2, ....$
$a^2 - a^1 = -3.2 - (-1.2) = -3.2 + 1.2 = -2.0$
$a^3 - a^2 = -5.2 - (-3.2) = -5.2 + 3.2 = -2.0$
$a^4 - a^3 = -7.2 - (-5.2) = -7.2 + 5.2 = -2.0$
i.e. $a_{k+1} - a_k$_ is the same everytime,
So, the given list of numbers form an AP with the common differenced $d = -2.0$
The next three terms are:
$-7.2 + (-2.0) = -9.2$
$-9.2 + (-2.0) = -11.2$
and$ -11.2 + (-2.0) = -13.2$
View full question & answer→Question 463 Marks
Is the given series: $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$ form an $AP$? If It forms an AP, then find the common difference d and write the next three terms.
AnswerAs per the question:
$a_1= 2$
$a_{2 =}\frac{5}{2}$
$a_3= 3$
$a_4= {7\over 2}$
now check the common difference (d)
${a_2} - {a_1} = \frac{5}{2} - 2 = \frac{1}{2}$
${a_3} - {a_2} = 3 - \frac{5}{2} = \frac{1}{2}$
${a_4} - {a_3} = \frac{7}{2} - 3 = \frac{1}{2}$
we can see that the common difference is the same everywhere, so the given series forms an AP.
now next three terms are: ${a_5} = \frac{7}{2} +\frac{1}{2}= 4$
${a_6}= 4+\frac{1}{2} = \frac{9}{2}$
${a_7}= \frac{9}{2} +\frac{1}{2} = 5$
View full question & answer→Question 473 Marks
Is the given series $2, 4, 8, 16, ......$ form an $AP$? If It forms an $AP$, then find the common difference d and write the next three terms.
AnswerIf $a_{k+1}- a_k$is same for different values of k, then the series is in the form of an AP.
here, we have $a_1= 2, a_2= 4, a_3= 8$ and $a_4 = 16$
$a_4- a_3 = 16 - 8 = 8$
$a_3- a_2 = 8 - 4 = 4$
$a_2- a_1= 4 - 2 = 2$
$Here, $a_{k+1}- a_k$_ i.e. the common difference is not same for all values of k
Hence, the given series does not form an AP.
View full question & answer→Question 483 Marks
The amount of money in the account every year, when ₹ $10000$ is deposited at compound interest at $8\ %$ per annum. Is this situation make an arithmetic progression and why?
AnswerAmount of money after 1 year $ = Rs10000\left( {1 + \frac{8}{{100}}} \right) = {a_1}$
Amount of money after 2 year $ = Rs10000{\left( {1 + \frac{8}{{100}}} \right)^2} = {a_2}$
Amount of money after 3 year $ = Rs10000{\left( {1 + \frac{8}{{100}}} \right)^3} = {a_3}$
Amount of money after 4 year $ = Rs10000{\left( {1 + \frac{8}{{100}}} \right)^4} = {a_4}$
${a_2} - {a_1} = Rs10000{\left( {1 + \frac{8}{{100}}} \right)^2} - Rs10000\left( {1 + \frac{8}{{100}}} \right)$
$ = Rs10000\left( {1 + \frac{8}{{100}}} \right)\left( {1 + \frac{8}{{100}} - 1} \right)$
$= 10000(1 + {8 \over {100}})({8 \over {100}})$
$a_3-a_2$
$=10000{\left( {1 + \frac{8}{{100}}} \right)^2} - 10000\left( {1 + \frac{8}{{100}}} \right)$
$ = 10000\left( {1 + \frac{8}{{100}}} \right)\left( {1 + \frac{8}{{100}} - 1} \right)$
$ = 10000\left( {1 + \frac{8}{{100}}} \right)\left( {\frac{8}{{100}}} \right)$
Since.${a_{^3}} - {a_2} \ne {a_2} - {a_1}$.It does not form AP.
View full question & answer→Question 493 Marks
The cost of digging a well after every metre of digging, when it costs ₹ $150$ for the first metre and rises by ₹ $50$ for each subsequent metre. Is this situation make an arithmetic progression and why?
AnswerCost of digging the well after 1 metre of digging $= Rs 150 = a_1$
Cost digging the well after 2 metres of digging
$= Rs 150 + Rs 50 = Rs 200 = a_2$_
Cost of digging the well after 3 metres of digging
$= Rs 200 + Rs 50 = Rs 250 = a_3$_
Cost of digging the well after 4 metres of digging
$= Rs 250 + Rs 50 = Rs 300 = a_4$_
and so on.
$a_2 – a_1 = Rs 200 – Rs 150 = Rs 50$
$a_3 – a_2 = RS 250 – Rs 200 = Rs 50$
$a_4– a_3 = Rs 300 – Rs 250 = Rs 50$
i.e.$ a_{k + 1} – a$ is the same everytime.
So this list of numbers forms an AP with the first term $a = Rs 150$
and the common difference $d = Rs 50.$
View full question & answer→Question 503 Marks
The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}{/tex} of the air remaining in the cylinder at a time. Is this situation make an arithmetic progression and why?
AnswerLet the volume of the cylinder be 16 litres($a_1$).
Air removed by pump = ${1 \over 4} \times 16 = 4~litres$
Air present after first removal = 16 - 4 = 12 litres($a_2$)
Air again removed = ${1 \over 4} \times 12 = 3litres$
Air present after second removal = 12 - 3 = 9 litres($a_3$)
The amount of air present in the cyinder is the series
$16,12,9.....$
$a_2- a_1= 12-16 = -4$
$a_3- a_2= 9-12 = -3$
Since the difference is not same. This is not A.P.
View full question & answer→Question 513 Marks
The taxi fare after each km when the fare is ₹ $15$ for the first km and ₹ $8$ for each additional km. Is this situation make an arithmetic progression and why?
AnswerTaxi fare for $1 km = Rs 15 = a_1$_
Taxi Fare for 2 kms
$= R s 15 + R s 8 = R s 23 = a _ { 2 }$
Taxi fare for 3 km s
$= R s 23 + R s 8 = R s 31 = a _ { 3 }$
Taxi fare for 4 kms
$= R S 31 + R s 8 = R S 39 = a _ { 4 }$
and so on
$a _ { 2 } - a _ { 1 } = R s .23 - R s .15 = R s 8$
$a _ { 3 } - a _ { 2 } = R s 31 - R s \cdot 23 = R s 8$
$a _ { 4 } - a _ { 3 } = R s 39 - R s 31 = R s 8$
So, the arithmetic progression formed is:-
i.e.$,a_{k+1}-a_k$_ is the same every time.
So, this list of numbers form an arithmetic
Progression with the first term $a = Rs 15$ and
the common difference $d = Rs 8.$
View full question & answer→Question 523 Marks
A sum of ₹ $1000$ is invested at $8\ %$ simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of $30$ years making use of this fact.
AnswerLet P be the principle, R rate of interest and $I_n$_ be the interest at the end of n year
We know that
$I_n = \frac { P R n } { 100 }$
$\left[ \text { Using : Interest } = \frac { P R T } { 100 } \right]$
A sum of ₹1000 is invested at $8\ %$ simple interest per annum.
Here, we have
$P = ₹1000,$ and $R = 8\ % $per annum
$\therefore$ I_n = ₹$\left( \frac { 1000 \times 8 \times n } { 100 } \right)$= ₹ 80n
Putting $n = 1,2,3,...,$ we have
$l_n = 80n$
$I_1 = 80 \times 1 = ₹80$
$I_2 = 80 \times 2 = ₹160$
$I_3 = 80 \times 3 = ₹240$
$I_4 = 80 \times 4 = ₹320$ and so on.
Since, $I_n$_ is a linear expression in n.
Therefore, the sequence of interest forms an A.P. with common difference 80.
Hence, the sequence of interests is an A.P.
Also, Interest at the end of 30 years $= I_{30}= 80n = ₹ (80 \times 30) = ₹ 2400$
View full question & answer→Question 533 Marks
Check whether 301 is a term of the given list of numbers: $5, 11, 17, 23,...?$
AnswerWe have :
$a_2-a_1=11-5=6, a_3-a_2=17-11=6, a_4-a_3=23-17=6$
As $a_{k+1}-a_k$ is the same for $k=1,2,3$, etc., so the given list of numbers are in AP.
Now, $a=5$ and $d=6$.
Let 301 be a term, say, the $n ^{\text {th }}$ term of this AP.
We know that
$a_n=a+(n-1) d$
So, $301=5+(n-1) \times 6$
i.e., $301=6 n-1$
So, $n=\frac{302}{6}=\frac{151}{3}$, since n is in the form of fraction , thus 301 is not the term of given AP
View full question & answer→Question 543 Marks
Determine the $AP$ whose 3rd term is $5$ and the $7th$ term is $9$.
AnswerWe have
$a_3 = a + (3 – 1) d = a + 2d = 5 ....(i)$
and $a_7 = a + (7 – 1) d = a + 6d = 9 .....(ii)$
Solution by substitution method: Now from equation (i), value of$ a = 5 - 2d .....(iii)$
put value of a from equation (iii) in equation (ii), we get
$5 - 2d + 6d = 9$
$4d = 9 - 5$
$4d = 4$
$d = 1$
now put value of d in equation (iii), we get
$a = 5 - 2\times1$
$a = 3$
Hence, the required AP is $3, 4, 5, 6, 7,...$
View full question & answer→Question 553 Marks
Which term of the A.P$ 21, 18, 15, . . .$ is $– 81$? Also, is any term $0$? Give reason for your answer.
AnswerHere, $a = 21, d = 18 – 21 = – 3$ and $a_n = – 81$, and we have to find n.
As $a_n = a + ( n – 1) d$,
we have$ – 81 = 21 + (n – 1)(– 3)$
$– 81 = 24 – 3n$
$– 105 = – 3n$
$So, n = 35$
Therefore, the 35th term of the given $A.P$ is $– 81.$
Next, we want to know if there is any n for which an $= 0$. If such an n is there, then
$21 + (n – 1) (–3) = 0,$
$i.e., 3(n – 1) = 21$
$i.e., n = 8$
So, the eighth term is $0.$
View full question & answer→Question 563 Marks
Elpis Technology is a TV manufacturer company. It produces smart TV sets not only for the Indian market but also exports them to many foreign countries. Their TV sets have been in demand every time but due to the Covid-19 pandemic, they are not getting sufficient spare parts especially chips to accelerate the production. They have to work in a limited capacity due to the lack of raw material.
They produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:
- the production in the 1st year (2)
- the production in the 10th year (1)
- the total production in first 7 years (1)
Question 573 Marks
If the sum of the first $14$ terms of an $A.P.$ is $1050$ and its first term is $10$ find its $20^{th}$ term.
AnswerGiven, $a = 10,$ and $S_{14} = 1050$
Let the common difference of the A.P. be d
$\therefore \quad S _ { n } = \frac { n } { 2 } [ 2 a + ( n - 1 ) d ]$
$\therefore \quad S _ { 14 } = \frac { 14 } { 2 } [ 2 \times 10 + ( 14 - 1 ) d ]$
1050 = $7 (20 + 13 d )$
$20 + 13d$ = $\frac{1050}{7}$
$20 + 13d = 150$
$13d = 150 - 20$
$13d = 130$
$d$ = $\frac{130}{13} = 10$
$a_{20} = a + (n - 1)d$
$= 10 + (20 - 1) 10$
$= 10 + 19 $\times$ 10$
$= 10 + 190$
$= 200$
Hence, $a_{20} = 200$
View full question & answer→