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Mathematics [2 Mark Question Answer]

Arithmetic and Geometric Progressions · ICSE STD 10 (ICSE - English Medium). 29 questions.

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[2 Mark Question Answer]

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29 questions · timed · auto-graded

Question 12 Marks
Find the sum of first $22$ terms, of an A.P. in which $d = 7$ and $a_{22}$ is $149$.
Answer
Sum of first 22 terms of an A.P. whose $d=7$
$
\begin{aligned}
& a_{22}=149 \text { and } n=22 \\
& 149=a+(n-1) d \\
& =a+21 \times 7 \\
& 149=a+147 \\
& \Rightarrow a=149-147=2 \\
& \therefore S_{22}=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{22}{2}[2 \times 2+(22-1)(7)] \\
& =11[4+21 \times 7] \\
& =11 \times[4+147] \\
& =11 \times 151 \\
& =1661 .
\end{aligned}
$
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Question 22 Marks
In an A.P. (with usual notations) : given a = 3, n = 8, S = 192, find d.
Answer
$\begin{aligned} & a =3, n =8, S =192 \\ & S _{ n }=\frac{n}{2}[2 a+(n-1) d] \\ & 192=\frac{8}{2}[2 \times 3+7 \times d] \\ & 192=4[6+7 d ] \\ & \Rightarrow \frac{192}{4}=6+7 d \\ & \Rightarrow 48=6+7 d \\ & \Rightarrow 7 d =48-6=42 \\ & d =\frac{42}{7}=6 \\ & \therefore d =6 .\end{aligned}$
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Question 32 Marks
In an $A.P$., the sum of its first n terms is $6n – n^2$. Find is $25th$ term.
Answer
$S_n = 6n – n²$
$T_{25} = ?$
$S_{(n–1)} = 6(n – 1) – (n – 1)^2$
$= 6n – 6 – (n^2 – 2n + 1)$
$= 6n – 6 – n^2 + 2n –1$
$= 8n – n^2 – 7$
$a_n = S_n – S_n – 1$
$= 6n – n^2 – 8n + n^2 + 7$
$= –2n + 7$
$a_{25}= –2(25) + 7$
$= –50 + 7$
$= –43.$
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Question 42 Marks
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given that,
$
\begin{aligned}
& a_2=14 \\
& a_3=18 \\
& d=a_3-a_2=18-14=4 \\
& a_2=a+d \\
& 14=a+4 \\
& a^2=10 \\
& S_n=\frac{n}{2}[2 a+(n-1) d] \\
& S_{51}=\frac{51}{2}[2 \times 10+(51-1) 4] \\
& =\frac{51}{2}[20+(50)(4)] \\
& =\frac{51(220)}{2} \\
& =51 \times 110 \\
& =5610
\end{aligned}
$
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Question 52 Marks
Find the 12th from the end of the A.P. – 2, – 4, – 6, …; – 100.
Answer
$
\begin{aligned}
& \text { A.P. }=-2,-4,-6, \ldots ;-100 \\
& a=-2, d=-4-(-2) \\
& =-4+2 \\
& =-2 \\
& I=-100 \\
& \therefore T_n=a+(n-1) d \\
& \Rightarrow-100=-2+(n-1) \times(-2) \\
& \Rightarrow-100=-2-2 n+2 \\
& \Rightarrow+2 n=100 \\
& \Rightarrow n=\frac{100}{2}=50
\end{aligned}
$
Let $mth$ term is the 12th term from the end
Then $m=I-(n-1) d$
$
\begin{aligned}
& =-100-(12-1) \times(-2) \\
& =-100+22 \\
& =-78
\end{aligned}
$
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Question 62 Marks
Is 0 a term of the A.P. 31,28, 25,…? Justify your answer.
Answer
A.P. $31,28,25, \ldots$
Here, $a=31, d=28-31=-3$
Let 0 be the nth term, then
$
\begin{aligned}
& T_n=a+(n-1) d \\
& 0=31+(n-1)(-3) \\
& 0=31-3 n+3 \\
& \Rightarrow 3 n=34 \\
& n=\frac{34}{3} \\
& =11 \frac{1}{3}
\end{aligned}
$
Hence 0 is not any term of the A.P.
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Question 72 Marks
Find whether 55 is a term of the A.P. 7, 10, 13, … or not. If yes, find which term is it.
Answer
A.P. $7,10,13, \ldots$
Here, $a=7, d=10-7=3$
Let 55 is the nth term, then
$
\begin{aligned}
& T _{ n }=a+( n -1) d \\
& \Rightarrow 55=7+( n -1) \times 3 \\
& \Rightarrow 55=7+3 n -3 \\
& \Rightarrow 3 n =55-7+3=51 \\
& \therefore n =\frac{51}{3}=17
\end{aligned}
$
$\therefore 55$ is a term of the given A.P. and it is 17 th term.
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Question 82 Marks
Check whether – 150 is a term of the A.P. 11, 8, 5, 2, …
Answer
A.P. is $11,8,5,2, \ldots$
Here, $a=11, d=8-11=-3$
Let $-150=n$, then
$
\begin{aligned}
& T_n=a+(n-1) d \\
& \Rightarrow-150=11+(n-1)(-3) \\
& \Rightarrow-150=3-3 n+11 \\
& \Rightarrow 3 n=3+150+11 \\
& =153+11 \\
& =164 \\
& n=\frac{164}{3} \\
& =54 \frac{2}{3}
\end{aligned}
$
No, -150 is not any terms of the A.P.
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Question 92 Marks
Which term of the A.P. $18,15 \frac{1}{2}, 13, \ldots$ is -47 ?
Answer
$
18,15 \frac{1}{2}, 13, \ldots \text { is }-47
$
Let nth term is -47
$
\begin{aligned}
& a =18, d =15 \frac{1}{2}-18=-2 \frac{1}{2}=\frac{-5}{2} \\
& \therefore-47= a +( n -1) d \\
& \Rightarrow-47=18+(n-1)\left(\frac{-5}{2}\right) \\
& \Rightarrow-47-18=\frac{-5}{2} n+\frac{5}{2} \\
& \Rightarrow-65-\frac{5}{2}=\frac{-5}{2} n \\
& \Rightarrow \frac{-135}{2}=\frac{-5}{2} n \\
& \therefore n =\frac{-135}{2} \times \frac{2}{-5}=27
\end{aligned}
$
$\therefore-47$ is 27 th term.
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Question 102 Marks
Find the nth term and the 12th term of the list of numbers:$ 5, 2, – 1, – 4, …$
Answer
$5, 2, -1, -4, …$
Here, $a = 5 d = 2 – 5 = -3$
$(i) T_n = a + (n – 1)d$
$= 5 + (n – 1) (-3)$
$= 5 – 3n + 3$
$= 8 – 3n$
$(ii) T_{12} = a + 11d$
$= 5 + 11(-3)$
$= 5 – 33$
$= -28$
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Question 112 Marks
Justify whether it is true to say that the following are the nth terms of an $ A.P. n^2 + 1$
Answer
$n^2 + 1$
Giving some difference values to n such as $1, 2, 3, 4,...$
$(1)^2 + 1 = 1 + 1 = 2$
$(2)^2 + 1 = 4 + 1 = 5$
$(3)^2 + 1 = 9 + 1 = 10$
$(4)^2 + 1 = 16 + 1 = 17$
We see that $a = 2,$
$d = 5 – 2 = 3$
$= 10 – 5 = 5$
$= 17 – 10 = 7$
The common difference is not same.
$\therefore $ No. It is not an $A.P$.
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Question 122 Marks
Justify whether it is true to say that the following are the nth terms of an A.P. 2n – 3
Answer
2n – 3
Giving the some difference values to n such as 1, 2, 3, 4, … then
2 x 1 – 3 = 2 – 3 = -1
2 x 2 – 3 = 4 – 3 = 1
2 x 3 – 3 = 6 – 3 = 3
2 x 4 – 3 = 8 – 3 = 5
We see that –1, 1, 3, 5,... are in A.P. whose
first term = –1
and
d = 1 – (–1)
= 1 + 1
= 2.
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Question 132 Marks
Ramkali saved Rs 5 in the first week of a year and then increased her savings by Rs 1.75. If in the rcth week, her weekly savings become Rs 20.75, find n.
Answer
$\begin{aligned} & \text { Savings in the first week }=\text { Rs } 5 \\ & \text { Increase every week }=\text { Rs } 1.75 \\ & \text { No. of weeks }=n \\ & \text { and } \\ & \text { last saving }=
20.75 \\ & \text { Here, } a=5 \text { and } d=1.75 \\ & \therefore I=20.75 \\ & \therefore T_n(l)=a+(n-1) d \\ & 20.75=5+(n-1) 1.75 \\ & \Rightarrow 20.75-5.00=1.75(n-1) \\ & 15.75=1.75(n-1) \\ & \frac{15.75}{1.75}=n-1 \\ & \Rightarrow n-1=9 \\ & \therefore n=9+1=10 \\ & \therefore n=10 .\end{aligned}$
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Question 142 Marks
A man starts repaying a loan as first instalment of $Rs 500$. If he increases the instalment by Rs $25$ every month, what,amount will he pay in the 30th instalment?
Answer
First instalment of loan $= Rs 500$
Increases $Rs 25$ every month
Here, $a = 500, d = 25$
Total instalments $(n) = 30$
We have to find $T_{30}$
$T_{30} = a + (n – 1 )d = a + 29d$
$= 500 + 29 x 25$
$= 500 + 725$
$= Rs 1225.$
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Question 152 Marks
Find the indicated terms in each of following $A.P.s: – 4, – 7, – 10, – 13, …, a_{25}, a_n$​​​​​​​
Answer
$– 4, – 7, – 10, – 13, …, a_{25}, a_n$
Here, $a = – 4, d = – 7 – (– 4)$
$= – 7 + 4$
$= – 3$
$a_{25} = a + (n – 1)d$
$= –4 + (25 – 1) x –3$
$= –4 + 24 x (–3)$
$= –4 – 72$
$= –76$
and
$an = a_n + (n – 1)d$
$= –4 + (n – 1) (–3)$
$= –4 – 3n + 3$
$= –1 – 3n$
$= –3n – 1.$
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Question 162 Marks
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Answer
Number between 10 and 300 , which when divided by 4 leave a remainder 3 will be 11, 15, 19, 23,..299
$
\begin{aligned}
& \text { Here, } a=11, d=15-11=4, I=299 \\
& \therefore T_n=1=a+(n-1) d \\
& 299=11+(n-1) \times 4 \\
& \Rightarrow 299-11=(n-1) 4 \\
& 4(n-1)=288 \\
& \Rightarrow n-1=\frac{288}{4}=72 \\
& \therefore n=72+1 \\
& =73 .
\end{aligned}
$
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Question 172 Marks
Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer
Number divisible by both 2 and 5 are $110,120,130, \ldots \ldots, 990$
Here $a=110, d=120-110=0$
$
\begin{aligned}
& a_n=990 \\
& \Rightarrow a+(n-1) d=990 \\
& \Rightarrow 110+(n-1)(10)=990 \\
& \Rightarrow(n-1)(10) \\
& =990-110 \\
& =880
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow(n-1)=\frac{880}{10}=88 \\
& \therefore n=88+1 \\
& =89
\end{aligned}
$
Hence, number between 101 and 999 which are divisible by both 2 and 5 are 89 .
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Question 182 Marks
How many two digit numbers are divisible by 3?
Answer
Two digits numbers divisible by 3 are
$
12,15,18,21, \ldots, 99
$
Here, $a=13, d=15-12=3$ and $I =99$
Let number divisible by 3 and $n$
$
\begin{aligned}
& \therefore T_n=1=a+(n-1) d \\
& 99=12+(n-1) \times 3 \\
& \Rightarrow 99-12=3(n-1) \\
& \Rightarrow 3(n-1)=87 \\
& \Rightarrow n-1=\frac{87}{3}=29 \\
& \therefore n=29+1 \\
& =30 .
\end{aligned}
$
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Question 192 Marks
If 8th term of an $A.P$. is zero, prove that its 38th term is triple of its 18th term.
Answer
$T_8 = 0$
To prove that $T_{38} = 3 x T_{18}$​​​​​​​
Let a be the first term and d be the common difference
$\therefore T_8 = a + 7d = 0$
$\Rightarrow a = –7d$
Now
$T_{38} = a + 37d$
$= –7d + 37d$
$= 30d$
and
$T_{18} = a + 17d$
$= –7d + 17d$
$= 10d$
It is clear that $T_{38}.$ is triple of $T_{18}.$
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Question 202 Marks
Find the 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.
Answer
$
\begin{aligned}
& T_{11}-T_7=24 \\
& a=12
\end{aligned}
$
Let $a$ be the first term and $d$ be the common difference, then
$
\begin{aligned}
& (a+10 d)-(a+6 d)=24 \\
& a+10 d-a-6 d=24 \\
& \Rightarrow 4 d=24 \\
& \Rightarrow d=\frac{24}{4}=6 \\
& a=12 \\
& \therefore T_{20}=a+19 d \\
& =12+19 \times 6 \\
& =12+114 \\
& =126 .
\end{aligned}
$
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Question 212 Marks
Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12
Answer
$
\begin{aligned}
& T _3=16 \\
& T _7- T _5=12
\end{aligned}
$
Let $a$ be the first term and $d$ be the common difference
$
\begin{aligned}
& T_3=a+2 d=16 ......(1)\\
& T_7-T_5=(a+6 d)-(a+4 d)=12 \\
& \Rightarrow a+6 d-a-4 d=12 \\
& \Rightarrow 2 d=12 \\
& \Rightarrow d=\frac{12}{2}=6
\end{aligned}
$
Substitute the value of $d$ in eq. (i), we get
$
\begin{aligned}
& \therefore a+2 \times 6=16 \\
& \Rightarrow a+12=16 \\
& \Rightarrow a=16-12=4 \\
& \therefore \text { A.P. is } 4,10,16,22,28, \ldots
\end{aligned}
$
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Question 222 Marks
Determine the A.P. whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Answer
$
\begin{aligned}
& \text { In an A.P., } \\
& T_5=19 \\
& T_{13}-T_8=20
\end{aligned}
$
Let $a$ be the first term and $d$ be the common difference
$
\begin{aligned}
& \therefore T_5=a+4 d=19......(1) \\
& T_{13}-T_8=(a+12 d)-(a+7 d) \\
& \Rightarrow 20=a+12 d-a-7 d \\
& \Rightarrow 20=5 d \\
& \Rightarrow d=\frac{20}{5}=4
\end{aligned}
$
Substitute the value of $d$ in eq. (i), we get
$
\begin{aligned}
& \therefore a+4 \times 4=19 \\
& \Rightarrow a+16=19 \\
& \Rightarrow a=19-16=3 \\
& \therefore \text { A.P. is } 3,7,11,15, \ldots
\end{aligned}
$
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Question 232 Marks
Which term of the A.P. 53, 48, 43,… is the first negative term?
Answer
Let $n$th term is the first negative term of the A.P. $53,48,43, \ldots$
Here, $a=53, d=48-53=-5$
$
\begin{aligned}
& \therefore T_n=a+(n-1) d \\
& =53+(n-1) \times(-5) \\
& =53-5 n+5 \\
& =58-5 n
\end{aligned}
$
$
\begin{aligned}
& 5 n=58 \\
& n=\frac{58}{5} \\
& =11 \frac{3}{5}
\end{aligned}
$
$\therefore 12$ th term will be negative.
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Question 242 Marks
Find the $A.P$. whose nth term is $7 – 3K$. Also find the 20th term.
Answer
$T_n = 7 – 3n$
Giving values $1, 2, 3, 4, …$ to n, we get
$T_1 = 7 – 3 x 1 = 7 – 3 = 4$
$T_2 = 7 – 3 x 2 = 7 – 6 = 1$
$T_3 = 7 – 3 x 3 = 7 – 9 = -2$
$T_4 = 7 – 3 x 4 = 7 – 12 = -5$
$T_{20} = 7 – 3 x 20 = 7 – 60 = -53$
$A.P$. is $4, 1, -2, -5, …$
$20$th term $= -53$
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Question 252 Marks
Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms : a, 2a + 1, 3a + 2, 4a + 3,...
Answer
a, 2a + 1, 3a + 2, 4a + 3,...
Here first term (a) = a
and common difference (d)
= 2a + 1 – a = a + 1
3a + 2 – 2a – 1 = a + 1
4a + 3 – 3a – 2 = a + 1
∵ Common difference is same.
∴ It is an A.P.
and three next terms are
5a + 4, 6a + 5, 7a + 6,...
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Question 262 Marks
Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms : a, 2a, 3a, 4a,...
Answer
a, 2a, 3a, 4a,...
Here first term (a) = a
Common difference (d) = 2a – a = a
3a – 2a = a
4a – 3a = a
∵ The common difference is same
∴ It is an A.P.
and next three terms are
5a, 6a, 7a.
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Question 272 Marks
Which of the following lists of numbers form an A.P.? If they form an A.P. , find the common difference $d$ and write the next three terms : $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots$
Answer
$
\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots
$
Here, $a =\sqrt{3}$
$
\begin{aligned}
& d=\sqrt{6}-\sqrt{3} \\
& =\sqrt{3} \times \sqrt{2}-\sqrt{3} \\
& =\sqrt{3}(\sqrt{2}-1) \\
& =\sqrt{9}-\sqrt{6} \\
& =3-\sqrt{2} \sqrt{3} \\
& =\sqrt{3}(\sqrt{3}-\sqrt{2})
\end{aligned}
$
$\because$ Common difference is not same.
$\because$ It is not an A.P.
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Question 282 Marks
Which of the following lists of numbers form an A.P.? If they form an A.P. ${ }^4$, find the common difference $d$ and write the next three terms : $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}$,..
Answer
$
3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots
$
Here, first term (a) $=3$
and $d =3+\sqrt{2}-2=\sqrt{2}$
$
\begin{aligned}
& 3+2 \sqrt{2}-3-\sqrt{2}=\sqrt{2} \\
& 3+3 \sqrt{2}-3+2 \sqrt{2}=\sqrt{2}
\end{aligned}
$
$\because$ Common difference is same
$\because$ It is an A.P.
and next three terms are
$
3+4 \sqrt{2}, 3+5 \sqrt{2}, 3+6 \sqrt{2}, \ldots
$
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Question 292 Marks
Which of the following lists of numbers form an A.P.? If they form an A.P. find the common difference $d$ and write the next three terms : $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32} \ldots$
Answer
$
\begin{aligned}
& \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots \\
& \Rightarrow \sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \ldots
\end{aligned}
$
Here, first term (a) $=\sqrt{2}$
and common difference (d)
$
\begin{aligned}
& =2 \sqrt{2}-\sqrt{2}=\sqrt{2} \\
& =3 \sqrt{2}-2 \sqrt{2}=\sqrt{2} \\
& =4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}
\end{aligned}
$
$\because$ The common difference is same
$\because$ It is an A.P.
and next three terms are
$
\sqrt{50}, \sqrt{72}, \sqrt{98}, \ldots
$
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