2 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip

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2 Marks Questions

Question 512 Marks

Which of the following sequences are A.P. ? If they are A.P. find the common difference.
$3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$

Answer

3, 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 .

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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, $d =0$.

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Question 532 Marks

Which of the following sequences are A.P. ? If they are A.P. find the common difference.
0, – 4, – 8, – 12, . . .

Answer

0, – 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.

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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, . . .

Answer

0.3, 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

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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.

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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, $d =\frac{1}{2}$.

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Question 572 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, 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.

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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.

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Question 592 Marks

Find the sum of first n natural numbers.

Answer

First $n$ 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 $\frac{n(n+1)}{2}$.

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Question 602 Marks

Which term of the following A.P. is 560 ?
$
2,11,20,29, \ldots
$

Answer

Given A.P. 2, 11, 20, 29,. .
Here $a=2, d=11-2=9$
$n^{\text {th }}$ term of this A.P. is 560 .
\[
\begin{array}{l}
t_{\mathrm{n}}=a+(n-1) d \\
\therefore 560=2+(n-1) \times 9 \\
\quad=2+9 n-9 \\
\therefore 9 n=567 \\
\therefore \quad n=\frac{567}{9}=63 \\
\therefore 63^{\text {rd }} \text { term of given A.P. is } 560 .
\end{array}
\]

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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
$

Answer

Given 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}
$

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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}
$

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Question 632 Marks

The first term a and common difference d are given. Find first four terms of A.P :

$a=-3, d=4$

Answer

Given $a=-3, d=4$
$
\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}
$

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Question 642 Marks

The first term a and common difference d are given. Find first four terms of A.P :

$a=200, d=7$

Answer

Given $a=200, d=7$
$
\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}
$

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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 $=-1,-\frac{3}{2},-2,-\frac{5}{2}, \ldots$.

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Question 662 Marks

Which of the following sequences are A.P ? If it is an A.P, find next two terms : $\frac{3}{2}, \frac{1}{2},-\frac{1}{2}, \ldots$

Answer

In 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 $-\frac{7}{2}$

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Question 672 Marks

Which of the following sequences are A.P ? If it is an A.P, find next two terms : $5,12,19,26, \ldots$

Answer

In 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

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Question 682 Marks

Which of the following sequences are A.P ? If it is an A.P, find next two terms : $2,-2,-6,-10, \ldots$

Answer

In 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 $(-14)+(-4)=-18$

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Question 692 Marks

Which of the following sequences are A.P ? If it is an A.P, find next two terms : $1,1,2,2,3,3, \ldots$

Answer

In 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.

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Question 702 Marks

How many terms are there in the A.P. 187, 194, 201, ..., 439 ?

Answer

37 terms

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Question 712 Marks

Find $t_{11}$ from the following A.P. 4, 9, 14,…

Answer

$t_{11}=54$

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Question 722 Marks

Find $n$, if the $n^{\text {th }}$ term of the following sequence is 68 .
$5,8,11,14, \ldots$

Answer

$n=22$

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2 Marks Questions - Page 2 - Maths STD 10 Questions - Vidyadip