2 Marks Questions

Question 12 Marks

Prove that if $A$ and $G$ be the A.M. and G.M. between two positive number then the numbers are $A \pm \sqrt{A^2-G^2}$

Answer

The equation having its roots as the given number in
$\begin{aligned} x^2-2 A x+G^2 & =0 \\ \Rightarrow x =\frac{2 A \pm \sqrt{4 A^2-4 G^2}}{2}\end{aligned}$
$\Rightarrow x=A \pm \sqrt{A^2-G^2}$
Hence Proved

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Question 22 Marks

Find the number of squares that can be formed on $8 \times 8$ chess board ?

Answer

The number of square in a $n \times n$ chess board will be $\sum n^2 ; n$ varying from 1 to $n$.
Now,
$
\begin{aligned}
\sum n^2 & =1^2+2^3+3^2+\ldots+n^2 \\
& =\frac{n(n+1)(2 n+1)}{6} \quad \ldots(i)
\end{aligned}
$
Put $n=8$ in eq (i), we get
$
\begin{aligned}
\Sigma 8^2 & =\frac{8 \times(8+1)(2 \times 8+1)}{6} \\
& =\frac{8 \times 9 \times 17}{6}=204
\end{aligned}
$

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Question 32 Marks

Write the $n^{\text {th }}$ term of the series$
\frac{3}{7.11^2}+\frac{5}{8.12^2}+\frac{7}{9.13^2}+\ldots
$

Answer

$\begin{array}{l} n^{\text {th }} \text { term of } 3,5,7, \ldots \text { is } 2 n+1 \\ n^{\text {th }} \text { term of } 7,8,9, \ldots \text { is } 6+n \\ ^{\text {th }} \text { term of } 11^2, 12^2, 13^2, \ldots \text { is }(10+n)^2 \\ \therefore n^{\text {th }} \text { term of the given series }=\frac{2 n+1}{(6+n)(10+n)^2}\end{array}$

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Question 42 Marks

If $a, b, c$ are in G.P, then show that $a^2+b^2, a b+b c$, $b^2+c^2$ are also in G.P.

Answer

Given that, $a, b, c$ are in G.P
$
\begin{array}{lrr}
\Rightarrow b^2=a c \\
\Rightarrow b^2-a c=0 \\
\Rightarrow \left(b^2-a c\right)^2=0 \\
\Rightarrow b^4+a^2 c^2-2 b^2 a c=0 \\
\Rightarrow a^2 b^2+b^2 c^2+2 b^2 a c=a^2 b^2+b^2 c^2+a^2 c^2+b^4 & \\
\text { (Adding } a^2 b^2+b^2 c^2 \text { both sides) } \\
\Rightarrow (a b+b c)^2=\left(a^2+b^2\right)\left(b^2+c^2\right) \\
\Rightarrow a^2+b^2, a b+b c, b^2+c^2 \text { are in G.P. }
\end{array}
$
Hence Proved

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Question 52 Marks

Find the sum of the G.P. 0.15,0.015,0.0015,. . . . up to 20 terms.

Answer

Here, $a=0.15, r=\frac{0.015}{0.15}=\frac{15}{1000} \times \frac{100}{15}$
$r=\frac{1}{10}<1 \text { and } n=20$
Now, $\quad S_n=\frac{a\left(1-r^n\right)}{1-r}$
$
\Rightarrow \quad S_{20}=\frac{0.15\left[1-\left(\frac{1}{10}\right)^{20}\right]}{1-\frac{1}{10}}=\frac{0.15\left[1-\frac{1}{10^{20}}\right]}{\frac{10-1}{10}}
$
$\begin{array}{l}=\frac{0.15 \times 10}{9}\left[1-\frac{1}{10^{20}}\right] \\ =\frac{15 \times 10}{900}\left[1-\left(\frac{1}{10}\right)^{20}\right]=\frac{1}{6}\left[1-(0.1)^{20}\right]\end{array}$

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Question 62 Marks

A man starts repaying a loan as first installments of ₹ 100 . If he increases the installment by ₹ 5 every month, what amount he will pay in the $30^{\text {th }}$ installment?

Answer

Given, a = 100 and d = 5
$\because \quad T_n=a+(n-1) d$
$\therefore \quad T_{30}=100+(30-1)(5)$
$\begin{array}{l}=100+29 \times 5 \\ =100+145 \\ =245\end{array}$
Therefore, he will pay ₹ 245 in the 30th installment.

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Question 72 Marks

If $a, b, c$ are in G.P., then show that $a^2, b^2, c^2$ are also in G.P.

Answer

Given, $a, b, c$ are in G.P.
$\therefore \quad b^2=a c$
On squaring both sides, we get
$\Rightarrow \quad\left(b^2\right)^2=(a c)^2$
$\Rightarrow \quad\left(b^2\right)^2=a^2 c^2$
$\Rightarrow \quad a^2, b^2, c^2$ are in G.P.
Hence Proved

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Question 82 Marks

If x, y, z are positive integers then find the value of the expression (x+y) (y+z) (z+x).

Answer

We know that,
$\begin{array}{lrl} & \text { A.M. } & >\text { G.M. } \\ \therefore & \frac{x+y}{2} & >\sqrt{x y}, \frac{y+z}{2}>\sqrt{y z} \\ \text { and } & \frac{z+x}{2} & >\sqrt{z x}\end{array}$
Multiplying the three inequalities, we get
$
\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{z+x}{2}>\sqrt{(x y)(y z)(z x)}
$
or, $(x+y)(y+z) \cdot(z+x)>8 x y z$
Hence, the value of $(x+y)(y+z)(z+x)>8 x y z$.

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Question 92 Marks

Write the value of $0 . \overline{3}$.

Answer

Let
$
\begin{array}{l}
x=0.3 \\
x=0.33333 \ldots \ldots \ldots \ldots \ldots . . \\
x=0.3+0.03+0.0003+.............
\end{array}
$
$
\therefore a=0.3, r=\frac{0.03}{0.3}=0.1, n=\infty
$
$\therefore$ Sum of infinite terms in G.P, $S=\frac{a}{1-r}$
$\begin{array}{ll}\therefore & x=\frac{0.3}{1-0.1}=\frac{0.3}{0.9}=\frac{1}{3} \\ \therefore & 0 . \overline{3}=\frac{1}{3}\end{array}$

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Question 102 Marks

The first terms of G.P. is 2 and sum to infinity is $6$, find common ratio.

Answer

Given, $a=2$ and $S_{\infty}=6$
$
\begin{array}{l}
\because \quad S_{\infty}=\frac{a}{1-r} \\
\therefore \quad 6=\frac{2}{1-r} \\
\Rightarrow \quad 6-6 r=2 \\
\Rightarrow \quad r=\frac{4}{6}=\frac{2}{3}
\end{array}
$

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Question 112 Marks

Which term of the G.P. $2,1, \frac{1}{2}, \frac{1}{4} \ldots$ is $\frac{1}{1024}$ ?

Answer

Given G.P. is $2,1, \frac{1}{2}, \frac{1}{4} \ldots$
First term $a=2$
and common ratio $=\frac{1}{2}$
Let $n^{\text {th }}$ term of G.P. be $\frac{1}{1024}$
$
\begin{array}{lrl}
\therefore & T_n & =a r^{n-1} \\
\Rightarrow & \frac{1}{1024} & =2\left(\frac{1}{2}\right)^{n-1}
\end{array}
$
$\begin{aligned} \Rightarrow & \frac{1}{2048} =\left(\frac{1}{2}\right)^{n-1} \\ \Rightarrow & \left(\frac{1}{2}\right)^{11} =\left(\frac{1}{2}\right)^{n-1} \\ \therefore & n-1 =11 \\ \text { or } & n =12\end{aligned}$

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Question 122 Marks

Which term of the sequence $25,24 \frac{1}{4}, 23 \frac{1}{2}$, $22 \frac{3}{4}, \ldots \ldots$. is the first negative term?

Answer

The given sequence is an AP with common difference,
$d=-\frac{3}{4}$ and first term, $a=25$.
Let the $n$th term of the given A.P. be the first negative term, then
$
\begin{array}{rlrl}
& a_n <0 \Rightarrow 25+(n-1)\left(-\frac{3}{4}\right) <0 \\
\Rightarrow & \frac{103}{4}-\frac{3 n}{4} <0 \\
\Rightarrow & 103-3 n <0 \\
\Rightarrow & 103 <3 n \\
\Rightarrow & 3 n >103 \\
\Rightarrow & n >34 \frac{1}{3}
\end{array}
$
Since, 35 in the least natural number satisfying
$n>34 \frac{1}{3} \Rightarrow n=35$.
Hence, $35^{\text {th }}$ term of the given sequence is the first negative term.

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Question 132 Marks

Find the 13 th and 14 th terms of the sequence defined by
$
a_n=\left\{\begin{array}{cc}
n^2, & \text { when } n \text { is even } \\
n^2+1, & \text { when } n \text { is odd }
\end{array}\right.
$

Answer

As 13 is odd, $a_{13}=n^2+1=(13)^2+1=169+1=170$ and as 14 is even, $a_{14}=n^2=(14)^2=196$

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Question 142 Marks

If sum of first $n$ term of an A.P. is $2 n^2+7 n$, write its $n ^{\text {th }}$ term.

Answer

Given,$S_n=2 n^2+7 n$
We know that, $\quad a_n=S_n-S_{n-1}$
$
\begin{aligned}
\therefore \quad a_n & =2 n^2+7 n-\left\{2(n-1)^2+7(n-1)\right\} \\
& =2 n^2+7 n-\left\{2\left(n^2-2 n+1\right)+7 n-7\right\} \\
& =2 n^2+7 n-\left(2 n^2-4 n+2+7 n-7\right) \\
& =2 n^2+7 n-\left(2 n^2+3 n-5\right) \\
& =2 n^2+7 n-2 n^2-3 n+5 \\
& =4 n+5
\end{aligned}
$

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Question 152 Marks

If in an A.P. 7th term is 9 and 9th term is 7, then find 16th term.

Answer

Let first term and common difference of A.P. is $a$ and $d$, respectively.
$
\begin{array}{rlrl}
\text { Given, } & T_7 =9=a+(7-1) d \\
\Rightarrow & 9 =a+6 d \quad \ldots(i)\\
\text { and } & T_9 =7=a+(9-1) d \\
\Rightarrow & 7 =a+8 d\quad \ldots(ii)
\end{array}
$
On solving equations (i) and (ii), we getNow,
$
\begin{aligned}
a & =15 \text { and } d=-1 \\
Now,\quad T_{16} & =a+(16-1) d
\end{aligned}
$
$\quad\quad\quad15+15(-1)=0
$

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Question 162 Marks

Which term of the sequence $3,10,17 \ldots .$. is 136 ?

Answer

The given sequence is an AP with common difference,
$d=-\frac{3}{4}$ and first term, $a=25$.
Let the $n$th term of the given A.P. be the first negative term, then
$
\begin{array}{rlrl}
& a_n <0 \Rightarrow 25+(n-1)\left(-\frac{3}{4}\right) <0 \\
\Rightarrow & \frac{103}{4}-\frac{3 n}{4} <0 \\
\Rightarrow & 103-3 n <0 \\
\Rightarrow & 103 <3 n \\
\Rightarrow & 3 n >103 \\
\Rightarrow & n >34 \frac{1}{3}
\end{array}
$
$\Rightarrow \quad n=20$
Since, 35 in the least natural number satisfying
$n>34 \frac{1}{3} \Rightarrow n=35$.
Hence, $35^{\text {th }}$ term of the given sequence is the first negative term.

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Applied Maths 2 Marks Questions

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