1 Marks Question

Question 11 Mark

Find the sum of odd numbers from 1 to 2001.

Answer

To find the number of terms in $1+3+5+\ldots \ldots$. 2001,
\[
\begin{aligned}
a_n & =a+(n-1) d \\
2001 & =1+(n-1) \times 2 \\
\Rightarrow \quad \frac{2000}{2} & =(n-1) \Rightarrow n=1001
\end{aligned}
\]
We know that sum of $n$ odd numbers is $n^2$.
$\therefore$ Sum of 1001 odd numbers $=(1001)^2$

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Question 21 Mark

Insert five GM.s between 3 and 192.

Answer

$\begin{array}{l}3, G_1, G_2, G_3, G_4, G_5, 192 \\\text { Number of terms }=7 \\\begin{array}{ll}\therefore \quad a_7 & =3 \times r^6=192=3 r^6 \\\Rightarrow \quad r & = \pm 2 .\end{array}\end{array}$
So, required 5 geometric means are :
$\pm 6, \pm 12, \pm 24, \pm 48, \pm 96$

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Question 31 Mark

Find 4th term from the end of the G.P. $3,6,12,24$…......3072.

Answer

4th term from end
\[
\begin{aligned}
a_4 & =l \cdot\left(\frac{1}{r}\right)^{4-1}=3072 \times\left(\frac{1}{2}\right)^{4-1} \\
& =3072 \times \frac{1}{8}=384 \text {}
\end{aligned}
\]

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Question 41 Mark

What is the formula to find the $n$th term of G.P. from the end?

Answer

$a_n=l \cdot\left(\frac{1}{r}\right)^{n-1}$

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Question 51 Mark

What is the 8th term of the G.P. $2,1, \frac{1}{2}, \frac{1}{4}, \ldots \ldots . ?$

Answer

$\begin{array}{l}2,1, \frac{1}{2}, \frac{1}{4} \ldots \ldots \ldots \\ a_n=a r^{n-1} \\ \therefore \quad \begin{aligned} a_8 & =(2)\left(\frac{1}{2}\right)^{8-1} \\ & =2 \times \frac{1}{2^7}=\frac{1}{2^6}=\frac{1}{64} \text { }\end{aligned}\end{array}$

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Question 61 Mark

If first term of G.P. is $a$, last term is $l$ and common ratio is $r$, then find its sum.

Answer

$S =\frac{a-l r}{1-r}, r \neq 1$

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Question 71 Mark

The sum of $n$ terms of the G.P. $3,6,12, \ldots \ldots . .$. is 381 . Find the value of $n$.

Answer

$3,6,12, \ldots \ldots \ldots \ldots$
First term $a=3$
Common ratio $r=2$
\[
\begin{aligned}
& S_n=\frac{a\left(r^n-1\right)}{r-1}, r>1 \therefore 381=\frac{3\left(2^n-1\right)}{2-1} \\
\Rightarrow & \frac{381}{3}=2^n-1 \Rightarrow 2^n=128=2^7 \\
\therefore & n=7 \text { }
\end{aligned}
\]

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Question 81 Mark

Write the first three terms of the sequence whose $n^{\text {th }}$ term is given by $a_n=\frac{n^2}{3^n}$

Answer

$\begin{array}{ll}\text { Here, } & a_n=\frac{n^2}{3^n} \\\therefore & a_1=\frac{1^2}{3^1}=\frac{1}{3} \\\text { } & a_2=\frac{2^2}{3^2}=\frac{4}{9} \\\text { and } & a_3=\frac{3^2}{3^3}=\frac{1}{3}\end{array}$
So, first three terms $=\frac{1}{3}, \frac{4}{9}, \frac{1}{3}$

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