Fill in the blanks.

Question 11 Mark

The value of $5^{\frac{1}{2}} \cdot 5^{\frac{1}{4}} \cdot 5^{\frac{1}{8}}$. . . . up to infinity is _________________

Answer

5, because
We have, $5^{\frac{1}{2}} \cdot 5^{\frac{1}{4}} \cdot 5^{\frac{1}{8}}$ . . . . $\infty$
$
\begin{aligned}
& =5^{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots \infty\right]} \\
& =5^{\left[\left(\frac{1}{2}\right) / 1-\left(\frac{1}{2}\right)\right]} \\
{[\because} & \left.\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots . . . \infty=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\right] \\
& =5^1=5 .
\end{aligned}
$

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Question 21 Mark

If in a G.P., $a_3+a_5=90$ and if $r=2$, then the first term of G.P. is _________________

Answer

$\frac{9}{2}$, because
Let $a$ be the first term of G.P.
Given,
$r=2$ and $a_3+a_5=90$
$\begin{array}{rlrl}\therefore\quad\quad\quad\quad a r^2+a r^4 =90 \\ \Rightarrow\quad\quad a(2)^2+a(2)^4 =90 \\ \Rightarrow \quad\quad\quad\quad 4 a+16 a =90 \\ \Rightarrow\quad\quad\quad\quad\quad\quad 20 a =90 \\ \Rightarrow \quad\quad\quad\quad\quad\quad\quad a=\frac{9}{2}\end{array}$

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Question 31 Mark

The third term of a G.P. is 4, the product of first five terms is _________________

Answer

$(4)^5$, because
Given, $T_3=4 \Rightarrow a r^2=4 \quad \ldots(i) $
Product of first five terms $=a \cdot a r \cdot a r^2 \cdot a r^3 \cdot a r^4$
$
\begin{array}{l}
=a^5 r^{10} \\
=\left(a r^2\right)^5 \\
=(4)^5 [using (i)]
\end{array}$

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Question 41 Mark

If $a, b$ and $c$ are in G.P., then the value of $\frac{a-b}{b-c}$ is equal to _________________

Answer

$\frac{a}{b}$ or $\frac{b}{c}$ because
Since, $a, b$ and $c$ are in G.P.
$
\begin{array}{l}
\therefore \quad \frac{b}{a}=\frac{c}{b}=r \quad \text { (common ratio) } \\
\Rightarrow \quad b=a r \text { and } c=b r \\
\Rightarrow \quad c=a r . r=a r^2 \\
\text { So, } \\
\frac{a-b}{b-c}=\frac{a-a r}{a r-a r^2}=\frac{a(1-r)}{a r(1-r)} \\
=\frac{1}{r}=\frac{a}{b}=\frac{b}{c}
\end{array}
$

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Question 51 Mark

If a, b, c are in A.P., then 2b = _________________

Answer

$a+c$, because
Given, $a, b, c$ are in A.P.
$
\begin{array}{ll}
\therefore & b-a=c-b \\
& {[\text { common difference are equal }]} \\
\Rightarrow & 2 b=a+c
\end{array}
$

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Question 61 Mark

In an AP 8, 11, 14, . . . the value of $S_n-S_{n-1}$ $=$ _________________ .

Answer

$3 n+5$, because
We know that, $S _n- S _{n-1}=a_n$, which is the $n^{\text {th }}$ term of AP.
Here, $a=8, d=3$
$
\begin{aligned}
\therefore \quad a_n=T_n & =a+(n-1) d \\
& =8+(n-1) 3 \\
& =3 n+5
\end{aligned}
$

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Question 71 Mark

The number of terms in the A.P. $7,10,13, . . ., 31$ is. _________________

Answer

9, because
Given, AP is 7, 10, 13 , . . . .31
Here, $a=7, l=31$ and $d=3$
Using formula, $l=a+(n-1) d$
$\Rightarrow \quad 31=7+(n-1) 3$
$\Rightarrow \quad 31-7=3 n-3$
$\Rightarrow \quad 24+3=3 n$
$\Rightarrow \quad n=9$

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Question 81 Mark

If the $n^{\text {th }}$ term of an A.P. is $6 n-7$, then its $50^{\text {th }}$ term is _________________

Answer

293, because
$
\begin{array}{ll}
\text { Given, } & T_n=6 n-7 \\
\therefore & T_{50}=6(50)-7=300-7=293
\end{array}
$

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