3 Marks Question

Question 13 Marks

Find the sum of the series
$1+(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots$

Answer

Let$
S_n=1+(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots
$to $n$ terms
Hence,
$
\begin{aligned}
(1-x) S_n= & (1-x)+(1-x)(1+x) \\
& +(1-x)\left(1+x+x^2\right) \\
& +(1-x)\left(1+x+x^2+x^3\right) \\
& +\ldots \text { to } n \text { terms }
\end{aligned}
$
or,
$\begin{array}{l}\begin{aligned}(1-x) S_n & =(1-x)+\left(1-x^2\right)+\left(1-x^3\right)\ldots to\ n\ terms \\ & =n-\left(x+x^2+x^3+x^4+\ldots\right) \text { to } n \text { terms } \\ & =n-\frac{x\left(1-x^n\right)}{(1-x)} \\ \text { Hence, } \quad \quad \quad S_n & =\frac{n}{1-x}-\frac{x\left(1-x^n\right)}{(1-x)^2}\end{aligned}\end{array}$

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Question 23 Marks

If the number $a^2, b^2, c^2$ are given to be in A.P. Show that $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in A.P.

Answer

$\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ will be in A.P.
if $\quad \frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$
i.e., if $\quad \frac{b-a}{(c+a)(b+c)}=\frac{c-b}{(a+b)(c+a)}$
i.e., if $\quad \frac{b-a}{b+c}=\frac{c-b}{a+b}$
i.e., if $b^2-a^2=c^2-b^2$
i.e., if $2 b^2=a^2+c^2$
i.e., So $a^2, b^2, c^2$ are in A.P.
Thus, $a^2, b^2, c^2$ are in A.P. $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in A.P.
Hence Proved.

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Question 33 Marks

In a G.P. , the $3^{\text {rd }}$ term is 24 and $6^{\text {th }}$ term is 192. Find the 10th term.

Answer

Let the first term of G.P. be $a$ and the common ratio be $r$.
$\begin{array}{lrl}\text { Given } & T_3 & =24 \\ \Rightarrow & a r^2 & =24 \quad \ldots(i) \end{array}$
$\begin{array}{lrl}\text { and } & T_6 & =192 \\ \Rightarrow & a r^5 & =192 \quad \ldots(ii) \end{array}$
Dividing eq. (i) by (ii), we get
$
\begin{array}{rlrl}
\frac{a r^5}{a r^2} =\frac{192}{24} \\
\Rightarrow \quad r^3 =8 \text { or } r=2
\end{array}
$
Putting the value of $r$ in eq. (i), we get
$
\begin{array}{rlrl}
a(2)^2 =24 \\
\Rightarrow\quad a =6 \\
\therefore T_{10} =a r^9
\end{array}
$
$\begin{array}{l}=6(2)^9 \\ =3072\end{array}$

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Question 43 Marks

If in an A.P. $\frac{a_7}{a_{10}}=\frac{5}{7}$, find $\frac{a_4}{a_7}$

Answer

Given, $\frac{a_7}{a_{10}}=\frac{5}{7}$
Let the first term and common difference of A.P. be ' $A$ ' and ' $D$ '. respectively.
$\therefore \quad \frac{A+6 D}{A+9 D}=\frac{5}{7}$
$
\begin{aligned}
\Rightarrow & 7 A+42 D =5 A+45 D \\
\Rightarrow & 2 A =3 D \\
\Rightarrow & A =\frac{3}{2} D \quad \ldots(i)
\end{aligned}
$
Now,
$
\begin{aligned}
\frac{a_4}{a_7} & =\frac{A+3 D}{A+6 D} \\
& =\frac{\frac{3}{2} D+3 D}{\frac{3}{2} D+6 D} \quad(using (i))\\
& =\frac{9 D}{15 D} \\
& =\frac{3}{5}
\end{aligned}
$

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Question 53 Marks

The first term of an A.P. is $a$ and the sum of the first $p$ terms is zero, show that the sum of its next $q$ terms is $\frac{-a(p+q) q}{p-1}$.

Answer

Let the common difference of the A.P. be $d$.
Given that $S_p=0$
$
\begin{aligned}
\Rightarrow & & \frac{p}{2}[2 a+(p-1) d] & =0 \\
\Rightarrow & & 2 a+(p-1) d & =0 \\
\Rightarrow & & d & =\frac{-2 a}{p-1}
\end{aligned}
$
Now, sum of next $q$ terms $=S_{p+q}-S_p$
$
\begin{array}{l}
=S_{p+q}-0 \\
=\frac{p+q}{2}[2 a+(p+q-1) d] \\
=\frac{p+q}{2}[2 a+(p-1) d+q d]
\end{array}
$
$
\begin{array}{l}
=\frac{p+q}{2}\left[0+\frac{q(-2 a)}{p-1}\right] \\
=\frac{-a(p+q) q}{p-1}
\end{array}
$
Hence proved

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Question 63 Marks

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Answer

Here, $a=5$ and $d=2$
Let he finished the job in $n$ days.
Then, $S_n=192 \quad S_n=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \quad 192=\frac{n}{2}[2 \times 5+(n-1) 2]$
$
\begin{array}{l}
\Rightarrow \quad 192=\frac{n}{2}[10+2 n-2] \\
\Rightarrow \quad 192=\frac{n}{2}[8+2 n] \\
\Rightarrow \quad 192=4 n+n^2 \\
\Rightarrow \quad n^2+4 n-192=0 \\
\Rightarrow \quad(n-12)(n+16)=0 \\
\Rightarrow \quad n=12,-16 \\
\Rightarrow \quad n=12 \quad[\because n \neq-16]
\end{array}
$

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Question 73 Marks

Show that the sum of $(m+n)^{\text {th }}$ and $(m-n)^{\text {th }}$ term of an A.P. is equal to twice of $m^{\text {th }}$ terms.

Answer

Let $a$ and $d$ be the first term and common difference of AP.
Then,
$
\begin{aligned}
T_{m+n}+T_{m-n}= & {[a+(m+n-1) d] } \\
& \quad+[a+(m-n-1) d] \\
& \quad\left[\because T_n=a+(n-1) d\right] 1 \\
= & 2 a+(m+n-1+m-n-1) d \\
= & 2 a+2(m-1) d \\
= & 2[a+(m-1) d] \\
= & 2 T_m
\end{aligned}
$
$\therefore \quad T_{m+n}+T_{m-n}=2 T_m$
Hence Proved.

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Applied Maths 3 Marks Question

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