5 Marks Questions

Question 15 Marks

Match the questions given under Column $I$ with their appropriate answers given under the Column $II.$

	Column I		Column II
$(a)$	$1^2+2^2+3^2+....+\text{n}^2$	$(i)$	$\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$(b)$	$1^3+2^3+3^3+....\text{n}^3$	$(ii)$	$\text{n}(\text{n}+1)$
$(c)$	$2+4+6+....+2\text{n}$	$(iii)$	$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$(d)$	$1+2+3+....\text{n}$	$(iv)$	$\frac{\text{n}(\text{n}+1)}{2}$

Answer

	Column I		Column II
$(a)$	$1^2+2^2+3^2+....+\text{n}^2$	$(iii)$	$\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$(b)$	$1^3+2^3+3^3+....\text{n}^3$	$(i)$	$\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$(c)$	$2+4+6+....+2\text{n}$	$(ii)$	$\text{n}(\text{n}+1)$
$(d)$	$1+2+3+....\text{n}$	$(iv)$	$\frac{\text{n}(\text{n}+1)}{2}$

Solution:

Let $\text{S}=1^2+2^2+3^2+....+\text{n}^2$

We have, $\text{n}^3-(\text{n}-1)^3=3\text{n}^2-3\text{n}+1;$
And by changing $n$ into $n - 1,$
$(\text{n}-1)^3-(\text{n}-2)^3=3(\text{n}-1)^2-3(\text{n}-1)+1;$
$(\text{n}-2)^2-(\text{n}-3)^3=3(\text{n}-2)^2-3(\text{n}-2)+1;$
$..........\\..........\\..........$
$3^3-2^3=3.3^2-3.3+1;$
$2^3-1^2=3.2^2-3.2+1;$
$1^2-0^2=3.1^2-3.1+1$
Hence, by addition, $\text{n}^3=3(1^2+2^2+3^2+....+\text{n}^2)-3(1+2+3+....+\text{n})+\text{n}$
$=3\text{s}-\frac{3\text{n}(\text{n}+1)}{2}+\text{n}$
$\Rightarrow3\text{S}=\text{n}^3-\text{n}+\frac{3\text{n}(\text{n}+1)}{2}=\text{n}(\text{n}+1)\Big(\text{n}-1+\frac{3}{2}\Big)$
$\Rightarrow\text{S}=\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$

Let $\text{S}=1^3+2^3+3^3+.....+\text{n}^3$

We have, $\text{n}^4-(\text{n}-1)^4=4\text{n}^3-6\text{n}^2+4\text{n}-1;$
$(\text{n}-1)^4-(\text{n}-2)^4=4(\text{n}-1)^3-6(\text{n}-1)^2+4(\text{n}-1)-1;$
$(\text{n}-2)^4-(\text{n}-3)^4=4(\text{n}-2)^3-6(\text{n}-2)^2+4(\text{n}-2)-1;$
$..........\\..........\\..........$
$3^4-2^4=4.3^3-6.3^2+4.3-1;$
$2^4-1^4=4.2^3-6.2^2+4.2-1;$ $1^4-0^4=4.1^3-6.1^2+4.1-1.$
Hence, by addition, $\text{n}^4=4\text{S}-6\big(1^2+2^2+....+\text{n}^2\big)+4(1+2+....+\text{n})-\text{n};$
$\therefore\ 4\text{S}=\text{n}^4+\text{n}+6\big(1^2+2^2+....+\text{n}^2\big)-4(1+2+....+\text{n})$ $=\text{n}^4+\text{n}+\text{n}(\text{n}+1)(2\text{n}+1)-2\text{n}(\text{n}+1)$
$=\text{n}({\text{n}+1})\big(\text{n}^2-\text{n}+1+2\text{n}+1-2\big)$
$=\text{n}(\text{n}+1)\big(\text{n}^2+\text{n}\big)$
$\therefore\ \text{S}=\frac{\text{n}^2(\text{n}+1)^2}{4}=\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$

$1+2+3+....+\text{n}=\frac{\text{n}(\text{n}+2)}{2}$

$2+4+6+....+2\text{n}=2(1+2+3+....+\text{n})$
$=2\times\frac{\text{n}}{2}(1+\text{n})=\text{n}(\text{n}+1)$

$1 + 2 + 3 + .... + n =$ Sum of n terms of $A.P$. with first term $'1'$ and common difference $'1'=\frac{\text{n}}{2}(1+\text{n})$

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Question 25 Marks

If $\theta_1,\theta_2,\theta_3,....\theta_\text{n}$ are in A.P., whose common difference is d, show that $\sec\theta_1\cdot\sec\theta_2+\sec\theta_2+\sec\theta_3+\dots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$

Answer

Since, $\theta_1,\theta_2,\theta_3,\cdots\theta_\text{n}$ are in A.P.
$\therefore\ \theta_2-\theta_1=\theta_3-\theta_2=\theta_\text{n}-\theta _{\text{n}-1}=\text{d}$
Now we have to prove that
$\sec\theta_1\cdot\sec\theta_2+\sec\theta_2\cdot\sec\theta_3+\cdots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}$
$=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}\text{L.H.S}.$
$\Rightarrow\frac{\sin\text{d}}{\sin\text{d}}\big[\sec\theta_1\cdot\sec\theta_2+\sec\theta_2\sec\theta_3+\cdots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}\big]$
Taking only $ \frac{\sin\text{d}[\sec\theta_1\cdot\sec\theta_2]}{\sin\text{d}}=\frac{\sin\text{d}\Big[\frac{1}{\cos\theta}\cdot\frac{1}{\cos\theta _2}\Big]}{\sin\text{d}}$
$=\frac{\sin(\theta_2-\theta_1)}{\sin\text{d}}\cdot\frac{1}{\cos\theta_1\cos\theta_2}$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\cos\theta_1\cos\theta_2}\Big]$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1}{\cos\theta_1\cos\theta_2}-\frac{\cos\theta_2\sin\theta_1}{\cos\theta_1\cos\theta_2}\Big]$
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1]$
Similarly we can solve other terms which will be
$=\frac{1}{\sin\text{d}}[\tan\theta_3-\tan\theta_2]$ and $\frac{1}{\sin\text{d}}[\tan\theta_4-\tan\theta_3]$
Here L.H.S. $=\frac{1}{\sin\text{d}}\Big[\tan\theta_2-\tan\theta_1+\tan\theta_3-\tan\theta_2+\cdots+\tan \theta_\text{n}-\tan\theta_{\text{n}-1}\Big]$
$=\frac{1}{\sin\text{d}}[-\tan\theta_1+\tan\theta_\text{n}]=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}\text{R.H.S.}$

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Question 35 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, first term $a = 5$ and the common difference $d = 2$ let the carpenter will take $n$ days to finish the job
$S_n = 192$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(2\text{a}+(\text{n}-1)\text{d}]$
$192=\frac{\text{n}}{2}[2\times5+(\text{n}-1)2]$
$\Rightarrow 192 = n[10 + 2n - 2]$
$\Rightarrow 384 = n(2n + 8)$
$\Rightarrow 384 = 2n2 + 8n$
$\Rightarrow 2n^2 + 8n - 384 = 0$
$\Rightarrow n^2 + 4n - 192 = 0$
$\Rightarrow n^2 + 16n - 12n - 192 = 0$
$\Rightarrow n(n + 16) - 12(n + 16) = 0$
$\Rightarrow (n - 12)(n + 16) = 0$
$\Rightarrow n = 12\ [\because\ \text{n}\neq-16]$
Hence$,$ the required number of days $ = 12.$

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Question 45 Marks

$1$f the sum of p terms of an $AP$. is $q$ and the sum of $q$ terms is $p$, then show that the sum of $p +q$ terms is $-(p + q)$. Also$,$ find the sum of first $p - q$ terms $($where$, p > q).$

Answer

Let first term and common difference of the $A.P.$ be $a$ and $d,$ respectively. Given$, S_p = q.$
$\Rightarrow\frac{\text{p}}{2}[2\text{a}+(\text{p}-1)\text{d}]=\text{q}$
$\Rightarrow2\text{a}+(\text{p}-1)\text{d}=\frac{2\text{q}}{\text{p}}\ ....(1)$
Also$,$
$\text{S}_\text{q}=\text{p}$
$\Rightarrow\frac{\text{q}}{2}[2\text{a}+(\text{q}-1)\text{d}]=\text{p}$
$\Rightarrow2\text{a}+(\text{q}-1)\text{d}=\frac{2\text{p}}{\text{q}}\ ....(1)$
On subtracting $Eq. (2)$ from $Eq. (1),$ we get
$(\text{p}-\text{q})\text{d}=\frac{2\text{q}}{\text{p}}-\frac{2\text{p}}{\text{q}}$ or $(\text{p}-\text{q})\text{d}=\frac{2(\text{q}^2-\text{p}^2)}{\text{pq}}$
$\therefore\ \text{d}=\frac{-2(\text{p + q})}{\text{pq}}\ ....(3)$
On substituting the value of d into $Eq. (1),$ we get
$2\text{a}+(\text{p}-1)\Big(\frac{-2(\text{p}+\text{q})}{\text{pq}}\Big)=\frac{2\text{q}}{\text{p}}$
$\Rightarrow\text{a}=\frac{\text{q}}{\text{p}}+\frac{(\text{p}+\text{q})(\text{p-1})}{\text{pq}}\ ....(4)$
Now$,$
$\text{S}_{\text{p + q}}=\frac{\text{p}+ \text{q}}{2}[2\text{a}+(\text{p}+\text{q}-1)\text{d}]$
$=\frac{\text{p + q}}{2}\Big[\frac{2\text{q}}{\text{p}}+\frac{2(\text{p + q})(\text{p}- 1)}{\text{pq}}-\frac{2(\text{p}+\text{q}-1)(\text{p}+\text{q})}{\text{pq}}\Big]$
$=(\text{p}+\text{q})\Big[\frac{\text{q}}{\text{p}}+\frac{(\text{p}+\text{q})(\text{p}-1-\text{p}-\text{q}+1)}{\text{pq}}\Big]$
$=(\text{p}+\text{q})\Big[\frac{\text{q}}{\text{p}}+\frac{(\text{p}+\text{q})(-\text{q})}{\text{pq}}\Big]$
$=(\text{p}+\text{q})\Big[\frac{\text{q}}{\text{p}}-\frac{\text{p}+\text{q}}{\text{p}}\Big]=-(\text{p + q})$
Also$,$
$\text{S}_{\text{p}-\text{q}}=\frac{\text{p}-\text{q}}{2}[2\text{a}+(\text{p}-\text{q}-1)\text{d}]$
$=\frac{\text{p}-\text{q}}{2}\Big[\frac{2\text{q}}{\text{p}}+\frac{2(\text{p}+\text{q})(\text{p}-1)}{\text{pq}}-\frac{(\text{p}-\text{q}-1)2(\text{p}+\text{q})}{\text{pq}}\Big]$
$=(\text{p}-\text{q})\Big[\frac{\text{q}}{\text{p}}+\frac{(\text{p}+\text{q})(\text{p}-1-\text{p}+\text{q}+1)}{\text{pq}}\Big]$
$=(\text{p}-\text{q})\Big[\frac{\text{q}}{\text{p}}+\frac{(\text{p}+\text{q})\text{q}}{\text{pq}}\Big]$
$=(\text{p}-\text{q})\Big[\frac{\text{q}}{\text{p}}+\frac{\text{p}+\text{q}}{\text{p}}\Big]=(\text{p}-\text{q})\frac{(\text{p}+2\text{q})}{\text{p}}$

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Question 55 Marks

In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Answer

As per the given information we have the following diagram:

Starting point = S
Distance travelled to bring the first photo = 24 + 24 = 48m
Distance travelled to bring the second photo = 2(24 + 4) = 56m
Distance travelled to bring the third photo = 2(24 + 4 - 4) = 64m
Therefore the siries will be = 48, 56, 64 ....
Which an A.P. in which a = 48, d = 56 - 48 = 8
We have to find the total distance to bring all the potatoes back, so, n = 20
$\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\text{S}_\text{n}=\frac{20}{2}[2\times48+(20-1)8]=10[96+152]$
$=10\times248=2480\text{m}$
Hence, the required distance = 2480m

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Question 65 Marks

If the $p^{th}$ and qth terms of a $G.P.$ are $q$ and $p$ respectively$,$ show that its $(p + q)^{th}$ term is $\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^{\frac{1}{\text{p}-\text{q}}}.$

Answer

Let the first term and common ratio of $G.P.$ are $a$ and $r,$ respectively.
Given that$, p^{th} term = q$
$\Rightarrow ar^{p-1} = q ....(1)$
and $q^{th} term = p$
$\Rightarrow ar^{q-1} = p ....(2)$
On dividing $Eq. (1)$ by $(2),$ we get,
$\frac{\text{ar}^{\text{p}-1}}{\text{ar}^{\text{q}-1}}=\frac{\text{q}}{\text{p}}$
$\Rightarrow\text{r}^{\text{p}-\text{q}}=\frac{\text{q}}{\text{p}}$
$\Rightarrow\text{r}=\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{1}{\text{p}-\text{q}}}$
On substituting the value of r in $Eq. (1),$we get
$\text{a}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{p}-1}{\text{p}-\text{q}}}=\text{q}$
$\Rightarrow\text{a}=\text{q}\Big(\frac{\text{p}}{\text{q}}\Big)^{\frac{\text{p}-1}{\text{p}-\text{q}}}$
$\therefore (p + q)^{th}$ term $, T_{p+q} = a . r^{p+q-1}$
$=\text{q}\Big(\frac{\text{p}}{\text{q}}\Big)^{\frac{\text{p}-1}{\text{p}-\text{q}}}\Big(\frac{\text{a}}{\text{p}}\Big)^{\frac{\text{p + q}-1}{\text{p}-\text{q}}}$
$=\text{q}\Big(\frac{\text{p}}{\text{q}}\Big)^{\frac{\text{p}-1}{\text{p}-\text{q}}-\frac{\text{p}+\text{q}-1}{\text{p}-\text{q}}}$
$=\text{q}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{q}}{\text{p}-\text{q}}}$
$=\frac{\text{q}^{\frac{\text{q}}{\text{p}-\text{q}}+1}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}}}$
$=\frac{\text{q}^{\frac{\text{p}}{\text{p}-\text{q}}}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}}}$
$=\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^{\frac{1}{\text{p}-\text{q}}}$

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Question 75 Marks

$(3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ....$ to:
$(1) \ n$ terms.
$(2) \ 10$ terms.

Answer

Given series:
$= (3^3 - 2^3) + (5^3 - 4^3) + (7^3 - 6^3) + ....$
$= (3^3 + 5^3 + 7^3 + ....) - (2^3+ 4^3 + 6^3+ ....)$
$= [3^3 + 5^3 + 7^3 + .... (2n + 1)^3] - [2^3 + 4^3 + 6^3 + .... (2n)^3]$
$\therefore T_n = (2n + 1)^3 - (2n)^3$
$= (2n + 1 - 2n) [(2n + 1)^2+ (2n + 1)(2n) + (2n)^2] [\because a^3 - b^3 = (a - b)(a^2 + ab + b^2)]$
$= 1 - [4n^2 + 1 + 4n + 4n^2 + 2n + 4n^2]$
$=12n^2 + 6n + 1$
$\text{S}_\text{n}=\sum\text{T}_\text{n}=12\sum\text{n}^2+6\sum\text{n}+\text{n}$$=12.\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{6\text{n}(\text{n}+1)}{2}+\text{n}$
$= 2n(n + 1)(2n + 1) + 3n(n + 1) + n$
$= n[2(n + 1)(2n + 1) + 3n(n + 1) + 1]$
$= n[2(2n^2 + 3n + 1) + 3n + 3 + 1]$
$= n[4n^2 + 6n + 2 + 3n + 4]$
$= n[4n^2 + 9n + 6]$
$= 4n^3+ 9n^2 + 6n$
$S_{10} = 4(10)^3 + 9(10)^2 + 6(10) = 4 \times 1000 + 900 + 60= 4000 + 960 = 4960$

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Question 85 Marks

If $p^{th}, q^{th}$ and $r^{th}$ terms of an $A.P.$ and $G.P.$ are both $a, b$ and $c,$ respectively$,$ show that $a^{b-c}. b^{c-a}. c^{a-b} = 1.$

Answer

Let $A$ and $d$ be the first term and common difference of $A.P.,$ respectively. Also, let $B$ and $R$ be the first term and common ratio of $G.P.,$ respectively.
It is given that,
$A + (p - 1)d = a ....(1)$
$A + (q - 1)d = b ....(2)$
$A + (r - 1)d = c ....(3)$
Also$, a = BR^{p-1} ....(4)$
$b = BR^{q-1} ....(5)$
$c = BR^{r-1} ....(6)$
On subtracting $Eq. (2)$ from $Eq. (1),$ we get
$a - b = d(p - q)$
On subtracting $Eq. (3)$ from $Eq. (2),$ we get
$b - c = d(q - r)$
On subtracting $Eq. (1)$ from $Eq. (3),$ we get
$c - a = d(r - p)$
$\therefore a^{b-c}. b^{c-a}. c^{a-b}$
$=(\text{Br}^{(\text{p}-1)})^{\text{d}(\text{q}-\text{r})}(\text{Br}^{(\text{q}-1)})^{\text{d}(\text{r}-\text{p})}(\text{Br}^{(\text{r}-1)})^{\text{d}(\text{p}-\text{q})}$
$=\text{B}^{\text{d}[(\text{q}-\text{r})+(\text{r}-\text{p})+(\text{p}-\text{q})]}\text{R}^{\text{d}[(\text{p}-1)(\text{q}-\text{r})+(\text{q}-1)(\text{r}-\text{p})+(\text{r}-1)(\text{p}-\text{q})]}$
$=\text{B}^0\text{R}^0=1$

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