5 Marks Questions

Question 15 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 25 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 ⇒ ar^p-1 = q ....(1)
and q^th term = p ⇒ 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 35 Marks

(3³ - 2³) + (5³ - 4³) + (7³ - 6³) + .... to:

n terms.
10 terms.

Answer

Given series:
= (3³ - 2³) + (5³ - 4³) + (7³ - 6³) + ....
= (3³ + 5³ + 7³ + ....) - (2³+ 4³ + 6³+ ....)
= [3³ + 5³ + 7³ + .... (2n + 1)³] - [2³ + 4³ + 6³ + .... (2n)³]
$\therefore$ T_n = (2n + 1)³ - (2n)³
= (2n + 1 - 2n) [(2n + 1)²+ (2n + 1)(2n) + (2n)²] $[\because$ a³ - b³ = (a - b)(a² + ab + b²)$]$
= 1 - [4n² + 1 + 4n + 4n² + 2n + 4n²]
=12n² + 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(2n² + 3n + 1) + 3n + 3 + 1]

= n[4n² + 6n + 2 + 3n + 4]

= n[4n² + 9n + 6]

= 4n³+ 9n² + 6n

S₁₀ = 4(10)³ + 9(10)² + 6(10) = 4 × 1000 + 900 + 60

= 4000 + 960 = 4960

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Question 45 Marks

1f the sum of p terms of an AP. is q and the sum of q terms isp, then show that the sum ofp +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

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
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(2\text{a}+(\text{n}-1)\text{d}]$
$192=\frac{\text{n}}{2}[2\times5+(\text{n}-1)2]$
⇒ 192 = n[10 + 2n - 2]
⇒ 384 = n(2n + 8)
⇒ 384 = 2n2 + 8n
⇒ 2n² + 8n - 384 = 0
⇒ n² + 4n - 192 = 0
⇒ n² + 16n - 12n - 192 = 0
⇒ n(n + 16) - 12(n + 16) = 0
⇒ (n - 12)(n + 16) = 0
⇒ n = 12 $[\because\ \text{n}\neq-16]$
Hence, the required number of days = 12.

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Question 65 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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Question 75 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 85 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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