M.C.Q (1 Marks)

MCQ 11 Mark

If $t_n$ denotes the $n^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + ...$ then $t_{50}$ is:

A
$49^2 - 1$
B
$49^2$
C
$50^2 + 1$
✓
$49^2 + 2$

Answer

Correct option: D.

$49^2 + 2$

$S_n = 2 + 3 + 6 + 11 + 18 + .... + t_{50}$
Using method of difference$,$ we get
$S_n= 2 + 3 + 6 + 11 + 18 + .... + t_{50} ....(1)$
And $S_n = 0 + 2 + 3 + 6 + 11 + .... + t_{49} + t_{50} ....(2)$
Subtracting eq. $(2)$ from eq. $(2),$ we get
$0 = 2 + 1 + 3 + 5 + 7 + .... -t_{50}$ terms
$\Rightarrow t_{50} = 2 + (1 + 3 + 5 + 7 + ....$ upto $49$ terms$)$
$\Rightarrow\text{t}_{50}=2+\frac{49}{2}[2\times1+(49-1)2]=2+\frac{49}{2}[2+96]$
$\Rightarrow2+\frac{49}{2}\times98=2+49\times49=49^2+2$
Hence$,$ the correct option is $(d).$

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MCQ 21 Mark

Let $S_n$ denote the sum of the cubes of the first n natural numbers and $s_n$ denote the sum of the first $n$ natural numbers. Then $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$ equals to:

✓
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
B
$\frac{\text{n}(\text{n}+1)}{2}$
C
$\frac{\text{n}^{2}+3\text{n}+2}{2}$
D
None of these.

Answer

Correct option: A.

$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

$\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\frac{\text{r}(\text{r}+1)}{2}}=\sum\limits^\text{n}_{\text{r}=1}\frac{\text{r}(\text{r}+1)}{2}=\frac{1}{2}\bigg[\sum\limits^\text{n}_{\text{r}=1}\text{r}^2+\sum\limits^\text{n}_{\text{r}=1}\text{r}\bigg]$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}{+1})(2\text{n}{+1})}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{1}{2}.\frac{\text{n}({\text{n}+1})}{2}\Big[\frac{2\text{n}{+1}}{3}+1\Big]$
$=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$
$=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

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MCQ 31 Mark

Let $S_n$ denote the sum of the first $n$ terms of an $\text{A.P.}$ If $S_{2n} = 3S_n,$ then $S_{3n} : S_n$ is equal to:

A
$4$
✓
$6$
C
$8$
D
$10$

Answer

Correct option: B.

$6$

$6$
Let the first term be a and common difference be d.
Then$,$
$\text{S}_{2\text{n}}=3\text{S}_{\text{n}}$
$\Rightarrow\frac{2\text{n}}{2}[2\text{a}+(2\text{n}-1)\text{d}]=\frac{3\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow2[2\text{a}+(2\text{n}-1)\text{d}]=3[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow4\text{a}+(4\text{n}-2)\text{d}=6\text{a}+(3\text{n}-3)\text{d}$
$\Rightarrow2\text{a}=(\text{n}+1)\text{d}$
Now$,$
$\frac{\text{S}_{3\text{n}}}{\text{S}_{\text{n}}}=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}=\frac{3[2\text{a}+(3\text{n}-1)\text{d}]}{[2\text{a}+(\text{n}-1)\text{d}]}$
$=\frac{3[(\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)\text{d}+(\text{n}-1)\text{d}]}=\frac{3[4\text{nd}]}{2\text{nd}}=6$

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MCQ 41 Mark

The minimum value of $4^{\text{x}}+4^{1-\text{x}},\text{x}\in\text{R}$ is:

A
$2$
✓
$4$
C
$1$
D
$0$

Answer

Correct option: B.

$4$

We know that $\text{AM}\geq\text{GM}$
$\therefore\ \frac{4^\text{x}+4^{1-\text{x}}}{2}\geq\sqrt{4^\text{x}.4^{1-\text{x}}}$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq2\sqrt{4^{\text{x}+1-\text{x}}}$
$4^\text{x}+4^{1-\text{x}}\geq2.2$
$\Rightarrow4^\text{x}+4^{1-\text{x}}\geq4$
Hence, the correct option is $(b).$

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MCQ 51 Mark

If $x, 2y$ and $3z$ are in $\text{A.P.}$ where the distinct numbers $x, y$ and $z$ are in $\text{G.P.,}$ then the common ratio of the $\text{G.P.}$ is:

A
$3$
B
$\frac{1}{3}$
C
$2$
✓
$\frac{1}{2}$

Answer

Correct option: D.

$\frac{1}{2}$

Since$, x, 2y$ and $3z$ are in $\text{A. P.,}$ we get
$\text{2y}=\frac{\text{x}+3\text{z}}{2}$
$\Rightarrow 4y = x + 3z$
Also$, x, y $and $z$ are ibn $\text{G.P.}$
Therefore$, y = xr$ and $z = xr^2.$
Where $'r\ '$ is the common ratio.
$\therefore 4xr = x + 3xr^2 [$Using $(1)]$
$\Rightarrow 4r = 1 + 3r^2$
$\Rightarrow 3r^2 - 4r + 1 = 0$
$\Rightarrow (3r - 1)(r - 1) = .0$
$\Rightarrow\text{r}=\frac{1}{3}$
$($For $r = 1; x, y, z$ are not distinct$)$

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MCQ 61 Mark

If $9$ times the $9^{th}$ term of an $A.P.$ is equal to $13$ times the $13^{th}$ term, then the $22^{nd} $ term of the $A.P.$ is:

✓
$0$
B
$22$
C
$198$
D
$220$

Answer

Correct option: A.

$0$

Let the first term and common difference of given $A.P.$ be a and d, respectively.
It is given that $9 \times t_9 = 13 \times t_{13}$
$\Rightarrow 9(a + 8d) = 13(a + 12d)$
$\Rightarrow 9a + 72d = 13a + 156d$
$\Rightarrow 4a + 84d = 0$
$\Rightarrow 4(a + 21d) = 0$
$\Rightarrow t_{22}= 0$

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MCQ 71 Mark

If in an $A.P., S_n = qn^2$ and $S_m = qm^2,$ where $S_r$ denotes the sum of $r$ terms of the $AP,$ then $S_q$ equals:

A
$\frac{\text{q}^3}{2}$
B
$mnq$
✓
$q^3$
D
$(m + n)q^2$

Answer

Correct option: C.

$q^3$

Given$,$
$Sn = qn^2$ and $Sm = qm^2$
$\therefore S_1 = q, S_2 = 4q, S_3 = 9q$ and $S_4 = 16q$
Now$, t_1 = q$
$\therefore t_2 = S_2 - S_1 = 4q - q = 3q$
$t_3 = S_3 - S_2 = 9q - 4q = 5q$
$t_4 = S_4 - S_3 = 16q - 9q = 7q$
So$,$ the $A.P. $is: $q, 3q, 5q, 7q, ....$
Thus$,$ first term is $q $and common difference is $3q - q = 2q.$
$\therefore\ \text{S}_\text{q}=\frac{\text{q}}{2}[2\times\text{q}+(\text{q}-1)2\text{q}]=\frac{\text{q}}{2}\times[2\text{q}+2\text{q}^2-2\text{q}]$
$=\frac{\text{q}}{2}\times2\text{q}^2=\text{q}^3$

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MCQ 81 Mark

The lengths of three unequal edges of a rectangular solid block are in $G.P.$ If the volume of the block is $216\ cm^3$ and the total surface area is $252\ cm^2,$ then the length of the longest edge is:

✓
$12\ cm$
B
$6\ cm$
C
$18\ cm$
D
$3\ cm$

Answer

Correct option: A.

$12\ cm$

$12\ cm.$
Let the length, breadth and height of rectangular solid block be $\frac{\text{a}}{\text{r}}, a$ and $ar,$ respectively.
$\therefore $ Volume $=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216\text{ cm}^3$
$\Rightarrow\text{a}^3=216=6^3$
$\Rightarrow\text{a}=6$
Also, Surface area $=2\Big(\frac{\text{a}}{\text{r}}.\text{a}+\text{a}.\text{ar}+\frac{\text{a}}{\text{r}}.\text{ar}\Big)=252$
$\Rightarrow2\text{a}^2\Big(\frac{1}{\text{r}}+\text{r}+1\Big)=252$
$\Rightarrow2\times36\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=252$
$\Rightarrow2(1+\text{r}^2+\text{r})=7\text{r}$
$\Rightarrow2\text{r}^2-5\text{r}+2=0$
$\Rightarrow(2\text{r}-1)(\text{r}-2)=0$
$\therefore\ \text{r}=\frac{1}{2},2$
For $\text{r}=\frac{1}{2}:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6\times2}{1}=12,$ Breadth $= a = 6$
Height $=\text{ar}=6\times\frac{1}{2}=3$
For $r = 2:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6}{2}=3,$ Breadth $ = a = 6$
Height $= ar = 6 \times 2 = 12$

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MCQ 91 Mark

If the third term of $G.P.$ is $4,$ then the product of its first $5$ terms is:

A
$4^3$
B
$4^4$
✓
$4^5$
D
None of these.

Answer

Correct option: C.

$4^5$

Given that:
$T_3 = 4$
$\Rightarrow ar^{3-1} = 4 \big[\because\ \text{T}_\text{n}=\text{ar}^{\text{n}-1}\big]$
$\Rightarrow ar^2 = 4$
Product of first $5$ terms $= a. ar. ar^2. ar^3. ar^4$
$= a^5r^{10}= (ar^{2)5} = (4)^5$
Hence$,$ the corrrect option is $(c).$

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MCQ 101 Mark

If the sum of n terms of an $A.P.$ is given by
$S_n = 3n + 2n^2,$ then the common difference of the $A.P.$ is:

A
$3.$
B
$2.$
C
$6.$
✓
$4.$

Answer

Correct option: D.

$4.$

Given that:
$S_n = 3n + 2n^2$
$S_1 = 3(1) + 2(1)2 = 5$
$S_2 = 3(2) + 2(4) = 14$
$S_1 = a_1 = 5$
$S_2 - S_1 = a_2 = 14 - 5 = 9$
$\therefore$ Common difference $d = a_2 - a_1 = 9 - 5 = 4$
Hence$,$ the correct option is $(d).$

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