M.C.Q (1 Marks)

MCQ 1511 Mark

If $a_1,a_2,a_2,….., a_{20}$ are $\text{AM’s}$ between $13$ and $67,$ then the maximum value of $a_1 a_2 a_3.... a_{20} …$ is equal to:

A
$(20)^{20}$
✓
$(40)^{20}$
C
$(60)^{20}$
D
$(80)^{20}$

Answer

Correct option: B.

$(40)^{20}$

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MCQ 1521 Mark

What is the third term of Fibonacci sequence?

A
$0$
B
$1$
✓
$2$
D
$3$

Answer

Correct option: C.

$2$

$a_1=1$ and $a_2=1$.
$a_n=a_{n-1}+a_{n-2}, n>2$
This is a recurrence relation which gives Fibonacci sequence.
$\Rightarrow a_3=a_1+a_2=1+1=2$

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MCQ 1531 Mark

Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ If $5^{th}$ and $8^{th}$ term of a $\text{GP}$ be $48$ and $384$ respectively, then the common ratio of $\text{GP}$ is $2.$
Reason $(R):$ If $18, x, 14$ are in $AP,$ then $x = 16.$

A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
C
$A$ is true; $R$ is false
D
$A$ is false; $R$ is true.

Answer

Correct option: B.

$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$

Assertion Let a be the first term and $r$ be the common ratio of the given $GP.$
According to the question,
$T_5= 48$
$\Rightarrow ar^4 = 48 ...(i)$
and $T8 = 384$
$\Rightarrow ar^7= 384 ...(ii)$
On dividing Eq. $(ii)$ by Eq. $(i)$, we get
$=\frac{\text{ar}^7}{\text{ar}^4}=\frac{384}{48}$
$\Rightarrow\text{r}^3=8$
$\Rightarrow\text{r}=2$
Reason $18, x, 14$ are in $\text{AP.}$
$\Rightarrow x - 18 = 14 - x$
$\Rightarrow 2x = 32$
$\Rightarrow x = 16$

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MCQ 1541 Mark

The two geometric means between the numbers $1$ and $64$ are:

A
$1$ and $64$
✓
$4$ and $16$
C
$2$ and $16$
D
$8$ and $16.$

Answer

Correct option: B.

$4$ and $16$

Let the two $\text{G.M.s}$ between $1$ and $64$ be $G_1$ and $G_2$.
Thus, $1, \mathrm{G}_1, \mathrm{G}_2$ and $64$ are in $\text{G.P.}$
$64=1\times\text{r}^3$
$\Rightarrow\text{r}=\sqrt[3]{64}$
$\Rightarrow\text{r}=4$
$\Rightarrow\text{G}_1=\text{ar}=1\times4=4$
And, $\text{G}_2=\text{ar}^2=1\times4^2=16$
Thus, $4$ and $16$ are the required $\text{G.M.s.}$

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MATHS M.C.Q (1 Marks)