Assertion (A) & Reason (B) MCQ

MCQ 11 Mark

Assertion (A): If the sum of first two terms of an infinite GP is 5 and each term is three times the sum of the succeeding terms, then the common ratio is $\frac{1}{4}$.
Reason (R): In an AP 3, 6, 9, 12........... the 10th term is equal to 33 .

A
Both A and R are true and R is the correct explanation of A .
B
Both $A$ and $R$ are true but $R$ is not the correct explanation of A.
✓
$A$ is true but $R$ is false.
D
$A$ is false but $R$ is true.

Answer

Correct option: C.

$A$ is true but $R$ is false.

(c) A is true but R is false.
Explanation: Assertion Let a be the first term and $\mathrm{r}(|\mathrm{r}|<1)$ be the common ratio of the GP.
$\therefore$ The GP is a, ar, $\mathrm{ar}^{2}, \ldots$
According to the question,
$\mathrm{T}_{1}+\mathrm{T}_{2}=5 \Rightarrow \mathrm{a}+\mathrm{ar}=5 \Rightarrow \mathrm{a}(1+\mathrm{r})=5$
and $\mathrm{T}_{\mathrm{n}}=3\left(\mathrm{~T}_{\mathrm{n}+1}+\mathrm{T}_{\mathrm{n}+2}+\mathrm{T}_{\mathrm{n}+3}+\ldots\right)$
$\Rightarrow a r^{n-1}=3\left(a^{n}+a r^{n+1}+\mathrm{ar}^{\mathrm{n}+2}+\ldots\right)$
$\Rightarrow \operatorname{ar}^{\mathrm{n}-1}=3 \mathrm{ar}^{\mathrm{n}}\left(1+\mathrm{r}+\mathrm{r}^{2}+\ldots\right)$
$\Rightarrow 1=3 \mathrm{r}\left(\frac{1}{1-r}\right)$
$\Rightarrow 1-\mathrm{r}=3 \mathrm{r}$
$\Rightarrow \mathrm{r}=\frac{1}{4}$
Reason: Given, 3, 6, 9, 12 ...
Here, $\mathrm{a}=3, \mathrm{~d}=6-3=3$
$\therefore \mathrm{T}_{10}=\mathrm{a}+(10-1) \mathrm{d}$
$=3+9 \times 3$
$=3+27=30$

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MCQ 21 Mark

Assertion (A): If each of the observations $x_{1}, x_{2}, \ldots, x_{n}$ is increased by a, where $a$ is a negative or positive number, then the variance remains unchanged.
Reason (R): Adding or subtracting a positive or negative number to (or from) each observation of a group does not affect the variance.

✓
Both $A$ and $R$ are true and $R$ is the correct explanation of A .
B
Both A and R are true but R is not the correct explanation of A.
C
A is true but R is false.
D
A is false but $R$ is true.

Answer

Correct option: A.

Both $A$ and $R$ are true and $R$ is the correct explanation of A .

(a) Both A and R are true and R is the correct explanation of A .
Explanation: Assertion: Let $\bar{x}$ be the mean of $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots, \mathrm{x}_{\mathrm{n}}$. Then, variance is given by
$\sigma_{1}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$
If $a$ is added to each observation, the new observations will be
$y_{i}=x_{i}+a$
Let the mean of the new observations be $\bar{y}$.
Then,
$\bar{y}=\frac{1}{n} \sum_{i=1}^{n} y_{i}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+a\right)$
$=\frac{1}{n}\left[\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} a\right]$
$=\frac{1}{n} \sum_{i=1}^{n} x_{i}+\frac{n a}{n}=\bar{x}+a$
i.e. $\bar{y}=\bar{x}+a$...(ii)Thus, the variance of the new observations is $\sigma_{2}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}+a-\bar{x}-a\right)^{2}$ [using Eqs. (i) and (ii)]
$=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=\sigma_{1}^{2}$
Thus, the variance of the new observations is same as that of the original observations.
Reason: We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance.

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Applied Maths Assertion (A) & Reason (B) MCQ

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