MATHS Assertion (A) & Reason (B) MCQ

4. A is false; R is true.

Solution:

Assertion Let a be the first term and r be the common ratio of the given GP. According to the question,

T₄ = (T₂)² and a = -3

$\because$ T₄ = (T₂)²

^$\therefore$ ar³  = (ar)²

⇒ -3r³ = (-3)² r²     [$\because$ a = -3]

⇒ r = -3

Now, T₇ = ar⁶ = -3(-3)⁶ = -3 × 729 = -2187

Reason Given AP is 6, 8, 10, ...

$\because\text{a} = \text{6}, \text{d} = 8 - 6 = 2$

$\therefore\text{s}_{10}=\frac{10}{2}[2\times6+(10-1)\times2]$

$=5[12+18]$

$=5\times30=150$

3. A is true; R is false

Solution:

Assertion We have, $\text{a}_\text{n}=\frac{\text{n}(\text{n}^2+5}{4}$

Now, we need to find x which is second term of the sequence, so put m = 2 in a_n.

$\therefore \text{a}_2=\frac{2(4+5)}{4}=\frac{18}{4}=\frac{9}{2}$

Reason We have, $\text{a}_\text{n} =(-1)^{\text{n}-1}\ 5^{\text{n}+1}$

^{$\therefore\text{a}_3=(-1)^{3-1}5^4=625$}

3. A is true; R is false

Solution:

Assertion

Let $\text{s}_\text{n}=\frac{3}{\sqrt{5}}+\frac{4}{\sqrt{5}}+\sqrt{5}...25\text{th}\text{ terms} $

 $\Rightarrow\text{s}_\text{n}=\frac{3}{\sqrt{5}}+\frac{4}{\sqrt{5}}+\frac{5}{\sqrt{5}}...25\text{th}\text{ terms} $

Clearly, the successive difference of the terms is same. So, RHS of the above series forms an AP, with first term,

$\text{a}=\frac{3}{\sqrt{5}}$ and common V5 difference,

$\text{d}=\frac{4}{\sqrt{5}}-\frac{3}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$

$\therefore\text{x}_{25}=\frac{25}{2}\Big[2\times\frac{3}{\sqrt{5}}+(25-1)\frac{1}{\sqrt{5}}\Big]$

$=25\Big[\frac{3}{\sqrt{5}}+\frac{12}{\sqrt{5}}\Big]$

$=25\times\frac{15}{5}\times\sqrt{5}$

$=75\sqrt{5}$

Reason Given, 27, x , 3 are in GP.

$\therefore\frac{\text{x}}{27}=\frac{3}{\text{x}}$

$\Rightarrow\text{x}^2=81\Rightarrow\text{x}=\pm9$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Given AP is 4, 8, 12, ...

$\therefore \text{a}=4,\text{d}=8-4=4 $

Now, $\text{s}_{20}=\frac{20}{2}[2\times4+(20-1)\times4]$

$=10[8+76]$

$=840$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion Given GP 4, 16, 64, ...

$\therefore\text{a}=4,\text{r}=\frac{16}{4}=4>1$

$\therefore\text{s}_6=\frac{4(4)^6-1)}{4-1}=\frac{4(4095)}{3}$

$=5460$

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

Solution:

Let S = 5 + 55 + 555 +... upto m terms

= 5(1 + 11 + 111 +... upton terms)

$=\frac{5}{9}(9+99+999+...\text{upton terms)} $

$=\frac{5}{9}[(10-1)+(100-1)+(1000-1)+...\text{upto n terms}] $

$=\frac{5}{9}[(10+100+1000+...\text{upto n terms}-(1+1+1+...\text{upto n term})]$

$=\frac{5}{9}\Big[\frac{10(10^\text{n}-1)}{10-1}-\text{n}\Big]$

$\Big[\because\text{sum of GP }=\frac{\text{a}(\text{r}^\text{n}-1)}{\text{r}-1},\text{r}>1\text{ and}\sum1=\text{n}\Big]$

$=\frac{5}{9}\Big[\frac{10(10^\text{n}-1)}{9}-\text{n}\Big]$

3. A is true; R is false

Solution:

Assertion We have, $\text{a}_\text{n} \frac{\text{n}^2}{2^\text{n}}$

Putting n = 7, $\text{a}_7=\frac{7^2}{2^7}=\frac{49}{128}$

Reason We have, $\text{a}_\text{n} \frac{\text{n}\text{(n}-2)}{\text{n}+3}$

Puttingn =20, $\text{a}_{20}=\frac{20(20-2)}{20+3}$

$\Rightarrow\text{a}_{20}=\frac{20\times18}{23}$

$=\frac{360}{23}$

1. A is true, R is true; R is acorrect explanation of A.

Solution:

Assertion If $-\frac{2}{7},\text{k},-\frac{7}{2}$ are in GP.

Then, $\frac{\text{a}_2}{\text{a}_1}=\frac{\text{a}_3}{\text{a}_2}$

$\Big[\because\text{ common ratio }\text{(r)}=\frac{\text{a}_2}{\text{a}_1}=\frac{\text{a}_3}{\text{a}_2}=\frac{\text{a}_4}{\text{a}_3}= ...\Big]$

$\therefore\frac{\text{k}}{\frac{-2}{7}}=\frac{\frac{-7}{2}}{\text{k}}$

$\Rightarrow\frac{7}{-2}\text{k}=\frac{-7}{2}\times\frac{1}{\text{k}}$

$\Rightarrow7\text{k}\times2\text{k}=-7\times(-2)$

$\Rightarrow14\text{k}^2=14$

$\text{k}^2=1\Rightarrow\text{k}=\pm1$

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

Solution:

Assertion Let a be the first term and r(r) <1) be the common ratio of the GP.

$\therefore$ The GP is a, ar, ar² , ...

 According to the question,

T₁ + T₂ = 5 ⇒ a + ar = 5 ⇒ a (l + r) = 5

and  T_n = 3 (T_n+1 + T_n+2  + T_n+3 +...)

⇒ ar^n-1 = 3 (arⁿ + arⁿ⁺¹ + arⁿ⁺² +...)

⇒ ar^n-1 = 3arⁿ (1 + r + r² +...)

$\Rightarrow1=3\text{r}\Big(\frac{1}{1-\text{r}}\Big)$

$\Rightarrow1-\text{r}=3\text{r}$

$\Rightarrow\text{r}=\frac{1}{4}$

Reason Given, 3, 6, 9, 12 ...

Here, a = 3, d = 6 - 3 = 3

$\therefore$ T₁₀ = a + (10 - 1)d

= 3 + 9 × 3

= 3 + 27 = 30

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

Solution:

Assertion Given AP is 16, 11, 6, ... Here,

$\text{a} =16,\text{d} =11-16=-5 $

$\because\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$

$\therefore\text{s}_{23}=\frac{23}{2}[2\times16+(23-1)(-5)]$

$=\frac{23}{2}[32+(22)(-5)]=\frac{23}{2}[32-110]$

$=\frac{23}{2}[-78]=-897$

Reason Given AP is x + y, x - y, x - 3y, ... Here, a = x + y

$\text{d} = (\text{x} - \text{y}) - (\text{x} + \text{y}) = -2\text{y}$

$\because\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$

$\therefore\text{s}_{22}\frac{22}{2}[2\times(\text{x}+\text{y})+(22-1)(-2)]$

$=11[2\text{x}+2\text{y}+(21)(-2\text{y})]$

$=11[2\text{x}+2\text{y}-42\text{y}]$

$=11[2\text{x}-40\text{y}]$

$=22[\text{x}-20\text{y}]$