MATHS M.C.Q (1 Marks)

3. q³.

Solution:

Given,

Sn = qn² and Sm = qm²

$\therefore$ S₁ = q, S₂ = 4q, S₃ = 9q and S₄ = 16q

Now, t₁ = q

$\therefore$ t₂ = S₂ - S₁ = 4q - q = 3q

t₃ = S₃ - S₂ = 9q - 4q = 5q

t₄ = S₄ - S₃ = 16q - 9q = 7q

So, the A.P. is: q, 3q, 5q, 7q, ....

Thus, first term is q and common difference is 3q - q = 2q.

$\therefore\ \text{S}_\text{q}=\frac{\text{q}}{2}[2\times\text{q}+(\text{q}-1)2\text{q}]=\frac{\text{q}}{2}\times[2\text{q}+2\text{q}^2-2\text{q}]$

$=\frac{\text{q}}{2}\times2\text{q}^2=\text{q}^3$

2. 6

Solution:

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).

4. 49² + 2.

Solution:

S_n = 2 + 3 + 6 + 11 + 18 + .... + t₅₀

Using method of difference, we get

S_n = 2 + 3 + 6 + 11 + 18 + .... + t₅₀ ....(1)

And S_n = 0 + 2 + 3 + 6 + 11 + .... + t₄₉ + t₅₀ ....(2)

Subtracting eq. (2) from eq. (2), we get

0 = 2 + 1 + 3 + 5 + 7 + .... -t₅₀ terms

⇒ t₅₀ = 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).

2. $\frac{1}{2}.$

Solution:

Since, x, 2y and 3z are in A. P., we get

$\text{2y}=\frac{\text{x}+3\text{z}}{2}$

⇒ 4y = x + 3z

Also, x, y and z are ibn G.P.

Therefore, y = xr and z = xr².

Where 'r' is the common ratio.

$\therefore$ 4xr = x + 3xr² [Using (1)]

⇒ 4r = 1 + 3r²

⇒ 3r² - 4r + 1 = 0

⇒ (3r - 1)(r - 1) = .0

$\Rightarrow\text{r}=\frac{1}{3}$

(For r = 1; x, y, z are not distinct)

3. 4⁵.

Solution:

Given that:

T₃ = 4

⇒ ar^3-1 = 4 $\big[\because\ \text{T}_\text{n}=\text{ar}^{\text{n}-1}\big]$

⇒ ar² = 4

Product of first 5 terms = a. ar. ar². ar³. ar⁴

= a⁵r¹⁰ = (ar²⁾⁵ = (4)⁵

Hence, the corrrect option is (c).

4. 4.

Solution:

Given that:

S_n = 3n + 2n²

S₁ = 3(1) + 2(1)2 = 5

S₂ = 3(2) + 2(4) = 14

S₁ = a₁ = 5

S₂ - S₁ = a₂ = 14 - 5 = 9

$\therefore$ Common difference d = a₂ - a₁ = 9 - 5 = 4

Hence, the correct option is (d).

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

Solution:

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

$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}{+1})(2\text{n}{+1})}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$

$=\frac{1}{2}.\frac{\text{n}({\text{n}+1})}{2}\Big[\frac{2\text{n}{+1}}{3}+1\Big]=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$

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

2. 6

Solution:

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,

$\frac{\text{S}_{3\text{n}}}{\text{S}_{\text{n}}}=\frac{\frac{3\text{n}}{2}[2\text{a}+(3\text{n}-1)\text{d}]}{\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]}=\frac{3[2\text{a}+(3\text{n}-1)\text{d}]}{[2\text{a}+(\text{n}-1)\text{d}]}$

$=\frac{3[(\text{n}+1)\text{d}+(3\text{n}-1)\text{d}]}{[(\text{n}+1)\text{d}+(\text{n}-1)\text{d}]}=\frac{3[4\text{nd}]}{2\text{nd}}=6$

1. 12cm.

Solution:

Let the length, breadth and height of rectangular solid block be$\frac{\text{a}}{\text{r}},$ a and ar, respectively.

$\therefore\ \text{Volume}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216\text{cm}^3$

$\Rightarrow\text{a}^3=216=6^3\Rightarrow\text{a}=6$

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 × 2 = 12