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Sequences and Series

Multiple Choice Questions

281.

If $\log_{x} (a), a^{\frac{x}{2}} and \log_{b} (x)$ are in GP, then x is equal to

$\log_{b} (\log_{a} (b))$
$\log (\log (a))$
$\log (\log (a)) + \log (\log (b))$
$\log_{a} (\log_{b} (a))$

282.

At the foot of a mountain the elevation of its summit is 45°. After ascending 2 km towards the mountain up an incline of 30°, the elevation changes to 60°. The height ofmountain will be

$\sqrt{3} - 1$
$1 - \sqrt{3}$
$1 + \sqrt{3}$
$\sqrt{3}$

283.

If $\frac{a + b}{1 - ab}, \frac{b + c}{1 - bc}$ are in AP, then $a, \frac{1}{b}$ and c are in

AP
GP
HP
None of these

284.

The number of terms in the series 105 + 103 + 101 + ... + 49 + 47 is

285.

The geometric mean of roots of the equation x² - 18x + 81 = 0 is

286.

The sum of n terms of the series 1 + 5 + 12 + 22 + 35 + ... is

$\frac{n^{2} (n + 1)}{8}$
$\frac{n^{2} (n + 1)}{6}$
$\frac{n^{2} (n + 1)}{2}$
None of these

287.

If geometric mean and harmonic mean of two numbers a and b are 16 and 64/5 respectively, then the value of a : b is

4 : 1
3 : 2
2 : 3
1 : 4

288.
If the sum offour numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18, then the numbers are

3, 9, 27, 81

4, 8, 16, 32

2, 6, 18, 54

None of these

4, 8, 16, 32

$Let four terms mn a GP be {ar}^{3}, ar, \frac{a}{r} and \frac{a}{r^{3}} According to the given condition, {ar}^{3} + ar + \frac{a}{r} + \frac{a}{r^{3}} = 60 . . . (i) and \frac{{ar}^{3} + \frac{a}{r^{3}}}{2} = 18 \Rightarrow {ar}^{3} + \frac{a}{r^{3}} = 36 . . . (ii) Now, from Eq (i), we have (ar + \frac{a}{r}) + {ar}^{3} + \frac{a}{r^{3}} = 60 \Rightarrow a (r + \frac{1}{r}) + 36 = 60 [∵ from Eq . (ii)] \Rightarrow a (r + \frac{1}{r}) = 24 . . . (iii)$

$On dividing Eq . (iii) by Eq . (ii), we get \frac{a (r^{3} + \frac{1}{r^{3}})}{a (r + \frac{1}{r})} = \frac{36}{24} \Rightarrow \frac{(r + \frac{1}{r}) (r^{2} + \frac{1}{r^{2}} - 1)}{r + \frac{1}{r}} = \frac{3}{2} \Rightarrow 2 (r^{2} + \frac{1}{1^{2}} - 1) = 3 \Rightarrow \frac{2 (r^{4} + 1 - r^{2})}{r^{2}} = 3 \Rightarrow 2 r^{4} + 2 - 2 r^{2} = 3 r^{2} \Rightarrow 2 r^{4} - 5 r^{2} + 2 = 0 \Rightarrow 2 r^{4} - 4 r^{2} - r^{2} + 2 = 0 \Rightarrow 2 r^{2} (r^{2} - 2) - 1 (r^{2} - 2) = 0 \Rightarrow (r^{2} - 2) (2 r^{2} - 1) = 0 \Rightarrow r^{2} = 2, 2 r^{2} = 1 \Rightarrow r = \pm \sqrt{2}, r = \pm \frac{1}{\sqrt{2}}$

$On putting r = \sqrt{2} in Eq . (iii), we get a (\sqrt{2} + \frac{1}{\sqrt{2}}) = 24 \Rightarrow a (\frac{2 + 1}{\sqrt{2}}) = 24 \Rightarrow 3 a = 24 a \sqrt{2} \Rightarrow a = 8 \sqrt{2} ∴ Senes becomes 8 \sqrt{2} {(\sqrt{2})}^{3}, 8 \sqrt{2} (\sqrt{2}), \frac{8 \sqrt{2}}{\sqrt{2}} and \frac{8 \sqrt{2}}{{(\sqrt{2})}^{3}} i . e ., 32, 16, 8 and 4 . If we take r = \frac{1}{\sqrt{2}}, we get the seris 4, 8, 16 and 32 .$

289.

iF a,b and c are in HP, then for any n $\in$ N, which one of the following is true ?

aⁿ + cⁿ < 2bⁿ
aⁿ + cⁿ > 2bⁿ
aⁿ + cⁿ = 2bⁿ
None of the above

290.

The sum of the series $\frac{1}{2 . 3} + \frac{1}{4 . 5} + \frac{1}{6 . 7} + . . .$ is

$\log (\frac{e}{2})$
$\log (\frac{2}{e})$
$\frac{e}{2}$
$\frac{2}{e}$

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Multiple Choice Questions

288. If the sum offour numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18, then the numbers are 3, 9, 27, 81 4, 8, 16, 32 2, 6, 18, 54 None of these

288.
If the sum offour numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18, then the numbers are

3, 9, 27, 81

4, 8, 16, 32

2, 6, 18, 54

None of these