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Sequences and Series

Multiple Choice Questions

121.

Let a, b and c be in AP and $|a| < 1, |b| < 1, |c| < 1$ . If x = 1 + a + a² + ... to $\infty$ , y = 1 + b + b² + ... to $\infty$ , z = 1 + c + c² + ... to $\infty$ , then x, y and z are in

AP
GP
HP
None of these

122.
The sum of the series
$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . .$ is

3e

$\frac{17}{6}$ e

$\frac{13}{6}$ e

$\frac{19}{6}$ e

$\frac{17}{6}$ e

$We have, T_{n} = \frac{1^{2} + 2^{2} + 3^{2} + . . . + n^{2}}{n!} = \frac{\sum n^{2}}{n!} = \frac{n (n + 1) (2 n + 1)}{6 n!} = \frac{1}{6} (\frac{2 n^{3} + 3 n^{2} + n}{n!}) = \frac{1}{6} (2 . \frac{n^{3}}{n!} + 3 . \frac{n^{2}}{n!} + \frac{n}{n!})$

$∴ Sum of the series = \frac{1}{6} (2 \sum_{n = 1}^{\infty} \frac{n^{3}}{n!} + 3 \sum_{n = 1}^{\infty} \frac{n^{2}}{n!} + \sum_{n = 1}^{\infty} \frac{n}{n!}) = \frac{1}{6} (2 \times 5 e + 3 \times 2 e + e) = \frac{1}{6} (10 e + 6 e + e) = \frac{17}{6} e$

123.

For a GP, a_n = 3(2ⁿ), ∀ n ∈ N. Find the common ratio

124.

If a, b, c are in HP, then $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ will be in

AP
GP
HP
None of these

125.

The sum of the coefficients of (6a - 5b)ⁿ, where n is a positive integer, is

1
- 1
2ⁿ
2^{n - 1}

126.

If a^x = b^y = c^z = d^u and a, b, c, d are in GP, then x, y, z, u are in

AP
GP
HP
None of these

127.

If x is numerically so small so that x² and higher power of x can be neglected, then ${(1 + \frac{2 x}{3})}^{\frac{3}{2}} . {(32 + 5 x)}^{- \frac{1}{5}}$ is approximately equal to