The correct answer is 40
Let G.P. be a1 = a, a2 = ar, a3 = ar2, ……
\(∵ 3a_2 + a_3 = 2a_4\)
\(⇒ 3ar + ar^2 = 2ar³\)
⇒ 2ar² - r - 3 = 0
∴ r = -1 or \(\frac{3}{2}\)
∵ a1 = a > 0 then r ≠ -1
Now,
\(a_2 + a_4 = 2a_3 + 1\)
\(ar + ar³ = 2ar² + 1\)
\(a ( \frac{3}{2} + \frac{27}{8} - \frac{9}{2} ) = 1\)
\(∴ a = \frac{8}{3}\)
\(∴ \frac{8}{3} ( \frac{3}{2} + \frac{27}{8} + \frac{81}{8} )\)
= 40
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
A geometric progression is the sequence, in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant to the previous term. It is represented by: a, ar1, ar2, ar3, ar4, and so on.
Important properties of GP are as follows:
If a1, a2, a3,… is a GP of positive terms then log a1, log a2, log a3,… is an AP (arithmetic progression) and vice versa