The sum of first four terms of a geometric progression (G. P) is 6 5 1 2 and the sum of their respective reciprocals is 6 5 1 8 . If the product of first three terms of the G.P is 1,and the third term is a, then 2a is………………

Let a, ar, ar2, ar3 are in G.P. a + ar + ar2 + ar3 = 6 5 1 2 1 a + 1 a r + 1 a r 2 + 1 a r 3 = 6 5 1 8 ( a r ) 2 r = 3 2 ⇒ a = 2 3 , r = 3 2 ∴ third term = ar2 = 2 3 * 9 4 = 3 2 ∴ α = 3 2 ⇒ 2 α = 3

The sum of first four terms of a geometric progression (G. P) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18} .$ If the product of first three terms of the G.P is 1,and the third term is a, then 2a is………………

2 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

7 months ago

Let a, ar, ar², ar³ are in G.P.

a + ar + ar² + ar³ = $\frac{65}{12}$

$\frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} + \frac{1}{a r^{3}} = \frac{65}{18}$

${(a r)}^{2} r = \frac{3}{2} \Rightarrow a = \frac{2}{3}, r = \frac{3}{2}$

$∴$ third term = ar² = $\frac{2}{3} * \frac{9}{4} = \frac{3}{2}$

$∴ α = \frac{3}{2} \Rightarrow 2 α = 3$

Upvote

Similar Questions for you

If |z + 1| = αz + β (i + 1) and  $z = \frac{1}{2}$ – 2i, find α + β.

$| \frac{1}{2} - 2 i + 1 | = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\sqrt[]{\frac{9}{4} + 4} = α (\frac{1}{2} - 2 i) + β (1 + i)$

$\frac{5}{2} = α (\frac{1}{2}) + β + i (- 2 α + β)$

$\frac{α}{2} + β = \frac{5}{2}$ ...(1)

–2α + β = 0 …(2)

Solving (1) and (2)

$\frac{α}{2} + 2 α = \frac{5}{2}$

$\frac{5}{2} α = \frac{5}{2}$

a =

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

Start with

(1) $\bar{E} : \frac{6!}{2!} = 360$

(2) $\bar{G E} : \frac{5!}{2!},$ $\bar{G N} : \frac{5!}{2!}$

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

${(1 + x)}^{11} =^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} + . . . . . +^{11} C_{11} x^{11}$

= \frac{2^{12} - 2 - 24}{12}

= \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6} = \frac{m}{n}

m + n = 2035 + 6 = 2041

Let $f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$ 

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ 

The range of fog(x) is

$f (x) = {\begin{matrix} 2 + 2 x, & x \in (- 1, 0) \\ 1 - \frac{x}{3}, & x \in [0, 3) \end{matrix}$

$g (x) = {\begin{matrix} x, & x \in [0, 1) \\ - x, & x \in (- 3, 0) \end{matrix}$ ->g(x) = |x|, x Î (–3, 1)

$f (g (x)) = {\begin{matrix} 2 + 2 | x |, & | x | \in (- 1, 0) \Rightarrow x \in ? \\ 1 - \frac{| x |}{3}, & | x | \in [0, 3) \Rightarrow x \in (- 3, 1) \end{matrix}$