Question

geometric progression
a3 = 5
a6 = 40
a1 = ?
Please exactly explain this to me

Answer 1

Answer:

$a_1 = \cfrac{5}{4}$

Step-by-step explanation:

Geometric Progression (G.P.):

If the common ratio between two consecutive terms of a series is equal, then that series is called a Geometric Progression (G.P.).

In general, G.P. is written as

$a, ar, ar^2, ar^3, ar^4, ..., ar^{n-1}$

where a = first term and r = common ratio.

So,

First term = a,

Second term = $a_2 = ar$

Third term = $a_3 = ar^2$

Fourth term = $a_4 = ar^3$

...

nth term = $a_n = ar^{n-1}$

General formula for its term, i.e.

$a_n = ar^{n-1}$

Now, given $a_3 = 5$, i.e., $\text{As } a_3 = ar^2$

$\therefore ar^2 = 5$

$\Rightarrow a = \cfrac{5}{r^2}$      ...(1)

And, $a_6 = 40$, i.e., as $a_6 = ar^5$

$\therefore ar^5 = 40$

$\Rightarrow a = \cfrac{40}{r^5}$      ...(2)

From Eqns. (1) and (2),

$\cfrac{5}{r^2} = \cfrac{40}{r^5}$

$\Rightarrow \cfrac{r^5}{r^2} = \cfrac{40}{5}$

$\Rightarrow r^{5-2} = 8$

$\Rightarrow r^3 = 2^3$

$\therefore r = 2$

Putting value r in Eqn. (1), we get

$a_1=a = \cfrac{5}{2^2} = \cfrac{5}{4}$          Ans.

Please mark my answer as BRAINLIEST.

Answer 2

Given : geometric progression a₃ =  5 , a₆  = 40

To find : a₁

Solution:

aₙ - nth term

Let say First term a₁  = a

common ratio  =  r

nth term in gp aₙ = arⁿ⁻¹

a₃ = ar³⁻¹ = ar² = 5

a₆ = ar⁶⁻¹ = ar⁵ = 40

ar⁵ /ar² = 40/5

=> r³ = 8

=> r³ = (2)³

=> r = 2

ar² = 5

=> a(2)² = 5

=>  4a = 5

=> a = 5/4

=> a₁  =  5/4

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