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3 years ago

Is -2, -12, -60, -120 a geometric sequence?

Mathematics

1 answer:

nlexa [21]3 years ago

7 0

given,

1st term(a)= -2

2nd term (t2) =-12

3rd term (t3) =-60

4th term(t4)= -120

here common ratio (r)=t2/t1=t4/t3

or, -12/-2= -120/-60

or,6=2

hence,no -12, -60, -120 aren't in a geometric sequence because their common ratio aren't same.

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ABC and EDC are straight lines. EA is parallel to DB. EC = 8.1 cm. DC = 5.4 cm. DB = 2.6 cm. (a) Work out the length of AE. cm (

harkovskaia [24]

By applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

See the image in the attachment for the referred diagram.

The two triangles, triangle AEC and triangle BDC are similar triangles.
Therefore, the ratio of the corresponding sides of triangles AEC and BDC will be the same.

This implies that:

AC/BC = EC/DC = AE/DB

Given:

$EC = 8.1 $ cm\\\\DC = 5.4 $ cm\\\\DB = 2.6 cm\\\\AC = 6.15 $ cm$

a. Find the length of AE:

EC/DC = AE/DB

Plug in the values

$\frac{8.1}{5.4} = \frac{AE}{2.6}$

Cross multiply

$5.4 \times AE = 8.1 \times 2.6\\\\5.4 \times AE = 21.06$

Divide both sides by 5.4

$AE = \frac{21.06}{5.4} = 3.9 $ cm$

b. Find the length of AB:

$AB = AC - BC$

AC = 6.15 cm

To find BC, use AC/BC = EC/DC.

Plug in the values

$\frac{6.15}{BC} = \frac{8.1}{5.4}$

Cross multiply

$BC \times 8.1 = 6.15 \times 5.4\\\\BC = \frac{6.15 \times 5.4}{8.1} \\\\BC = 4.1$

Thus:

$AB = AC - BC$

Substitute

$AB = 6.15 - 4.1\\\\AB = 2.05 $ cm$

Therefore, by applying the knowledge of similar triangles, the lengths of AE and AB are:

a. $\mathbf{AE = 3.9 $ cm}\\\\$

b. $\mathbf{AB = 2.05 $ cm} \\\\$

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Answer:

see explanation

Step-by-step explanation:

To calculate the first 3 terms substitute n = 1, 2, 3 into the n th term rule

$a_{1}$ = (3 × 1) + 2 = 3 + 2 = 5

$a_{2}$ = (3 × 2) + 2 = 6 + 2 = 8

$a_{3}$ = (3 × 3) + 2 = 9 + 2 = 11

------------------------------------------------------------------

substitute n = 10 into the n th term rule

$a_{10}$ = (3 × 10) + 2 = 30 + 2 = 32

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