Ineed on these other 2 sequences.

bear in mind that one can always get the common ratio by simply dividing any of the terms by the one before it namely, say 18÷12 or 27÷18.

591.09375

1/4

Step-by-step explanation:

"Find the sum of the first 8 terms of the geometric sequence that begins: 12, 18, 27,..."

The sum of the first n terms of a geometric sequence is:

S = a (1 − r^n) / (1 − r)

where a is the first term and r is the common ratio.

Here, the first term is 12, so a = 12.  The common ratio is 18/12 = 1.5.  So the sum of the first 8 terms is:

S = 12 (1 − 1.5^8) / (1 − 1.5)

S = 591.09375

"Find the sum of the infinite geometric sequence that begins 1/12, 1/18, 1/27,..."

The sum of an infinite geometric sequence is:

S = a / (1 − r)

Here, the first term is 1/12, so a = 1/12.  The common ratio is (1/18) / (1/12) = 2/3.  So the infinite sum is:

S = 1/12 / (1 − 2/3)

S = 1/4

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.