Find a1, for the given geometric series. round to the nearest hundredth if necessary. sn= 88,560, r= 2.2, n= 6a. 8,765.73b. 2,477.6c. 945.65d. 14,754.5

* The value of a1 = 945.65 ⇒ answer c

Step-by-step explanation:

* Lets revise the geometric series

- There is a constant ratio between each two consecutive numbers

- Ex:

# 5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

# 5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric Progression:

- U1 = a  ,  U2  = ar  ,  U3  = ar²  ,  U4 = ar³  ,  U5 = ar^4

- Un = ar^n-1, where a is the first term , r is the constant ratio

 between each two consecutive terms, n is the position

 of the term

- The sum of first n terms of a Geometric series is calculate

  from Sn = [a1 (1 - r^n)]/(1 - r) , where a1 is the first term, r is the

  common ratio and n is the number of the terms

* Lets solve the problem

∵ Sn = 88,560

∵ r = 2.2

∵ n = 6

∵ Sn = [a1 (1 - r^n)]/(1 - r)

∴  88,560 = [a1 (1 - 2.2^6)]/(1 - 2.2) ⇒ simplify up and down

∴ 88,560 = [a1 (-112.379904)]/(-1.2) ⇒ simplify the fraction

∴ 88,560 = a1 (93.64992) ⇒ divide both sides by 93.64992

∴ a1 = 945.6494998 ≅ 945.65

* The value of a1 = 945.65

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.