Carl is buying supplies for a party bags of chips cost three dollars per bag and sodas cost two dollars per can he has $24 to sp
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Answer:
700 lei
Step-by-step explanation:
You have a geometric sequence for which you know the common ratio and the last term.
There are only three terms, so you can solve the problem in either of two ways.
1. The brute force method (easiest for only a few terms)
Each term is half the one before it, so each term is double the one after it.
3rd term = 100 lei
2nd term = 200
1st term = 400
Total = 700 lei
Oana spent 700 lei
2. Using formulas (best for longer sequences)
The general formula for your sequence is
aₙ = a₁rⁿ⁻¹
For your sequence,
a₃ = 100; r = 0.5
(a) Calculate a₁
Set the last term equal to the general formula.
a₃ = a₁(0.5)ⁿ⁻¹
100 = a₁(0.5)² = 0.25a₁
a₁ = 100/0.25 = 400
(b) Calculate the sum
The general formula for the sum of a geometric sequence is
Oana spent 700 lei.
First off, let's convert the decimal to a fraction, notice, we have two decimals, so we'll use in the denominator, a 1 with two zeros then, two decimals, two zeros, thus 
now, we know then the ratio dimensions for the new photograph,
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{7}{4}\implies \cfrac{4+3}{4}\implies \cfrac{4}{4}+\cfrac{3}{4}\implies 1+\boxed{\cfrac{3}{4}}\impliedby \textit{perimeter is }\frac{3}{4}\textit{ larger} \\\\\\ \stackrel{areas'~ratio}{\cfrac{s^2}{s^2}}\implies \cfrac{3^2}{4^2}\implies \cfrac{9}{16}\impliedby \textit{area is }\frac{9}{16}\textit{ larger than original}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B7%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%2B3%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B4%7D%2B%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%201%2B%5Cboxed%7B%5Ccfrac%7B3%7D%7B4%7D%7D%5Cimpliedby%20%5Ctextit%7Bperimeter%20is%20%7D%5Cfrac%7B3%7D%7B4%7D%5Ctextit%7B%20larger%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bareas%27~ratio%7D%7B%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5E2%7D%7B4%5E2%7D%5Cimplies%20%5Ccfrac%7B9%7D%7B16%7D%5Cimpliedby%20%5Ctextit%7Barea%20is%20%7D%5Cfrac%7B9%7D%7B16%7D%5Ctextit%7B%20larger%20than%20original%7D)
now, we know then the ratio dimensions for the new photograph,