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Natasha2012 [34]
2 years ago
15

1.) Find the nth term of the sequence 1, 3, 7, 13, 21, 31… 2.) The three consecutive terms of an exponential sequence are the se

cond,sixth and seventh terms of a linear sequence. Find the common ratio of the sequence.
Mathematics
1 answer:
trasher [3.6K]2 years ago
3 0

Answer:

1) xₙ = n² - (n - 1)

2) 1/4

Step-by-step explanation:

1)    1 = 1² - (1-1)

3 = 2² - (2-1)

7 = 3² - (3-1)

13 = 4² - (4-1)

...........

xₙ = n² - (n - 1)

2) three consecutive terms of an exponential sequence: x, rx , r²x

x: 2nd term of linear sequence

rx: 6th term of linear sequence      rx = x + 4d

r²x: 7th term of linear sequence     r²x = rx + d     d = r²x - rx

rx = x + 4 * (r²x - rx) = x + 4r²x - 4rx

4xr² -5rx + x = 0

x(4r² -5r + 1) = 0

x (4r - 1) ( r - 1) = 0

x = 0 or r = 1/4 or r = 1

if either x = 0 or r = 1      d will be equals to "0" everything became 0

So the only reasonable answer is r = 1/4

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6x+4.5y=3037.50\\6(600-y)+4.5y=3037.50\\3600-6y+4.5y=3037.50\\-1.5y=3037.50-3600\\-1.5y=-562.5\\y=\frac{-562.5}{-1.5}\\y=375

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