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 second, sixth and seventh terms of a linear sequence. Find the common ratio of the sequence.

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