Answer:
1
Explanation:
The nth term of an exponential sequence is expressed as ar^n-1
The nth term of a linear sequence is expressed as Tn = a + (n-1)d
a is the first term
r is the common ratio
d is the common difference
n is the number of terms
Let the three consecutive terms of an exponential sequence be a/r, a and ar
second term of a linear sequence = a +d
third term of a linear sequence = a + 2d
sixth term of a linear sequence = a + 5d
Now if the three consecutive terms of an exponential sequence are the second third and sixth terms of a linear sequence, this is expressed as;
a/r = a + d ..... 1
a = a + 2d ..... 2
ar = a+ 5d .... 3
From 2: a = a + 2d
a-a= 2d
0 = 2d
d = 0/2
d = 0
Substitute d = 0 into equation 1:
From 1: a/r = a + d
a/r = a+0
a/r = a
Cross multiply
a = ar
a/a = r
1 = r
Rearrange
r = 1
Hence the common ratio of the exponential sequence is 1