tsmith93 wrote:The first term of sequence I is 2, and each subsequent term is 2 more than the preceding term. The first term of sequence II is 2 and each subsequent term is 2 times the preceding term. What is the ratio of the 32nd term of quwuence II to the 32nd term of sequence I?

Sequence 1-2,4,6,8...........
Sequence 2 -2,4,8,16,........


For sequence 1, 32 th term = 2+31*2 = 64 = 2^6[A.P.]
For sequence 2 ,32 th term = 2^32[G.P.]

Hence,ratio of (32th term sequence II/32th term sequence I) = (2^32)/(2^6)
=2^26

I hope the explanation is clear.

tsmith93 wrote:The first term of sequence I is 2, and each subsequent term is 2 more than the preceding term. The first term of sequence II is 2 and each subsequent term is 2 times the preceding term. What is the ratio of the 32nd term of quwuence II to the 32nd term of sequence I?

I
first term =2
each subsequent term is 2 more than the preceding term so cd = 2
32nd term = 2+(32-1)2 = 64
II

first term =2
each subsequent term is 2 times the preceding term so cr =2
32nd term = 2 (2)^32-1 = (2)^32
ratio 64/ (2)^32 = (2)^6 /(2)^32 =[spoiler](2)^-26 Ans[/spoiler]

tsmith93 wrote:The first term of sequence I is 2, and each subsequent term is 2 more than the preceding term. The first term of sequence II is 2 and each subsequent term is 2 times the preceding term. What is the ratio of the 32nd term of quwuence II to the 32nd term of sequence I?

n th term is first series = a + (n-1)*d = 2+ (n-1)*2 [ It is in Arithmetic progression, a - first term, d common difference]
= 2*n => 32nd term = 64 = 2^6

n th term in the sequence II = a*r^n-1 = 2*2^n-1 [It is in GP, a first term, r common ratio]

= 2^n => 32nd term = 2^32


the ratio = 2^32/2^6 = 2^26