1. Use the formula of general term to find the common difference of the new sequence. Then, show that the new sequence is also an arithmetic sequence. then the sequence kT(1)+m, kT(2)+m, kT(3)+m,... can be obtained. 18. Arithmetic and Geometric Sequences then the sequence kT(1), kT(2), kT(3),... can be obtained. Let Q(n) be the general term of the new sequence, ∵ kd is a constant ∴ kT(1)+m, kT(2)+m, kT(3)+m,... is also an arithmetic sequence Remove the square brackets Factorize More about the properties of arithmetic and geometric sequences: (a)If T(1), T(2), T(3),... is an arithmetic sequence, then kT(1)+m, kT(2)+m, kT(3)+m,... is also an arithmetic sequence. = kT(n+1)+m － then Q(n+1) － Q(n) = [kT(n+1)+m] － [kT(n)+m] = k[T(n+1) － T(n)] = kd [kT(n)+m] kT(n) － m and m is added, then d = T(n+1) － T(n). Let d be the common difference of the arithmetic sequence T(1), T(2), T(3),..., Each term of the sequence is multiplied by k,

2. If the common difference of the arithmetic sequence a 1, a 2, a 3,... is d, prove that 3a 1 +5, 3a 2 +5, 3a 3 +5,... is an arithmetic sequence, and hence, find its common difference. The common difference of the arithmetic sequence a 1, a 2, a 3,... is d, then the sequence 3a 1 +5, 3a 2 +5, 3a 3 +5,... can be obtained. Each term is multiplied by 3, and 5 is added, then d = a n+1 － a n. According to the property of arithmetic sequence, 3a 1 +5, 3a 2 +5, 3a 3 +5,... is also an arithmetic sequence. The common difference of the new sequence = 3a n +1 +5 － 3a n － 5 = 3(a n +1 － a n ) = 3d 18. Arithmetic and Geometric Sequences More about the properties of arithmetic and geometric sequences: (a)If T(1), T(2), T(3),... is an arithmetic sequence, then kT(1)+m, kT(2)+m, kT(3)+m,... is also an arithmetic sequence. E.g. = (3a n+1 +5) － (3a n +5)

3. Use the formula of general term to find the common ratio of the new sequence. Then, show that the new sequence is also a geometric sequence. Each term of the sequence is multiplied by k (k  0), then the sequence kT(1), kT(2), kT(3),... can be obtained. Let Q(n) be the general term of the new sequence, ∴ kT(1), kT(2), kT(3),... is also a geometric sequence Reduce the fraction The common ratio of the geometric sequence = r then Q(n) Q(n) Q(n+1) = kT(n) kT(n+1) = kT(n) kT(n+1) = T(n) T(n) T(n+1) 18. Arithmetic and Geometric Sequences More about the properties of arithmetic and geometric sequences: (b) If T(1), T(2), T(3),... is a geometric sequence, then kT(1), kT(2), kT(3),... is also a geometric sequence (where k  0). Let r be the common ratio of the geometric sequence T(1), T(2), T(3),..., then r =. T(n) T(n) T(n+1)

4. (i)Prove that 4, 16, 64, 256, 1 024,... is a geometric sequence. Hence, find the general term T(n) of the sequence. ∴ 4, 16, 64, 256, 1 024,... is a geometric sequence Prove: T(1) T(2) = 4 16 = 4 T(2) T(3) = 16 64 = 4 T(3) T(4) = 64 256 = 4 T(1) T(2) T(3) = T(4) = = 4 ∴ T(n) = 4(4) n － 1 = 4 n The common ratio of the geometric sequence 18. Arithmetic and Geometric Sequences E.g. More about the properties of arithmetic and geometric sequences: (b) If T(1), T(2), T(3),... is a geometric sequence, then kT(1), kT(2), kT(3),... is also a geometric sequence (where k  0).

5. (ii) If each term of the sequence in part (i) is multiplied by 2, then a new sequence is obtained. Find the general term Q(n) of the new sequence. ∴ Q(n) = 8(4) n － 1 = 2(4) n Each term is multiplied by 2, then the sequence 8, 32, 128, 512, 2 048,... can be obtained. According to the property of geometric sequence, Q(1) Q(2) Q(3) = Q(4) = Q(5) = 4 = The common ratio of the geometric sequence The common ratio of the new sequence is 4. 8, 32, 128, 512, 2 048,... is also a geometric sequence. 18. Arithmetic and Geometric Sequences E.g. More about the properties of arithmetic and geometric sequences: (b) If T(1), T(2), T(3),... is a geometric sequence, then kT(1), kT(2), kT(3),... is also a geometric sequence (where k  0).