急!F6 Arithmetic Sequences 1q12

19.
(a)
Let T1 = a
Then, T1, T2, T3, ..., Tn
= a, a+(1/2), a+2(1/2), ..., a+(n-1)(1/2)

3T1­, 3T2, 3T3, ... 3Tn
= 3a, 3[a+(1/2)], 3[a+2(1/2)], ..., 3[a+(n-1)(1/2)]
= 3a, 3a+(3/2), 3a+2(3/2), ..., 3a+(n-1)(3/2)

Hence, it is an arithmetic sequence.
The common difference = 3/2

(b)
3T1+2­, 3T2+2, 3T3+2, ... 3Tn+2
= 3a+2, 3[a+(1/2)]+2, 3[a+2(1/2)]+2, ..., 3[a+(n-1)(1/2)]+2
= (3a+2), (3a+2)+(3/2), (3a+2)+2(3/2), ..., (3a+2)+(n-1)(3/2)

Hence, it is an arithmetic sequence.
The common difference = 3/2

(c)
3-T1, 3-T2, 3-T3, ..., 3-Tn
= 3-a, 3-[a+(1/2)], 3-[a+2(1/2)], ..., 3-[a+(n-1)(1/2)]
= (3-a), (3-a)+(-1/2), (3-a)+2(-1/2), ..., (3-a)+(n-1)(-1/2)

Hence, it is an arithmetic sequence.
The common difference = -1/2

(d)
Tn, Tn-1, ..., T2, T1
= a+(n-1)(1/2), a+(n-2)(1/2), ..., a+(1/2), a
= [a+(n-1)(1/2)], [a+(n-1)(1/2)]+(-1/2), ....., [a+(n-1)(1/2)]+(n-2)(-1/2), [a+(n-1)(1/2)]+(n-1)(-1/2)

Hence, it is an arithmetic sequence.
The common difference = -1/2


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20.
The first term, T1
= S1
= 3(1)² + 2(1)
= 5

The 6th term,T6
= S6 - S5
= [3(6)² + 2(6)] - [3(5)² + 2(5)]
= 120 - 85
= 35