13若將等差數列a1 https://www.flickr.com/photos/101292575@N06/21846684444/in/dateposted-public/?

設原數列為：a, a+d, a+2d, ..., a+29d
則新數列是：2a-5, 2a+2d-5, 2a+4d-5, ..., 2a+58d-5
原數列的項差是 d，總和是 (30/2)(a + a+29d)，即 15(2a + 29d)
新數列的項差是 2d，總和是
(30/2)(2a-5 + 2a+58d-5)
= 15(4a + 58d - 10)
= 30(2a + 29d) - 150

所以答案是 (B)：新數列的總和比原數列總和的2倍少150。

11. An = a1 + (n-1)d

a100 = a1 + 99d

a99 = a1 + 98d

a98 = a1 + 97d

a2 = a1 + d

a3 = a1 + 2d

a100 + a99 + a98 - a1 - a2 - a3

= 3a1 + (99 + 98 + 97 - 1 - 2)d - 3a1

= (97*3 + 3 - 3)d

= 97*3*d

= 873

d = 873/97*3

= 873/291

= 3


a84 - a24

= 83d - 23d

= 60d

= 180

= (D)



12. An = a1 + (n-1)d

=> an = 1 + (n-1)*3; bn = 2 + (n-1)*4

=> an = bn

=> 3*(n-1) = 1 + 4*(n-1)

=> 3*n - 3 = 4*n - 3

=> 3n = 4n

=> d = [3,4] = 12


c10 = 10 + 9d

= 10 + 9*12

= 118

= (D)




13. Sn = Σan = n*a1 + n(n-1)d/2

Tn = Σ(2*an - 5)

= 2Σan - 5Σ1

= 2*Sn - 5*n


T30 = 2*S30 - 5*30

= 2*S30 - 150

= (B)