Geometric Sequence part 1

2012-08-31 7:20 pm

1(a) express the sum of the first n terms of the geometric sequence 1, 1/5 ,
1/25, 1/125,... in terms of n

1(b) express the sum of the first n terms of the geometric sequence 1, 1/4,1/16,1/64,... in terms of n

1(c) hence find 1+ 1/4 -1/5 +1/16 -1/25 +1/64 -1/125 +...+ 2n term

2.three numbers form an arithmetic sequence. the sum of the 1st term and the 3rd term is 22. when the 1st, 2nd and 3rd terms increased by 3, 31 and 275 respectively, The resulting numbers form a geometric sequence
.
2a) find the 2nd term of the arithmetic sequence

2b)find the 1st and 3rd terms of the arithmetic sequence

3 it is given that a,-8,b form a geometric sequence and -8,b,a form an arithmetic sequence. where a and b are positive numbers

3a) find the value of a and b
3b) find the sum to infinity of the geometric sequence a , -8, b,....
3c) find the sum to infinity of all the negative terms of the geometric sequence a,-8,b...

回答 (1)

myisland8132

2012-09-01 6:14 pm

✔ 最佳答案

1(a) [1 - (1/5)^n]/[1 - (1/5)] = (5/4)[1 - (1/5)^n](b) [1 - (1/4)^n]/[1 - (1/4)] = (4/3)[1 - (1/4)^n](c) 1+ 1/4 -1/5 +1/16 -1/25 +1/64 -1/125 +...+ 2n term= 1 + (1 + 1/4 + 1/16 + 1/64 + ... + n term)- (1 + 1/5 + 1/25 + 1/125 + ... + n term)= 1 + (5/4)[1 - (1/5)^n] - (4/3)[1 - (1/4)^n]= 11/12 + (4/3)(1/4)^n - (5/4)(1/5)^n2(a) a + a + 2d = 22...(1)(a + 3)(a + 2d + 275) = (a + d + 31)^2...(2)From (1) d = 11 - aSo, (a + 3)(297 - a) = 1764a^2 - 294a + 873 = 0a = 3, d = 8second term is 11(b) First term is 3 and third term is 193(a) ab = 64 and a - 8 = 2bSom b(2b + 8) = 64 => b = 4, a = 16(b) S = 16 - 8 + 4 - 2= 16/(1 + 1/2)= 32/3(c) S = (-8) + (-2) + (-1/2) + ...= -8/(1 + 1/4)= -32/5

👍 0 👎 0