1 (a) which term of the sequence 8-6i, 7-4i, 6-2i, .... Is purely imaginary? (b)

Question

1 (a) which term of the sequence 8-6i, 7-4i, 6-2i, .... Is purely imaginary? (b) find the common difference of an A.P whose first term is 100 and the sum whose first six terms is five times the sum of the next six terms. (c) if the first term of an A.P is 2 and the sum of the first five terms is equal to one-fourth of the sum of the next five terms, find the sum of first 30 terms. (d) if S1, S2, S3 be the sum of n, 2n, 3n terms respectively of an A.P, prove that S3= 3(S2-S1). (e) if logx, log(2^x-1) and log(2^x+3) are in A.P, then find the value of x. (f) the sum of three numbers in A.P is -3 and their products is 8. Find the numbers.

Explanation / Answer

a) As you can see in the sequence 8-6i, 7-4i, 6-2i, ...., the realpart of the terms are in a sequence 8, 7, 6, ...     The imaginary parts, are in a sequence, -6i,-4i, -2i.     For the term to be purely imaginary, the realpart must be zero.     ---     8, 7, 6, 5, 4, 3, 2, 1, 0     In this sequence, 0 is the 9th term.     So, we find the 9th term in the imaginary partsequence.     -6i, -4i, -2i, 0, 2i, 4i, 6i, 8i, 10i     Since 10i is the 9th term, the pure imaginaryterm in the original sequence is 0+10i=10i b) Let d=common difference     In an arithmetic progression, given n terms eachwith common difference d with the first term a, the sum,     S=(n/2)(2a+(n-1)d)     Another, if the first term of an AP with commondifference d is a, then the kth term is a+(k-1)d     So, since a=100, then the 7th term is100+(7-1)d=100+6d     According to the problem,     sum of 1st to 6th terms=5(sum of the 7th to 12thterms)    (6/2)(2(100)+(6-1)d)=5(6/2)(2(100+6d)+(6-1)d)     Since 6/2 is a commvon factor, we can removeit.     2(100)+(6-1)d=5(2(100+6d)+(6-1)d)     200+5d=5((200+12d)+5d)     200+5d=5(200+17d)     200+5d=1000+85d     -80d=800     d=-10 c) sum of 1st to 5th terms=1/4(sum of 6th to ten terms)     The first term of the left hand side of theequation is 2.     Let d=common difference     Then, the first term of the right hand sequence(6th term) is 2+(6-1)d=2+5d     So,    (5/2)(2(2)+(5-1)d)=(1/4)(5/2)(2(2+5d)+(5-1)d)     Since 5/2 is a common factor, we can removeit.     2(2)+(5-1)d=(1/4)(2(2+5d)+(5-1)d)     4+4d=(1/4)((4+10d)+4d)     4+4d=(1/4)(4+14d)     4+4d=(1/2)(2+7d)     8+8d=2+7d    d=-6    So,    sum of 1st to 30th terms    =(30/2)(2(2)+(30-1)(-6))    =15(4+29(-6))    =15(4-174)    =15(-170)    =-2550 c) I'm confused with the given. I do not know if S1 is the sum ofthe first n terms or what. d) If logx, log(2x-1) and log(2x+3) are in APthen    log(2x+3)-log(2x-1)=log(2x-1)-logx     So,    log((2x+3)/(2x-1))=log((2x-1)/x)    (2x+3)/(2x-1)=(2x-1)/x    x(2x+3)=(2x-1)2 e) Let d=common difference.           x=2ndterm     Such that the sequence is of the form, x-d, x,x+d.     According to the problem:     (x-d)+x+(x+d)=-3     3x=-3     x=-1     So, the sequence is now of the form -1-d, -1,-1+d     According to the problem:     (-1-d)(-1)(-1+d)=8     d2-1=8    d2=9    d=±3    So, if d=3, the sequence is -1-3, -1, -1+3 or -4, -1,2          if d=-3, thesequence is -1+3, -1, -1-3 or 2, -1, -4

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