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# Ques1) 1 -2+3 - 4 + 5 - 6 .................... - 2n is equals to(a) n       (b) n(n+1)/2     (c)  -n(n+1)/2              (d)    noneQues2) Let a1,a2,a3,..... be in A.P and ap, aq, ar  be in G.P.  Then show that ap / aq = (q-p) / (r - q).Ques3) If an >1 for all n belonging to N,log a2 a1 + log a3 a2 + ..........+ log an an-1 + log a1 an has the min value:-(a) 1    (b) 2        (c)  0   (d) none

24 Points
12 years ago

Hi Tushar

Please post different questions in a different post, i.e., one question per post.

A1}

1 -2+3 - 4 + 5 - 6 .................... - 2n

=

[1 +2+3 + 4 + 5 + 6 .................... + 2n] - 2[2+ 4  + 6 .................... + 2n]

=

2n(2n+1)/2  -  4(1 +2+3 + 4 + 5 + 6 .................... +n)

=

n(2n+1)  -  4n(n+1)/2

=

n[2n+1 - (2n+2)]

=

n[-1]

=

-n

Method 2:

1 -2+3 - 4 + 5 - 6 .................... - 2n

=

(1 -2)+(3 - 4)+(5 - 6)+.............+[(2n-1)- 2n].................................As you can see, these are n pairs of consecutive terms

=

(-1)+(-1)+(-1)+.........n Times

=

-n

Regards

Rajat