Ques1) 1 -2+3 - 4 + 5 - 6 .................... - 2n is equals to

(a) n (b) n(n+1)/2 (c) -n(n+1)/2 (d) none

Ques2) Let a₁,a₂,a₃,..... be in A.P and a_p, a_q, a_r be in G.P. Then show that ap / aq = (q-p) / (r - q).

Ques3) If a_n >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

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