If a1, a2,...a50 are in GP, then (a1-a3+a5-...+a49)/(a2-a4+a6-...+a50) is equal to (1) 0 (2) 1 (3) a1/a2 (4) 2 a25/a24

If a1, a2,…a50 are in GP, then (a1-a3+a5-…+a49)/(a2-a4+a6-…+a50) is equal to (1) 0 (2) 1 (3) a1/a2 (4) 2 a25/a24

Solution:

Let a be the first term and r be the common ratio of the GP.

a1 = a

a2 = ar

a3 = ar2

an = arn-1

(a1-a3+a5-…+a49)/(a2-a4+a6-…+a50) = (a-ar2+ar4+…ar48)/(ar-ar3+ar5+…ar49)

= (a-ar2+ar4+…ar48)/r(a-ar2+ar4+…ar48)

= 1/r

= a/ar

= a1/a2

Hence option (3) is the answer.

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