Question

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Answer 1

a1a2

a25a24

Answer :

c

Solution :

Let a2=ra1,a3=r2a1,…, so on

∴a1−a3+a5−....+a49a2−a4+a6−...+a50=a1−r2a1+r4a1−...r48a1a1r−r3a1+r5a1−...+r49a1  

a1(1−r2+r4−...+r48)a1r(1−r2+r4−...+r48)1r=a1a2

Answer 2

Answer:

try this once

Step-by-step explanation:

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