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Question

Which of the following series forms an AP?

(i)a2d,ad,a,a+d,a+2d

(ii)a,a+d,a+2d,a+3d,a+4d

(iii)a3d,ad,a+d,a+3d


A

(i) and (iii)

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B

(i) and (ii)

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C

(ii) and (iii)

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D

(i),(ii) and (iii)

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Solution

The correct option is D

(i),(ii) and (iii)


Consider each series (i)a2d,ad,a,a+d,a+2d

Difference between first two consecutive terms =ad(a2d)=d

Difference between third and second consecutive terms =a(ad)=d

Difference between fourth and third consecutive terms =a+2d(a+d)=d

Since a2 a1=a3 a2=a4 a3
So, the sequence is in AP
Similarly, (ii) has common diffrence of d and (iii) has commonn difference of 2d and all are in A.P.


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