    Question

# Which of the following series forms an AP? (i)a–2d,a–d,a,a+d,a+2d (ii)a,a+d,a+2d,a+3d,a+4d (iii)a–3d,a–d,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)a–2d,a–d,a,a+d,a+2d Difference between first two consecutive terms =a–d–(a–2d)=d Difference between third and second consecutive terms =a–(a–d)=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.  Suggest Corrections  1      Similar questions  Related Videos   Introduction to AP
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