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Question

 Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
 


Solution

Given a = 5,

Sum of n terms, Sn=n2[2a+(n1)d]

Sum of first n terms, S4=42[2×5+(41)d]

S4=2[10+3d]=20+6d----(1)

Sum of next 4 terms =S8S4

So, S8=82[2×5+(81)d]

S8=4[10+7d]=40+28d----(2)

According to the question,

S4=S8S42

2×S4=S8S4

3×S4=S8

3×(20+6d)=40+28d

60+18d=40+28d

28d18d=6040

10d=20d=2

Common difference = 2 

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