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ATQ, Total sum of 8 consecutive odd numbers = 20 X 8 = 160 According to Arithmetic progression: Sum = n/2[2a+(n-1)×d] [where n = number of terms, 'a' = irst term and 'd' = common dierence] Or, 160 = 8/2[2a+(8-1)×2] Or, 40 = 2a + 7 × 2 Or, 26 = 2a So, 'a' = 13 Last number = a + (n - 1) × d = 13 + (8 - 1) × 2 = 27 Previous odd number to these numbers = 13 - 2 = 11 And, next odd number to these numbers = 27 + 2 = 29 So, required average = (160+11+29)/10 = 200 ÷ 10 = 20 Alternate Solution Since, the number that are included is equally more and less than sum of the given 8 numbers, therefore, there eect will cancel out and the average will remain sum.
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