If the arithmetic mean of the first 20 terms of an arithmetic progression is 25 and the arithmetic mean of its first 40 terms is 55 , then the sum of the 21^ st to the 60^ th terms (inclusive) of this progression is equal to :

Step-by-step solution

Let the first term be a and the common difference be d . The sum of the first n terms is S_n = n 2 [2a + (n-1)d] . The arithmetic mean of the first n terms is S_n n = a + n-1 2 d . For n = 20 , the mean is a + 9.5d = 25 . For n = 40 , the mean is a + 19.5d = 55 . Subtracting the first equation from the second gives 10d = 30 d = 3 . Substituting d = 3 into the first equation yields a + 28.5 = 25 a = -3.5 . The sum of the 21^ st to the 60^ th terms is S₆₀ - S₂₀ . S₆₀ = 60 2 [2(-3.5) + 59(3)] = 30[-7 + 177] = 30(170)