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Find the sum of first 25 terms of an AP whose nth term is given by Tn equals 7 -3n?
The general expression for the nth term of an AP is : a + (n - 1)d, where a is the first term and d is the common difference.
Putting n = 1, then T1 = 7 - 3 = 4, which is the value of a.
Then 7 - 3n = 4 + (n - 1)d : 3 - 3n = (n - 1)d : -3(n - 1) = (n - 1) d : d = -3.
The formula for the sum of the AP to the nth term is : Sn = n/2[2a + (n - 1)d]
Therefore S25 = 25/2[2x4 + (25-1)x-3] = 25/2[8+(24 x -3)] = 25/2[8-72] = 25/2 x - 64 = -800
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What are the first ten terms of a sequence whose first term is -10 and whose common difference is -2.?
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Find the sum of first 25 terms of an AP whose nth term is given by Tn equals 2 -3n?
The nth term of the arithmetic progression (AP) is given by ( T_n = 2 - 3n ). To find the first 25 terms, we need to determine the first term ( T_1 ) and the common difference ( d ). The first term is ( T_1 = 2 - 3(1) = -1 ), and the common difference is ( d = T_2 - T_1 = (2 - 3(2)) - (-1) = -3 - (-1) = -2 ). To find the sum of the first 25 terms, we use the formula ( S_n = \frac{n}{2} (2a + (n-1)d) ), where ( n = 25 ), ( a = -1 ), and ( d = -3 ). Thus, [ S_{25} = \frac{25}{2} (2(-1) + (25-1)(-3)) = \frac{25}{2} (-2 - 72) = \frac{25}{2} \times -74 = -925. ] Therefore, the sum of the first 25 terms is (-925).
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