In an A.P 41st term is 50, then the sum of 81 terms of that A.P is:
Here 41st terms is 50 So, a 41 = 50 ⇒ [a +(41 – 1)d] = 50 ⇒ a + 40d = 50 …………….(i) S81 = ? ⇒ Sn = (n/2) [2 x a + (n – 1)d] = 0 ⇒ S81 = (81/2)[2a + (81 – 1)d] ⇒ 81[a + 40d]……..eq(ii) Now putting the value from eq(i) in eq(ii) then ⇒ S81 = 81[50] = 4050
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