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Find the 20 terms of ap whose second term is 14 and third |
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Answer» Answer: common DIFFERENCE (d) = 18-14 = 4 the second TERM = a+d = 14 so, a = 14-4 = 10 SUM number of TERMS (n) = 51 a = 10 d = 4 Sum=n/2 [2a + (n-1)d] = 51/2 [20 + (50 x 4)] = 51/2 [20+200] = 51 x 110 = 5610 therefore, the sum of 51 terms of AP is 5610 |
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