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oftare22. An arithmetic progression consists of 21 terms, the sumterms in the middle is 129 & the last three 237. Find the AP |
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Answer» Given total number of terms in an AP = 21. Given that the sum of three terms in the middle is 129. = > T10 + T11 + T12 = 129 = > a + 9d + a + 10d + a + 11d = 129 = > 3a + 30d = 129 = > a + 10d = 43 ----- (1) ----------------------------------------------------------------------------------------------------------- Given that sum of last three terms is 237. = > T19 + T20 + T21 = 237 = > a + 18d + a + 19d + a + 20d = 237 = > 3a + 57d = 237 = > a + 19d = 79 ----- (2) ---------------------------------------------------------------------------------------------------------------- On solving (1) & (2), we get = > a + 10d = 43 = > a + 19d = 79 --------------------- -9d = -36 d = 4. substitute d = 4 in (1), we get = > a + 10d = 43 = > a + 10(4) = 43 = > a + 40 = 43 = > a = 43 - 40 = > a = 3. Therefore, The first term is 3 and Common difference = 4. The required AP is 3,7,11.... |
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