Find the sum of all the 11 terms of an Ap whose middle most term is 30.
given n =11middle term = (a1 + a11)/2 = 30a1 + a11 = 60 ………. iSn = n/2(a1 + a11) ……… ii = 11/2 (60) … from i = 11 x 30 = 330
Find the sum of all the 11 terms of an AP whose middle most term is 30
Find the sum of all the 11 terms of an AP whose middle most term is 30
Let a be the first term and d be the common difference of the given A.P.Clearly, in\xa0an A.P. consisting of 11 terms,\xa0{tex} \\left( \\frac { 11 + 1 } { 2 } \\right) ^ { t h }{/tex}\xa0i.e. 6th term is the middle term.{tex}\\text{ it is given that the middle term =30}{/tex} So a+5d=30 …..(1){tex}S_{11}=\\frac{11}{2}(2a+10d){/tex}\xa0= 11(a+5d) But a+5d=30 from (1)Hence S\u200b\u200b\u200b\u200b\u200b\u200b11\xa0= 11 × 30= 330