How many terms of the series 54,51,48...be taken so that their sum is513?explain the double answer

How many terms of the series 54,51,48...be taken so that their sum is5

1.	How many terms of the series 54,51,48...be taken so that their sum is513?explain the double answer
Answer» Given series is 54, 51, 48,...Here, a = 54, d = 51 - 54 = - 3Let number of terms be n, then their sumSn\xa0= 513{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { n } { 2 }{/tex}\xa0[2a + (n - 1) d] = 513{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { n } { 2 }{/tex}\xa0[2\xa0{tex}\\times{/tex}\xa054 + (n - 1) (- 3)] = 513{tex}\\Rightarrow{/tex}\xa0n (108 - 3n + 3) = 513\xa0{tex}\\times{/tex}\xa02{tex}\\Rightarrow{/tex}\xa0- 3n2\xa0+ 111n = 1026{tex}\\Rightarrow{/tex}\xa0- (3n2\xa0- 111n + 1026) = 0{tex}\\Rightarrow{/tex}\xa0n2\xa0- 37n + 342 = 0{tex}\\Rightarrow{/tex}\xa0(n - 18) (n - 18) = 0\xa0{tex}\\Rightarrow{/tex}\xa0n = 18 or 19Here, the common difference is negative and 19th\xa0term= 54 + (19 - 1)\xa0{tex}\\times{/tex}\xa0(- 3) = 0So, the sum of 18 terms as well as that of 19 terms is 513.