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Is -150 a term of AP 17,12,7,2 |
| Answer» \xa0The given term of arithmatic progression is17, 12, 7, 2, ......Here, a = 17, d = 12 - 17 = - 5 where a is first term and d is common differenceSuppose an = -150a + (n - 1)d = -150 {tex}\\Rightarrow{/tex}\xa017 + (n - 1)(-5) = -150{tex}\\Rightarrow{/tex}\xa0(n - 1)(-5) = -150 - 17 {tex}\\Rightarrow{/tex}\xa0(n - 1)(-5) = -167{tex}\\Rightarrow{/tex}\xa0n - 1 =\xa0{tex}\\frac { 167 } { 5 }{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0n =\xa0{tex}\\frac { 167 } { 5 }{/tex} + 1{tex}\\Rightarrow{/tex}\xa0n =\xa0{tex}\\frac { 167 + 5 } { 5 } = \\frac { 172 } { 5 } = 34 \\frac { 2 } { 5 }{/tex}wheren is not a whole number.{tex}\\therefore{/tex}\xa0-150 is not a term of the A.P. | |