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| 1. |
The 6th and 17th terms of an A.P. are 19 & 41,find the 40th term? |
| Answer» We have nth\xa0term of an A.P is {tex}{ T }_{ n }=a+\\left( n-1 \\right) d{/tex}Now\xa0{tex}{ T }_{ 6 }=19\\Rightarrow a+5d=19.............(i)\\\\ { T }_{ 17 }=41\\Rightarrow a+16d=41..........(ii){/tex}Subtracting equation(i) from(ii), we get{tex}11d=22\\Rightarrow d=\\frac { 22 }{ 11 } =2{/tex}Now from equation (i) we get\xa0{tex}a+10=19\\Rightarrow a=9{/tex}Hence we get\xa0{tex}{ T }_{ 40 }=a+39d=9+39\\left( 2 \\right) =9+78=87{/tex}So the 40th\xa0term is 87.\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0 | |