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| 1. |
For what value of k will The consecutive terms 2k+1,3k+3 and 5 k - 1 form in AP |
| Answer» If {tex}2k + 1, 3k + 3, 5k -1{/tex} are in A.P.then {tex} (5k - 1) - (3k+ 3)= (3k + 3) - (2k + 1){/tex}or, {tex}5k- 1 - 3k- 3 = 3k + 3 - 2k -1{/tex}or, {tex}2k - 4 = k + 2{/tex}or, {tex}2k - k = 4 + 2{/tex}or {tex}k = 6{/tex} | |