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In the sequence 1, 2, 2, 3, 3, 3, 4, 4,4,4,....., where n consecutive terms have the value n, the 150 term is |
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Answer» Let the 150th term = n Then, `1+2+3+"...."+(n-1)lt150lt1+2+3+"....."+n` `implies ((n-1)n)/(2)lt150lt(n(n+1))/(2)` `impliesn(n-1)lt300 lt n(n+1)` Taking first two members `impliesn(n-1)lt300 implies n^(2)-n- 300 lt0` `implies (n-(1)/(2))lt300+(1)/(4)` `implies 0 lt n lt (1)/(2)+(sqrt(1201))/(2)` `implies 0 lt n lt 17.8 " " "......"(i)` and taking last two members, `n(n=1)gt300` `implies (n+(1)/(2))^(2)gt300+(1)/(4)` `therefore " " ngt -(1)/(2)+(sqrt1201)/(2)` `implies " "ngt 16.8" " "....."(ii)` From Eqs. (i)and (ii), we get 16.8ltnlt17.8 `implies " "n=17 ` |
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