

InterviewSolution
Saved Bookmarks
1. |
The largest term common to the sequences `1, 11 , 21 , 31 , to100`terms and `31 , 36 , 41 , 46 , to100`terms is`381`b. `471`c. `281`d. none of theseA. 281B. 381C. 471D. 521 |
Answer» Correct Answer - D Sequence `1,11,21,31,"…"` has common difference =10 and sequence `31,36,41,46,"…"` has common difference =5. Hence, the sequence with common terms has common dofference LCM of 10 and 5 which is 10. The first common term os 31. Hence, sequence is `31,41,51,61,71,".."" " "....(i)"` Now, `t_(100)` of forst sequence `=1+(100-1)10=991` and `t_(100)` of second sequence `=31+(100-1)5=526` Value of largest common term lt 526 ` :. t_(n) " of Eq. " (i) is 31+(n-1)10=10n+21` `t_(50)=10xx50+21=521` is the value of largest common term. Aliter Let mth term of the jfirst sequence be equal to the nth term of the second sequence, then `1+(m-1)10=31+(n-1)5` `implies 10m-9=5n+26 implies 10m-35=5n` `implies 2m-7=n le 100 implies 2m le 107` `implies m le 53(1)/(2)` `:.` Largest value of m=53 `:.` Value of largest term `=1+(53-1)10=521` |
|