The largest term common to the sequence 1,11,21,31,….to 100 terms and 31,36,41,46,…… to 100 tetms is

Options

A. 281
B. 381
C. 471
D. 521

Answer

D

Explanation

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`