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. 381
B. 471
C. 281
D. none of these

Answer

D

Explanation

Let `m^(th)` term of first A.P. be same as the `n^(th)` term of the second A.P. then,
`1+(m-1)xx10=31+(n-1)xx5`
`rArr" "2m=n+7`
`rArr" "(n+7)/(2)=(m)/(1)=lamda("say")`
`rArr" "n=2lamda-7andm=lamda`
`rArr" "1le2lamda-7le100and1lelamdale100" "[because1lem,nle100]`
`rArr" "4lelamdale53(1)/(2)and1lelamdale100`
`rArr" "lamda=4,5,6,7, . . . .,53`
`rArr" "n=1,3,5, . . . . .,99andm=4,5,6, . . .,53`
Hence, there are 50 terms common in the two A.P's the largest term is 53rd term of first A.P. on 99th term of second A.P. and its value is
`31+98xx5=521`