Correct Answer - C
Soppose `m^(th)` term of first sequence be same as the `n^(th)` term of the second sequences. Then,
`2^(m-1)=1+(n-1)xx3rArr2^(m-1)=3n-2rArr2^(m-2)+1=(3n)/(2)` . . . (i)
We have, `1lem,nle100`
`:." "(3n)/(2)le150`
`rArr" "2^(m-2)+1le150`
`rArr" "2^(m-2)le149rArrm-2le7rArrnle9rArrm=1,2, . . . ,9`.
From (i), we have
`n=(2)/(3)(2^(m-2)+1)`
We observe that `n inN` for m = 1,3,5,7,9.
Hence, there are five common terms in the sequences.
