Question

Let $r$ be the number of identical terms in the two A.P's. Form the sequence of identical terms, it will be an A.P, then the $r^{th}$ term of this A.P make $t_r\le$ the smaller of the last term of the two A.P's. The number of terms common to two A.P's $3,7,11,....,407$ and $2,9,16,...,709,$ is

• A. $14$
• B. $21$
• C. $28$
• D. $35$

Answer 1

Answer: (A)

$14$

Sequence $3,7,11,...407$ has common difference$=4$ and sequence $2,9,16,..709$ has common difference$=7$
Hence, the sequence with common terms has common difference LCM of $4$ and $7$ which is $28$.
The first common term is $23$.Hence, the sequence is $23,51,79,..387$ which has $14$ terms