Question

We will first chech the last twems of att 3 series .
[
begin{array}{ll}
2,5,8, & l_{1}=2+(150-1) 3=449
3,7,11, & 1,=3+(100-1) 4=399
end{array}
]
( 7,9,11, ldots, ) to 180 terms ( quad l_{3}=7+(180-1) 2=365 )
So the largest naor common term wicll be less than of equal to 365 . Firstly, you can start to check with 565 . as it is greatest among the eat options remaining (except 399 as it is greater than 365)
[
365=3+(n-1) 4 quad Rightarrow quad n eq 80.50 .365 text { is } X
]
Now check for 359 . ( 359=3+(n-1) 4 Rightarrow n=90 )
[
text { ex } 359=2+(n-1) 3 Rightarrow n=120
]
Qfourse for 3 rd series 359 satisfies as it's odd. oo Largest Common Jerm = 3 ( 59(B) rightarrow B i n g 0 ) (i

# The largest term common to the sequence 2, 5, 8, 11, .... to 150 terms and 3, 7, 11, 15, ..... to 100 terms and 7,9, 11, ... to 180 terms is (A) 349 (B) 359 (C) 399 (D) 365 TI e ftbefint 15 tarcia

Solution