A.P question solve.........

Common difference of first AP = 2

No of terms = 61

Common difference of second AP = 6

First common term = 21

AP formed by common terms,

First term= 21
Common Difference = 6

Now, since First AP ends at 135.

Note, in the first AP common terms appear at positions 4, 7, 10 and so on.
So, the 61st term will also be in the required AP.

So, the last term of the common AP is 135

135 = 21 + (n-1)6

Therefore, n=20 .

is the answer 135???

if we need the common difference of the common tems of the a.p.'s...
we just need the the L.C.M. of the common difference of the 2 a.p.'s..

Guys the answeris 135, Joydoot is absolutely right but Ashish is absolutely wrong...

How can the last term be 20...

soln : --

The first A.P is 15,17,19,............,135
so we get,

a = 15, d = 2 and n = 61

Again,

The second A.P is 21,27,33,...........,561
so we get,

a = 21, d = 6, n = 91

Now the first common term by trial is
21 so the common difference will be the L.C.M of the common difference
of both the A.Ps and the L.C.M is 6

and so,...
the common A.P is...

21,27,33,............

Now, 21 + (n - 1) x 6 < 135
=> n < 20
thus, n = 20

hence the last term is 21 + (19).(6) = 21 + 114 = 135 Ans......

or, the last term is 135...

Smart ...... [you edited that].......