Consider two A.P.s such that A.P.₁ has 1st term 1, common difference 5 and total terms 101 and A.P.₂ has 1st term 9, common difference 7 and total terms 71 then the number of common terms which appear in both these A.P.s is

To find the number of common terms between the two given Arithmetic Progressions (A.P.s), we can analyze each sequence step-by-step.

Let the first arithmetic progression be A.P.₁:
First term (a₁) = 1
Common difference (d₁) = 5
Number of terms (n₁) = 101

The general term of A.P.₁ is given by:
$t_r=1+(r-1)·5=5r-4$
where 1 ≤ r ≤ 101.

The last term of A.P.₁ is:
$L_1=1+(101-1)·5=1+500=501$

Let the second arithmetic progression be A.P.₂:
First term (a₂) = 9
Common difference (d₂) = 7
Number of terms (n₂) = 71

The general term of A.P.₂ is given by:
$T_s=9+(s-1)·7=7s+2$
where 1 ≤ s ≤ 71.

The last term of A.P.₂ is:
$L_2=9+(71-1)·7=9+490=499$

Any common term must be less than or equal to the smaller of the two last terms. Thus, if X is a common term:
$X\le \min (501,499)=499$

Let us find the first common term by equating the general terms or listing the elements:
A.P.₁ terms: 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, ...
A.P.₂ terms: 9, 16, 23, 30, 37, 44, 51, ...

By inspection, the first common term is 16.

The common terms themselves form a new arithmetic progression, A.P._c:
First term (a_c) = 16
The common difference (d_c) of the common terms is the Least Common Multiple (LCM) of the common differences of the two individual progressions:
$d_c=\mathrm{LCM}(d_1,d_2)=\mathrm{LCM}(5,7)=35$

Let there be N terms in this common progression. The N-th term is:
$T_N=16+(N-1)·35$

Since the common terms cannot exceed 499, we set up the inequality:
$16+(N-1)·35\le 499$
Subtracting 16 from both sides:
$(N-1)·35\le 483$
Dividing by 35:
$N-1\le \frac{483}{35}$
$N-1\le 13.8$
$N\le 14.8$

Since N must be an integer, the maximum value for N is 14. Therefore, there are 14 common terms in both arithmetic progressions.