Question

Consider two AP's $S_1$ & $S_2$ such that $S_1 : \{ \text{First term} = 1, \text{Common difference} = 5, \text{no. of terms} = 101 \}$
$S_2 : \{ \text{First term} = 9, \text{Common difference} = 7, \text{no. of terms} = 71 \}$
Find the number of common terms in both AP's.

Answer 1

Step 1: Write the general terms of both APs.
$A_n = 1 + (n-1)5 = 5n - 4$. Last term $A_{101} = 5(101) - 4 = 501$.
$B_m = 9 + (m-1)7 = 7m + 2$. Last term $B_{71} = 7(71) + 2 = 499$.


Step 2: Find the first common term.
Equate $5n - 4 = 7m + 2 \implies 5n = 7m + 6$.
For $m=1, 5n = 13$ (No)
For $m=2, 5n = 20 \implies n=4$.
So the first common term is $A_4 = 5(4) - 4 = 16$.


Step 3: Find the common difference of common terms.
The common terms themselves form an AP with common difference $D = \text{LCM}(d_1, d_2) = \text{LCM}(5, 7) = 35$.


Step 4: Find the number of terms.
Let the common AP be $C_k = 16 + (k-1)35$.
The last common term must be $\le$ the smaller of the two last terms, which is 499.
$16 + (k-1)35 \le 499$
$(k-1)35 \le 483$
$k-1 \le 13.8$
$k \le 14.8$.
Since $k$ is an integer, the number of common terms is 14.