Arithmetic Progression problem 1 by Dhaval Furia

Let the $n$ ' $^{th}$ term of the first sequence be equal to the $p$ ' $^{th}$ term of the second. Then

$4n+11=5p+9\implies p=\dfrac{4n+2}{5}$ . Hence the terms $t_2,t_7,t_{12},...,t_{97}$ of the first sequence match with the terms $T_2,T_6,T_{10},...$ of the second.

Hence, in all, $\lfloor {\dfrac{97-2}{5}}\rfloor +1$ or $\boxed {20}$ terms of them are common.

Terms in the first progression have form 15+4 k (for k=0,1, 2, ..., 100). Terms in the second progression have form 14+5 m (m=0, 1, ..., 90). Setting these two equal yields:

15 + 4 k = 14+ 5 m.

Expressed in modular arithmetic, this is equivalent to:

4k = -1 (mod 5), or

4k = 4 (mod 5), or

k = 1 (mod 5).

This is equivalent to k ending in 1 or 6. There are 20 such k's between 0 and 100.