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Answer :
Let the first term of both arithmetic progressions be $a$. For the first arithmetic progression (AP), we are given:
- Number of terms: $n_1 = 21$
- Common difference: $d_1 = 9$
The last term of an AP is given by:
$$
L = a + (n - 1)d.
$$
For the first AP, the last term is:
$$
L = a + (21 - 1) \times 9 = a + 20 \times 9 = a + 180.
$$
For the second AP, the common difference is $d_2 = 4$ and it has an unknown number of terms $n_2$. Its last term is:
$$
L = a + (n_2 - 1) \times 4.
$$
Since both APs have the same first and last terms, we equate the expressions for the last term:
$$
a + (n_2 - 1) \times 4 = a + 180.
$$
Subtracting $a$ from both sides, we obtain:
$$
(n_2 - 1) \times 4 = 180.
$$
Now, solve for $n_2 - 1$:
$$
n_2 - 1 = \frac{180}{4} = 45.
$$
Finally, add $1$ to find $n_2$:
$$
n_2 = 45 + 1 = 46.
$$
Thus, the second arithmetic progression has $\boxed{46}$ terms.
- Number of terms: $n_1 = 21$
- Common difference: $d_1 = 9$
The last term of an AP is given by:
$$
L = a + (n - 1)d.
$$
For the first AP, the last term is:
$$
L = a + (21 - 1) \times 9 = a + 20 \times 9 = a + 180.
$$
For the second AP, the common difference is $d_2 = 4$ and it has an unknown number of terms $n_2$. Its last term is:
$$
L = a + (n_2 - 1) \times 4.
$$
Since both APs have the same first and last terms, we equate the expressions for the last term:
$$
a + (n_2 - 1) \times 4 = a + 180.
$$
Subtracting $a$ from both sides, we obtain:
$$
(n_2 - 1) \times 4 = 180.
$$
Now, solve for $n_2 - 1$:
$$
n_2 - 1 = \frac{180}{4} = 45.
$$
Finally, add $1$ to find $n_2$:
$$
n_2 = 45 + 1 = 46.
$$
Thus, the second arithmetic progression has $\boxed{46}$ terms.
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