5 d = 20 ... d = 4
T5 + 11d = 3 T5 ... T5 = 22
T1 = T5 - 4d
T5 + 11d = 3 T5 ... T5 = 22
T1 = T5 - 4d
T12 = 20+T7,
a +11d = 20 + a + 6d,
5 d = 20, d = 4.
T16 = 3T5,
a + (16 - 1)d = 3[a+(5-1)d],
a + (16 - 1)4 = 3[a+(5-1)4],
a+60 =3a+48,
3a-a = 60 - 48.
2a = 12
a= 6
a = 6, d = 4
Let's denote the first term as 'a' and the common difference as 'd'.
We are given two conditions:
Condition 1: The 16th term of the AP is three times the 5th term.
Using the formula for the nth term of an AP:
nth term (Tn) = a + (n - 1)d
According to the first condition, we have:
T16 = 3 * T5
Substituting the formulas for the terms, we have:
a + (16 - 1)d = 3 * (a + (5 - 1)d)
Simplifying this equation gives:
a + 15d = 3a + 12d
Rearranging the terms, we get:
2a = 3d
Condition 2: The 12th term is 20 more than the 7th term.
Again, using the formula for the nth term of an AP:
T12 = T7 + 20
Plugging in the formulas for the terms, we have:
a + (12 - 1)d = a + (7 - 1)d + 20
Simplifying this equation gives:
a + 11d = a + 6d + 20
Rearranging the terms, we get:
5d = 20
Now we have a system of equations:
2a = 3d (from Condition 1)
5d = 20 (from Condition 2)
Solving the second equation for d, we find that d = 4.
Substituting this value of d back into the first equation, we can solve for a:
2a = 3d
2a = 3 * 4
2a = 12
a = 6
Therefore, the first term (a) is 6 and the common difference (d) is 4.