The 16th term of an AP is three times the 5th term. If the 12th term is 20 more than the 7th term, find the first term and the common difference.
Tn=a +(n-1)d
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
5 d = 20 ... d = 4
T5 + 11d = 3 T5 ... T5 = 22
T1 = T5 - 4d
To find the first term and the common difference of an arithmetic progression (AP), we need to use the given information.
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.