The 5th term of an arithmetic progression is 3times of 2nd term and 12th term exceeds 2times of 6th term by 1. Find the 16th term
a5 = 3 a2
a12 = 2 a6 + 1
For AP:
an = a1 + ( n - 1 ) d
a1 = initial term of an arithmetic progression
d = common difference of successive members
a2 = a1 + ( 2 - 1 ) d
a2 = a1 + 1 * d
a2 = a1 + d
a5 = a1 + ( 5 - 1 ) d
a5 = a1 + 4 d
a6 = a1 + ( 6 - 1 ) d
a6 = a1 + 5 d
a12 = a1 + ( 12 - 1 ) d
a12 = a1 + 11 d
Now:
a5 = 3 a2
a5 = 3 ( a1 + d )
a5 = 3 a1 + 3 d
a5 = a5
a1 + 4 d = 3 a1 + 3 d Subtract a1 to both sides
a1 + 4 d - a1 = 3 a1 + 3 d - a1
4 d = 2 a1 + 3 d Subtract 3 d to both sides
4 d - 3 d= 2 a1 + 3 d - 3 d
d = 2 a1
a12 = 2 a6 + 1
a1 + 11 d = 2 ( a1 + 5 d ) + 1
a1 + 11 d = 2 a1 + 10 d + 1 Subtract a1 to both sides
a1 + 11 d - a1 = 2 a1 + 10 d + 1 - a1
11 d = a1 + 10 d + 1 Subtract 10 d to both sides
11 d - 10 d = a1 + 10 d + 1 - 10 d
d = a1 + 1
d = d
2 a1 = a1 + 1 Subtract 1 to both sides
2 a1 - 1 = a1 + 1 - 1
2a1 - 1 = a1 Subtract a1 to both sides
2a1 - 1 - a1 = a1 - a1
a1 - 1 = 0 Add 1 to both sides
a1 - 1 + 1 = 0 + 1
a1 = 1
d = 2 a1
d = 2 * 1
d = 2
an = a1 + ( n - 1 ) d
a16 = a1 + ( 16 - 1 ) d
a16 = a1 + 15 d
a16 = 1 + 15 * 2
a16 = 1 + 30
a16 = 31
a+4d = 3(a+d)
a+11d = 2(a+5d)+1
Solve for a and d, then
T16 = a+15d
To find the 16th term of the arithmetic progression, we need to determine the common difference (d) first.
Let's assume the first term (a1) of the arithmetic progression is 'a' and the common difference is 'd'.
Given that the 5th term is 3 times the 2nd term, we can express this as:
a + 4d = 3(a + d)
Expanding and simplifying the equation, we get:
a + 4d = 3a + 3d
4d - 3d = 3a - a
d = 2a -- (Equation 1)
Now, we can find the 12th term (a12) by using the formula for the nth term of an arithmetic progression:
a12 = a + 11d
Given that the 12th term exceeds 2 times the 6th term by 1, we can express this as:
a + 11d = 2(a + 5d) + 1
Expanding and simplifying the equation, we get:
a + 11d = 2a + 10d + 1
a - d = 1 -- (Equation 2)
Now we have two equations:
d = 2a -- Equation 1
a - d = 1 -- Equation 2
Solving these two equations, we can find the values of 'a' and 'd'.
From Equation 1, we have:
d = 2a
Substituting this value in Equation 2, we get:
a - 2a = 1
-a = 1
a = -1
Now that we have the value of 'a', we can substitute it back into Equation 1 to find 'd':
d = 2a
d = 2(-1)
d = -2
Therefore, the first term (a1) is -1 and the common difference (d) is -2.
To find the 16th term (a16), we can use the formula for the nth term of an arithmetic progression:
a16 = a1 + (n - 1)d
Substituting the values, we get:
a16 = -1 + (16 - 1)(-2)
a16 = -1 + 15(-2)
a16 = -1 - 30
a16 = -31
Hence, the 16th term of the arithmetic progression is -31.