the sum of the first and second terms of ap is 4 and the tenth term is 19. find the sum of the 5th and 6th
a + a+d = 4
a+9d = 19
a=1
d=2
So, now add T5 and T6
To find the sum of the 5th and 6th terms of an arithmetic progression (AP), we need to first determine the common difference (d) between the terms.
Let's denote the first term of the AP as "a" and the common difference as "d".
Given that the sum of the first and second terms is 4:
a + (a + d) = 4
2a + d = 4 --- Equation 1
We are also given the tenth term of the AP as 19:
a + 9d = 19 --- Equation 2
We have two equations with two variables (a and d) that need to be solved simultaneously.
First, let's rearrange Equation 1 to express "d" in terms of "a":
d = 4 - 2a
Next, substitute the value of "d" from the first equation into the second equation:
a + 9(4 - 2a) = 19
a + 36 - 18a = 19
-17a + 36 = 19
-17a = 19 - 36
-17a = -17
a = -17 / -17
a = 1
Now that we have the value of "a", we can substitute it into Equation 1 to compute the common difference "d":
2(1) + d = 4
2 + d = 4
d = 4 - 2
d = 2
Therefore, the first term (a) is 1, and the common difference (d) is 2.
To find the sum of the 5th and 6th terms:
5th term = a + 4d (using the formula: nth term = a + (n-1)d)
6th term = a + 5d
Substituting the values of "a" and "d" into the formulas:
5th term = 1 + 4(2) = 1 + 8 = 9
6th term = 1 + 5(2) = 1 + 10 = 11
Finally, to find the sum of the 5th and 6th terms: 9 + 11 = 20.
Therefore, the sum of the 5th and 6th terms of the arithmetic progression is 20.