The sum of the first and second terms of an A.P us 4 and the 10th term is 19. Find the sum of the 5th and 6th terms.
2 a + d = 4
a + 9 d = 19 ... 2 a + 18 d = 38
subtracting equations ... 17 d = 34 ... d = 2 ... a = 1
5th plus 6th ... (a + 4 d) + (a + 5 d) = 2 a + 9 d = ?
2a+9d=2×1+9×2
=2+18=20
To find the sum of the 5th and 6th terms of the arithmetic progression (A.P.), we first need to determine the common difference between the terms.
In an arithmetic progression, the difference between consecutive terms is constant. Let's represent the common difference with the letter "d".
Given that the sum of the first and second terms is 4, we can write the equation:
First Term + Second Term = 4
Since the first term of the A.P. can be represented as "a" and the second term as "a + d", we can rewrite the equation:
a + (a + d) = 4
Simplifying this equation, we get:
2a + d = 4 ----(1)
Next, we are given that the 10th term of the A.P. is 19. Using the formula for the nth term of an A.P, which is given by:
nth term = a + (n - 1)d,
where "n" is the position of the term in the A.P., we can write the equation for the 10th term:
a + (10 - 1)d = 19
a + 9d = 19 ----(2)
Now, we need to solve equations (1) and (2) simultaneously to find the values of "a" and "d".
Subtracting equation (1) from equation (2), we get:
(a + 9d) - (2a + d) = 19 - 4
-a + 8d = 15
8d - a = 15 ----(3)
We have two equations: 2a + d = 4 and 8d - a = 15.
To solve these equations, we can use the method of elimination or substitution. Here, we will use substitution.
Rearranging equation (3) to express "a" in terms of "d", we get:
a = 8d - 15
Substituting this value of "a" into equation (1), we have:
2(8d - 15) + d = 4
16d - 30 + d = 4
17d = 34
d = 2
Now, we can substitute the value of "d" back into equation (1) or (2) to find the value of "a". Let's use equation (1):
2a + d = 4
2a + 2 = 4
2a = 2
a = 1
We have found that the first term "a" is 1 and the common difference "d" is 2.
Using the formula for the nth term of an A.P., we can calculate the values of the 5th and 6th terms:
5th term = a + (5-1)d = 1 + 4(2) = 9
6th term = a + (6-1)d = 1 + 5(2) = 11
Finally, we can find the sum of the 5th and 6th terms:
Sum = 9 + 11 = 20
Therefore, the sum of the 5th and 6th terms of the given A.P. is 20.