if the sum of the 3rd and 8th terms of an ap is 7 and the sum of 7th and 14th term is -3 ,find of 10th term
To find the 10th term of an arithmetic progression (AP), we need to use the given information about the sums of different terms.
First, let's assign variables to the terms of the AP:
The 3rd term = a + 2d (since "a" is the first term and "d" is the common difference)
The 8th term = a + 7d
The 7th term = a + 6d
The 14th term = a + 13d
Given that the sum of the 3rd and 8th terms is 7, we can write the equation:
(a + 2d) + (a + 7d) = 7
Similarly, the sum of the 7th and 14th terms is -3, which gives us another equation:
(a + 6d) + (a + 13d) = -3
Now we have a system of two equations with two variables, "a" and "d". We can solve this system of equations simultaneously to find the values of "a" and "d".
1. Expand the equations:
2a + 9d = 7
2a + 19d = -3
2. Subtract the second equation from the first equation:
(2a + 9d) - (2a + 19d) = 7 - (-3)
-10d = 10
3. Divide both sides of the equation by -10 to solve for "d":
d = -1
4. Substitute the value of "d" back into one of the original equations to solve for "a":
2a + 9(-1) = 7
2a - 9 = 7
2a = 7 + 9
2a = 16
a = 8
Now we have found the values of "a" and "d" in the AP: a = 8 and d = -1.
To find the 10th term, we use the formula:
10th term = a + (10-1)d = 8 + 9(-1) = 8 - 9 = -1.
Therefore, the 10th term of the arithmetic progression is -1.
t is the 3rd term
t + t + 5d = 7 ... 2t + 5d = 7
t + 4d + t + 11d = -3 ... 2t + 15 d = -3
subtracting equations (to eliminate t) ... 10d = -10 ... d = -1
substituting ... 2t + 5(-1) = 7 ...2t = 12 ... t = 6
10th term ... 3rd term plus 7 differences ... 6 + 7(-1) = ?