The 9th term of an AP is less than twice the 3rd term, if the common difference is 2, find the value of the first term
Let the first term be a. Then the 3rd term is a + 4 and the 9th term is a + 16.
According to the problem, we have:
a + 16 < 2(a + 4)
a + 16 < 2a + 8
a > 8
Therefore, the first term is greater than 8. However, we cannot determine the exact value of the first term without additional information.
To find the value of the first term, let's write out the terms of the arithmetic progression (AP).
Let's assume the first term is 'a' and the common difference is 2.
The terms of the AP can be written as:
a, a+2, a+4, a+6, a+8, a+10, a+12, a+14, a+16, ...
We know that the 9th term is less than twice the 3rd term, so we can set up the following inequality:
a + 8 < 2(a + 4)
Now, let's solve this inequality step-by-step to find the value of 'a'.
Step 1: Distribute 2 on the right-hand side of the inequality.
a + 8 < 2a + 8
Step 2: Subtract 'a' from both sides of the inequality.
8 < a + 8
Step 3: Subtract '8' from both sides of the inequality.
0 < a
Therefore, 'a' must be greater than 0.
Hence, the value of the first term of the arithmetic progression is greater than 0.