19th term of the A.P.: 10, 7, 4, ….. is
-44
The question asks us to find the 19th term of the given Arithmetic Progression (A.P.): 10, 7, 4, …
An Arithmetic Progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference.
In the given A.P.: 10, 7, 4, …
Let's calculate the common difference:
The common difference $d$ is indeed -3.
The formula to find the $n$th term of an Arithmetic Progression is:
\( a_n = a_1 + (n-1)d \)
Where:
We need to find the 19th term, so \(n=19\). We have \(a_1 = 10\) and \(d = -3\). Substituting these values into the formula:
\( a_{19} = a_1 + (19-1)d \)
\( a_{19} = 10 + (18)(-3) \)
Now, perform the multiplication:
\( 18 \times -3 = -54 \)
Substitute this back into the equation for \(a_{19}\):
\( a_{19} = 10 + (-54) \)
\( a_{19} = 10 - 54 \)
Finally, perform the subtraction:
\( a_{19} = -44 \)
So, the 19th term of the A.P. 10, 7, 4, ... is -44.
| Concept | Description | Formula |
|---|---|---|
| Arithmetic Progression (A.P.) | A sequence where the difference between consecutive terms is constant. | |
| First Term | The initial term of the sequence. | \(a_1\) |
| Common Difference | The constant difference between consecutive terms. | \(d = a_n - a_{n-1}\) |
| nth Term | The term at a specific position \(n\) in the sequence. | \(a_n = a_1 + (n-1)d\) |
Related to an Arithmetic Progression is an Arithmetic Series, which is the sum of the terms of an A.P.
The sum of the first \(n\) terms of an A.P., denoted as \(S_n\), can be calculated using the following formulas:
\( S_n = \frac{n}{2} (a_1 + a_n) \)
\( S_n = \frac{n}{2} [2a_1 + (n-1)d] \)
These formulas are useful for finding the sum of a certain number of terms in an A.P.
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