7, 19, 31, 43, ...
To determine the common difference of an Arithmetic Progression (A.P.) series, we use the formula for the common difference \(d\) of an A.P., which is given by subtracting the first term from the second term, i.e., \(d = a_2 - a_1\).
Given the series: 7, 19, 31, 43, ...
- The first term (\(a_1\)) is 7.
- The second term (\(a_2\)) is 19.
Now, calculate the common difference \(d\):
Using the formula\(d = a_2 - a_1\), we substitute the known values:
\(d = 19 - 7\)
This simplifies to:
\(d = 12\)
Therefore, the common difference of the given A.P. series is 12.
Looking at the options provided:
- Option 1: 19
- Option 2: 5
- Option 3: 12 (Correct Option)
- Option 4: 7
The correct answer, as calculated, is indeed 12. Thus, Option 3 is correct.
This common difference confirms that each term in the sequence after the first is obtained by adding 12 to the previous term, verifying the series is indeed an Arithmetic Progression.
Similar Questions
- What would be the $178^{\text{th}}$ term in the following series?
2, 5, 8, 11, 14, ... - The $5^{\text{th}}$ and the $17^{\text{th}}$ terms of an A. P. series are 52 and 184 respectively. Find the common difference of the series.
Important Questions from Progression (Notes)
- The sum of 16 terms of the series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + .....$ is :
- The $5^{\text{th}}$ and $9^{\text{th}}$ terms of an arithmetic progression are 7 and 13 respectively. What is the $15^{\text{th}}$ term?
- Find the sum of the G.P.:
$5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms. The value of $1 + (\frac{1}{2^1} + \frac{1}{3}) + (\frac{1}{2^2} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + ... + (\frac{1}{2^9} + ... + \frac{1}{1023})$ lies between
- A ball is dropped from a height of 100 m. The ball after each bounce rises vertically by half its previous height (This means at the first bounce it rises by 50 m, by 25 m at the second bounce and so on). What is the vertical distance travelled by the ball between the first and the fifth bounces?
