Answer

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**Hint**

The distance between each number in an arithmetic series differs frequently. It is known as the common difference because it is both the difference between each number in the sequence and the same, and common, to all of the numbers. An arithmetic progression is a series of differences. It can be determined easily by subtracting the second term from the first term in the arithmetic sequence, the third term from the second term, or any two successive numbers in the sequence.

Common Disparities can be both beneficial and bad. The distance between succeeding terms in an arithmetic series is always the same.

**Formula use:**

The geometric progression is \[a,b,c\]

\[{b^2} = ac\].

The N th term of the series is \[a + (n - 1)d\]

The A.P series with terms is assumed as

\[a,a + d,a + 2d,a + 3d\]

**Complete step-by-step solution**

The A.P series with terms is assumed as

\[a,a + d,a + 2d,a + 3d\]

The first term of the series is a.

The N th term of the series is \[a + (n - 1)d\]

\[{T_2} = a + d\]

\[{T_{10}} = a + 9d\]

\[{T_{34}} = a + 33d\]

The series of numbers are in G.P

Assume \[f,g,h\] is in G.P. then, \[{g^2} = fh\]

Hence, \[T_{10}^2 = {T_{2}} \times {T_{34}}\]

\[ = > {(a + 9d)^2} = (a + d) \times (a + 33d)\]

By expanding the equation, it becomes

\[ = > {a^2} + 81{d^2} + 18ad = {a^2} + 33{d^2} + 34ad\]

\[ = > 48{d^2} = 16d\]

Then , the value of d becomes\[ = > d = \frac{1}{3}\]

So, the common difference between the terms is calculated as \[\frac{1}{3}\]

**Therefore, the correct option is B.**

**Note**

The difference between two successive words is the most frequent difference in an arithmetic sequence. Given that each term has a common difference, this is an arithmetic sequence. An assortment of integers is referred to be a geometric sequence if each succeeding element is produced by multiplying the previous value by a fixed factor. There is a typical difference between words that follow one another. They all share the same common ratio between words.

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