All questions of Arithmetic Progressions for RRB NTPC/ASM/CA/TA Exam

Given Information:
- The fourth term of an arithmetic progression (A.P.) is zero.

Formula for nth term of an A.P.:
- The nth term of an A.P. is given by: \(t_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Finding the Common Difference (d):
- Given that the fourth term (\(t_4\)) is zero, we have: \(t_4 = a + 3d = 0\)
- Solving for \(d\), we get: \(d = -\frac{a}{3}\)

Finding the Ratio \(t_{25}/t_{11}\):
- The ratio of the 25th term to the 11th term of an A.P. can be expressed as: \(\frac{t_{25}}{t_{11}} = \frac{a + 24d}{a + 10d}\)
- Substituting \(d = -\frac{a}{3}\) into the ratio, we get: \(\frac{a - 8a}{a - \frac{10a}{3}} = \frac{-7a}{\frac{2a}{3}} = -\frac{21}{2}\)

Final Answer:
- The ratio \(t_{25}/t_{11}\) simplifies to \(-\frac{21}{2}\), which is equivalent to \(-3\frac{1}{2}\) or \(-3.5\).
- Therefore, the correct answer is option 'B' (3).

Given Information:
- a = -3
- a(2) = 4

Calculation:
To find a(21) in the series, we need to first determine the pattern in the series.

Pattern in the Series:
- The difference between a and a(2) is 4 - (-3) = 7
- This suggests that there is an increment of 7 in each term of the series.

Finding a(21):
- To find a(21), we need to calculate the increment from a to a(21).
- The increment from a to a(21) = 7 * (21 - 1) = 7 * 20 = 140
- Now, to find a(21), we add this increment to the value of a.
- a(21) = -3 + 140 = 137
Therefore, the value of a(21) in the series is 137, which corresponds to option 'C'.