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Question

What would be the $178^{\text{th}}$ term in the following series?
2, 5, 8, 11, 14, ...

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
533

Identify the Series Type

The given series is 2, 5, 8, 11, 14, ...

This is an arithmetic series because the difference between consecutive terms is constant.

  • First term ($a_1$): $2$
  • Common difference ($d$): $5 - 2 = 3$

Arithmetic Series Formula

The formula to find the $n^{\text{th}}$ term ($a_n$) of an arithmetic series is:

$ a_n = a_1 + (n-1)d $

Calculate the 178th Term

We need to find the $178^{\text{th}}$ term, so $n = 178$.

Substitute the values into the formula:

$ a_{178} = 2 + (178-1) \times 3 $

$ a_{178} = 2 + (177) \times 3 $

$ a_{178} = 2 + 531 $

$ a_{178} = 533 $

Conclusion

The $178^{\text{th}}$ term in the series is 533.

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Similar Questions

  1. What is the common difference of the following A.P. series?
    7, 19, 31, 43, ...
  2. 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)

  1. The sum of 16 terms of the series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + .....$ is :
  2. 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?
  3. Find the sum of the G.P.:
    $5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms.
  4. 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

  5. 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?
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