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Question

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.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
11

A.P. Series Common Difference Calculation

We are given the details of an Arithmetic Progression (A.P. series):

  • The $5^{\text{th}}$ term ($a_5$) is 52.
  • The $17^{\text{th}}$ term ($a_{17}$) is 184.

We need to find the common difference ($d$) of this A.P. series.

Applying the A.P. Formula

The formula for the n-th term of an A.P. is:

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

Where $a_1$ is the first term and $d$ is the common difference.

Using this formula, we can set up two equations based on the given information:

  1. For the $5^{\text{th}}$ term ($n=5$):

    $a_5 = a_1 + (5-1)d$

    $52 = a_1 + 4d \quad \text{(Equation 1)}$

  2. For the $17^{\text{th}}$ term ($n=17$):

    $a_{17} = a_1 + (17-1)d$

    $184 = a_1 + 16d \quad \text{(Equation 2)}$

Solving for the Common Difference

To find the common difference ($d$), we can subtract Equation 1 from Equation 2:

$ (a_1 + 16d) - (a_1 + 4d) = 184 - 52 $

Simplify the equation:

$ a_1 + 16d - a_1 - 4d = 132 $

$ 12d = 132 $

Now, isolate $d$ by dividing both sides by 12:

$ d = \frac{132}{12} $

$ d = 11 $

Therefore, the common difference of the A.P. series is 11.

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Similar Questions

  1. What would be the $178^{\text{th}}$ term in the following series?
    2, 5, 8, 11, 14, ...
  2. What is the common difference of the following A.P. series?
    7, 19, 31, 43, ...

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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