Sum of first n terms of an Arithmetic series

Sum of the first n terms of the sequence is calculated by

{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]

Example

If Sudha save some money every month in her piggy bank, then how much money will be there in her piggy bank after 12 months, if the money is in the sequence of 100, 150, 200, 250, ….respectively?

Solution

Given sequence is-

100, 150, 200, 250 …

a = 100 (first term)

d = 50 (common difference)

n = 12 (as we have to calculate money of 12 months)

\begin{array}{l}{S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]\\ = \frac{{12}}{2}\left[ {2(100) + (12 - 1)50} \right]\\ = \frac{{12}}{2}\left[ {200 + 550} \right]\\ = \frac{{12}}{2}(750)\\ = 6\left( {750} \right)\\ = 4500\end{array}

So the money collected in her bank in 12 months is Rs. 4500.

But when we have finite Arithmetic Progression or we know the last term of the sequence then the sum of all the given terms of the progression will be calculated by

{S_n} = \frac{n}{2}\left[ {a + 1} \right]

Where l = a + (n – 1)d i.e. the last term of the finite Arithmetic Progression.

EXAMPLES

  1. What is the common difference of an A.P. in which{a_{21}} - {a_7} = 98?
    Solution:
    \begin{array}{l}{a_{21}} - {a_7} = 98\\\therefore (a + 20d) - (a + 6d) = 98\\20d - 6d = 98\\14d = 98\\d = 7\end{array}
  1. Which term of the progression 4, 9, 14, 19, are 109?
    Solution:
    Here, d = 9 - 4 = 14 - 9 = 19 - 14 = 5

.
Given: First term,

\begin{array}{l}a = 4,d = 5,{a_n} = 109\\\therefore {a_n} = a + (n - 1)d\\\therefore 109 = 4 + (n - 1)5\\ \Rightarrow 109 - 4 = (n - 1)5\\ \Rightarrow 105 = 5(n - 1)\\ \Rightarrow n - 1 = 105/5 = 21\\ \Rightarrow n = 21 + 1 = 22\end{array}

∴ 109 is the 22nd term

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