Question

How do you find the number of terms \[n\] given the sum \[{s_n} = 375\] of the series \[ - 10 + ( - 5) + 0 + 5 + 10 + ...?\]

Answer 1

Hint: In the given question we have to find the number of terms of the series \[ - 10 + ( - 5) + 0 + 5 + 10 + ...\] where the result of this series will be \[375\]. Now you can do one by one addition of the terms but that would consume a lot of time so here you can use sum formula of arithmetic series because it clearly visible that the series given to us is an arithmetic progression with the common difference \[d = 5\] so simple you can use the formula:
\[{s_n} = \dfrac{n}{2}(2a + (n - 1)d)\], where
\[{s_n} = \]Sum of \[n\]terms
\[a = \]First term of the series
\[d = \]Common difference
\[n = \]Number of terms
Now put the values we have in the above given formula and find the value of \[n\].

Complete step by step solution:
In the given question it is visible that the given series is an arithmetic progression with a common difference of \[5\].
Now we can use the formula which is used to find the sum of \[n\] terms of arithmetic progression which is given by:
\[{s_n} = \dfrac{n}{2}(2a + (n - 1)d).........(i)\], where
\[{s_n} = \] Sum of \[n\]terms
\[a = \] First term of the series
\[d = \] Common difference
\[n = \] Number of terms
Now according to question we have:
\[
  a = - 10 \\
  d = 5 \\
  {s_n} = 375 \\
 \]
Now putting these value in \[(i)\] we get:
\[
  375 = \dfrac{n}{2}(2( - 10) + (n - 1)5) \\
  375 = \dfrac{n}{2}( - 20 + 5n - 5) \\
 \]
Further simplifying we get:
\[
  375 = \dfrac{n}{2}(5n - 25) \\
  750 = n(5n - 25) \\
  750 = 5{n^2} - 25n \\
  0 = 5{n^2} - 25n - 750 \\
 \]
Now we have to solve this quadratic equation i.e.
\[
  0 = 5({n^2} - 5n - 150) \\
  0 = (n - 15)(n + 10) \\
 \]
Now \[n = 15\] or \[n = - 10\]
We know that the number of terms cannot be negative so \[n = - 10\] is rejected.

Hence, the number of terms are \[15\].

Note: Here you should learn the formula to find sum of \[n\] terms of and arithmetic progression also you should lean the formula to find \[{n^{th}}\] term of an arithmetic progression which is \[{T_n} = a + (n - 1)d\]. In this formula too \[a = \] First term of the series, \[d = \] Common difference.