How many terms of the arithmetic series \[24{\text{ }} + {\text{ }}21{\text{ }} + {\text{ }}18{\text{ }} + {\text{ }}15{\text{ }} + .....\] be taken continuously so that their sum is $ - 351 $ .

Hint:In order to deal with this question first we will evaluate the common difference from the given series , in this question sum of series is given so we will directly use the formula of sum of nth series which is mentioned in the solution.

Complete step-by-step answer:
Given series is \[24{\text{ }} + {\text{ }}21{\text{ }} + {\text{ }}18{\text{ }} + {\text{ }}15{\text{ }} + .....\]
Sum of series is $ - 351 $
Here first term of series is $ 24 $
Or $ a = 24 $
Now we will calculate the common difference
As we know that
Common difference $ (d) $ = second term \[ - \] first term = third term \[ - \] second term
Or $ d = {a_{_2}} - {a_1} = {a_3} - {a_2} $
Substitute values in above formula we have
\[d{\text{ }} = {\text{ }}21 - {\text{ }}24{\text{ }} = {\text{ 18}} - 21\; = {\text{ - }}3\]
As we know that formula of sum of \[n\] number of series is given as
\[{S_n} = \dfrac{n}{2}\left[ {(2a + (n - 1)d)} \right]\]
Substitute values of \[a,d\] and \[{S_n}\] in above formula we have
\[
   - 351 = \dfrac{n}{2}\left[ {(48 + (n - 1)( - 3))} \right] \\
   \Rightarrow ( - 351) \times 2 = n[(48 + (n - 1)( - 3))] \\
   \Rightarrow - 702 = n[(48 - 3(n - 1))] \\
   \Rightarrow - 702 = n[(48 - 3n + 3)] \\
   \Rightarrow - 702 = n[(51 - 3n)] \\
   \Rightarrow - 702 = 51n - 3{n^2} \\
   \Rightarrow - 3{n^2} + 51n + 702 = 0 \\
   \Rightarrow 3{n^2} - 51n - 702 = 0 \\
   \Rightarrow 3({n^2} - 17n - 234) = 0 \\
   \Rightarrow ({n^2} - 17n - 234) = \dfrac{0}{3} \\
   \Rightarrow ({n^2} - 17n - 234) = 0 \\
 \]
Further by solving quadratic equation we get
 $
  ({n^2} - 17n - 234) = 0 \\
   \Rightarrow {n^2} - 26n + 9n - 234 = 0 \\
   \Rightarrow n(n - 26) + 9(n - 26) = 0 \\
   \Rightarrow (n - 26)(n + 9) = 0 \\
  {\text{Either }}(n - 26) = 0{\text{ or }}(n + 9) = 0 \\
  {\text{So }}n = 26{\text{ or }}n = - 9 \\
  $
Here $ n $ , being the number of terms needed, can not be negative. Thus, \[26\] terms are needed to get the sum \[ - 351\].

Note:The sum of \[n\] number of A.P. is given by the formula which is equal to the given expression\[{S_n} = \dfrac{n}{2}\left[ {(2a + (n - 1)d)} \right]\]. Where \[{S_n}\] represent the sum of \[n\] number of series , \[a\] is the first term of the A.P. and \[d\] is the common difference of the A.P.