Question

Find the next term of the sequence 3, 8, 14, 21, 29,….
(A) 35
(B) 36
(C) 37
(D) 38

Answer 1

Hint: Identify the pattern the subsequent numbers in the series are following. The difference between any two consecutive terms is 1 more than the difference between the previous two consecutive terms. Use this pattern to find the ${n^{{\text{th}}}}$ term of the sequence. Then put $n = 6$ to find the $6^{th}$ term.

Complete step-by-step answer:
According to the question, the given sequence is 3, 8, 14, 21, 29,….. We have to determine the next term in the sequence.
At first, we will identify the pattern the terms in the sequence are following. On close observation, we can see that the difference between any two consecutive terms is 1 more than the difference between the previous two consecutive terms.
The first term in the sequence is 3 and the difference between the first two consecutive terms is 5. Using these information, we can deduce an expression for general term for the sequence:
$ \Rightarrow {t_n} - {t_{n - 1}} = 5 + \left( {n - 2} \right){\text{ }}\forall {\text{ }}t > 1$ and ${t_1} = 3$
\[ \Rightarrow {t_n} = \left\{ {\begin{array}{*{20}{c}}
  {{t_{n - 1}} + 5 + \left( {n - 2} \right){\text{ for }}t > 1} \\
  {3{\text{ for }}t = 1}
\end{array}} \right.\]
Since we have to determine the 6th term of the sequence, we will put $n = 6$ in the above general term expression. On doing so, we’ll get:
$ \Rightarrow {t_6} = {t_5} + 5 + \left( {6 - 2} \right){\text{ }}.....{\text{(1)}}$
And from the sequence given in the question, we already know that the 5th term is 29 i.e. ${t_5} = 29$. Putting this in equation (1), we’ll get:
$
   \Rightarrow {t_6} = 29 + 5 + 4 \\
   \Rightarrow {t_6} = 38
 $
Thus the next term of the sequence is 38.

D is the correct option.

Note: In a sequence, if the difference between any two consecutive terms is the same then the sequence is called an arithmetic sequence. In the above sequence, we can see that the difference between two consecutive terms will form an arithmetic sequence.
$
   \Rightarrow 8 - 3,{\text{ }}14 - 8,{\text{ }}21 - 14,{\text{ }}29 - 21 \\
   \Rightarrow 5,{\text{ }}6,{\text{ }}7,{\text{ }}8 \\
 $
The first term of this sequence is 5 and the common difference is 1. This information can also be used to form an expression for the general term of the original sequence. The resultant expression will be same as determined above:
\[ \Rightarrow {t_n} = \left\{ {\begin{array}{*{20}{c}}
  {{t_{n - 1}} + 5 + \left( {n - 2} \right){\text{ for }}t > 1} \\
  {3{\text{ for }}t = 1}
\end{array}} \right.\]