Question

How many terms are there in the AP \[1,3,5.....73,75\] ?

A. \[28\]
B. \[30\]
C. \[36\]
D. \[38\]

Answer 1

Hint- In order to deal with this question first we will obtain the value of first term and common difference with the help of given AP further to find the number of terms we will directly use the formula of nth term of AP which is mentioned in the solution.

Complete step-by-step answer:
Given AP is \[1,3,5.....73,75\]\[{a_n}\]
Here first term \[a_1\] is $1$
Now to find the common difference we will use the below formula
\[d{\text{ }} = {\text{ }}a_2 - a_1{\text{ }} = {\text{ }}a_3 - a_2\]
Where \[a1\] is the first term , \[a_2\] is the second term and \[a_3\]is the third term of AP
In given AP second and third is \[3\] and \[5\] respectively
So now \[d{\text{ }} = {\text{ }}3 - 1{\text{ }} = {\text{ }}5 - 3{\text{ }} = {\text{ }}2\]
Therefore common difference is \[2\]
Now we will evaluate the number of terms of AP by using the following formula
\[{a_n}{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}\left( {n - 1} \right){\text{ }}d\]
Where \[{a_n}\] is the last term of AP,
\[n\] is the number of term,
\[d\] is the common difference
and \[a\] is the first term
By substituting all the values we have
\[75{\text{ }} = {\text{ }}1{\text{ }} + {\text{ }}\left( {n - 1} \right){\text{ }} \times 2\]
Further by solving above equation we get
\[\begin{array}{*{20}{l}}
  {75 - 1{\text{ }} = {\text{ }}\left( {n - 1} \right){\text{ }} \times 2} \\
  \; \\
  {74{\text{ }} = {\text{ }}\left( {n - 1} \right){\text{ }} \times 2}
\end{array}\]
By dividing both the side of above equation by \[2\]
\[\begin{array}{*{20}{l}}
  {74/2{\text{ }} = {\text{ }}\left( {n - 1} \right)} \\
  \; \\
  {37{\text{ }} = {\text{ }}n{\text{ }} - 1} \\
  {n{\text{ }} = {\text{ }}38}
\end{array}\]
Hence, the number of terms in the given AP is \[38\] and the correct option is D.

Note- An arithmetic progression or arithmetic sequence in mathematics is a sequence of numbers, such that the difference between the successive terms is constant. Difference here represents the second term minus first term. By \[{a_n}{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}\left( {n - 1} \right){\text{ }}d\] we can find the \[{n^{th}}\]term of the AP where \[n\] is the number of term, \[d\] is the common difference and \[a\] is the first term and is the \[{n^{th}}\]term of the AP.