Question

Determine an A.P. whose 3rd term is 16 and the difference of 7th term and 5th term is 12.

Answer 1

Hint: Here, we will find the common difference by usinng the formula of \[n\]th term of the arithmetic progression A.P., that is, \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference. Apply this formula, and then substitute the value of \[a\],\[d\] and \[n\] in the obtained equation to find the required A.P.

Complete step by step answer:

We are given that the 3rd term is 16 and the difference of 7th term and 5th term is 12.
We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
\[ \Rightarrow {a_7} - {a_5} = 12\]
Using the formula of \[n\]th term of the arithmetic progression A.P., that is, \[{a_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference, in the above equation, we get
\[
   \Rightarrow \left( {a + 6d} \right) - \left( {a + 4d} \right) = 12 \\
   \Rightarrow a + 6d - a - 4d = 12 \\
   \Rightarrow 2d = 12 \\
 \]
Dividing the above equation by 2 on both sides, we get
\[
   \Rightarrow \dfrac{{2d}}{2} = \dfrac{{12}}{2} \\
   \Rightarrow d = 6 \\
 \]
Using the above value of \[d\] in the 3rd term \[{a_3} = a + 2d\], we get
\[
   \Rightarrow {a_3} = a + 2\left( 6 \right) \\
   \Rightarrow {a_3} = a + 12 \\
 \]
Taking \[{a_3} = 16\] in the above equation, we get
\[ \Rightarrow 16 = a + 12\]
Subtracting the above equation by 12 on both sides, we get
\[
   \Rightarrow 16 - 12 = a + 12 - 12 \\
   \Rightarrow 4 = a \\
   \Rightarrow a = 4 \\
 \]
Hence the first term of A.P. is 4.
Substituting these values of \[a\] and \[d\] in the above formula for the second term of the arithmetic progression, we get
\[
   \Rightarrow {a_2} = 4 + 6 \\
   \Rightarrow {a_2} = 10 \\
 \]
Hence, the A.P. is 4, 10, 16, …

Note: In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula of sum, \[S = \dfrac{n}{2}\left( {a + l} \right)\], where \[l\] is the last term, but have the to find the value of \[{a_n}\] , so it will be wrong. We can also find the value of \[n\]th term by find the value of \[{S_n} - {S_{n - 1}}\], where \[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where \[a\] is the first term and \[d\] is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the \[{a_n}\] is the \[n\]th term in the arithmetic progression.