Question

How do you find the Arithmetic means of the sequence 3,_,_,_,_,_,27 ?

Answer 1

Hint: In order to determine the terms of the given arithmetic sequence, find out the first term $ a $ which will be equal to $ 3 $ and the common difference by subtracting any two consecutive terms from the sequence, which is not known. Now put these values along with $ n = 7 $ (because five terms were given blank previously) in the formula of the nth term of AP, $ {a_n} = a + (n - 1)d $ to obtain the value of common difference.

Complete step-by-step answer:
Clearly, the given sequence is an Arithmetic Progression (A.P.).
As we know the $ nth $ term of an A.P. is $ {a_n} = a + (n - 1)d $
where $ a $ is the first term, $ d $ is the common constant difference
So, the $ 7th $ term of the A.P. will be $ {a_7} $
In our sequence first term $ a = 3 $
According to the question we have to find the value of terms which would be $ {a_3} $ , $ {a_3} $ , $ {a_4} $ , $ {a_5} $ , $ {a_6} $ .
Now putting the values of $ n,a\,and\,d $ in the $ nth $ term of A.P. we get
 $
  {a_n} = a + (n - 1)d \\
  {a_7} = 3 + (7 - 1)(d) \\
  27 = 3 + (6)(d) \\
  27 - 3 = 6d \\
  24 = 6d \\
  \dfrac{{24}}{6} = \dfrac{{6d}}{6} \\
  4 = d \\
   = > d = 4 \;
 $
Therefore, $ d $ (common difference) is equal to $ 4 $ .
Now, for next term that is $ n = 2 $ , put the values of $ n,a\,and\,d $ in the $ nth $ term of A.P. we get:
 $
 \Rightarrow {a_2} = a + d \\
  {a_2} = 3 + 4 \\
  {a_2} = 7 \;
 $
Now, for next term that is $ n = 3 $ , put the values of $ n,{a_2}\,and\,d $ in the $ nth $ term of A.P. we get:
 $
 \Rightarrow {a_3} = {a_2} + d \\
  {a_3} = 7 + 4 \\
  {a_3} = 11 \;
 $
Now, for next term that is $ n = 4 $ , put the values of $ n,{a_3}and\,d $ in the $ nth $ term of A.P. we get:
$
 \Rightarrow {a_4} = {a_3} + d \\
  {a_4} = 11 + 4 \\
  {a_4} = 15 \;
 $
Now, for next term that is $ n = 5 $ , put the values of $ n,{a_4}\,and\,d $ in the $ nth $ term of A.P. we get:
 $
 \Rightarrow {a_5} = {a_4} + d \\
  {a_5} = 15 + 4 \\
  {a_5} = 19 \;
 $
Now, for next term that is $ n = 6 $ , put the values of $ n,{a_5}\,and\,d $ in the $ nth $ term of A.P. we get:
 $
 \Rightarrow {a_6} = {a_5} + d \\
  {a_6} = 19 + 4 \\
  {a_6} = 23 \;
 $
Hence, the Arithmetic means of the sequence 3,_,_,_,_,_,27 are $ 7,11,15,19,23 $ .
So, the correct answer is “ $ 7,11,15,19,23 $ ”.

Note: 1.Don’t forgot to cross-check your answer.
2.The difference between any two consecutive terms in an A.P. is always the same and if it is not the same, then the given series is not an A.P.
3. $ (n - 1) $ is nothing but the position of term in the sequence.