Question

Find the number of terms in the arithmetic progression 18, 27 ….99, where 18 is the first term and common difference is 9.

Answer 1

Hint: The given sequence is in arithmetic progression, which means every term starting from the second term is obtained by adding a fixed value to its previous term and the fixed value is called common difference. The last term of the progression is 99. Nth term of an arithmetic progression can be obtained using $ {T_n} = a + \left( {n - 1} \right)d $ . So equate 99 to this formula and obtain the value of n.

Complete step-by-step answer:
We are given an arithmetic progression 18, 27…99 where 18 is the first term and with common difference 9.
The first term of the sequence is 18, $ a = 18 $ and here n is equal to 1.
The second term of the sequence is 27, $ a + d = 27 $ and here n is equal to 2.
The last term of the sequence is 99 and here we do not know the value of n. That is what we have to find.
So,
 $
  a + \left( {n - 1} \right)d = 99 \\
  a = 18,d = 9 \\
\Rightarrow 18 + \left( {n - 1} \right)9 = 99 \\
\Rightarrow \left( {n - 1} \right)9 = 99 - 18 \\
\Rightarrow n - 1 = \dfrac{{81}}{9} \\
\Rightarrow n = 9 + 1 \\
  \therefore n = 10 \;
  $
So there are 10 terms in the given arithmetic sequence; they are 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
Therefore, n=10.

Note: Another approach
We are given an arithmetic progression and we have to find the no. of terms present in it.
No. of terms of an arithmetic progression can be calculated using $ \dfrac{{{T_n} - {T_1}}}{d} + 1 $ , where $ {T_n} $ is the last term, $ {T_1} $ is the first term and d is the common difference.
The first term of the sequence is 18, the last term of the sequence is 99 and the common difference is 9.
On substituting these values in the formula, we get
 $
  n = \dfrac{{{T_n} - {T_1}}}{d} + 1 \\
  {T_n} = 99,{T_1} = 18,d = 9 \\
\Rightarrow n = \dfrac{{99 - 18}}{9} + 1 \\
\Rightarrow n = 9 + 1 = 10 \;
  $
Therefore, there are 10 terms totally in the sequence.