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If $1 + 6 + 11 + 16 + .......... + x = 148$ , then the value of x is
$\left( a \right)36$
$\left( b \right)35$
$\left( c \right) - 36$
$\left( d \right)$None of these

Answer Verified Verified
Hint: Use formula of nth term of an A.P ${a_n} = a + \left( {n - 1} \right)d$ and also use sum of n term of an A.P ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where a is a first term, d is a common difference and n is number of terms.

Complete step-by-step answer:
Given series, $1 + 6 + 11 + 16 + .......... + x = 148$
First we check type of series,
Common difference, d=6-1=11-6=16-11=5
So, We can see given series form an A.P with first term, a=1 and common difference, d=5.
Now, last term of an A.P is ${a_n} = x$ .So, we apply formula of nth term of an A.P $
  {a_n} = a + \left( {n - 1} \right)d \\
   \Rightarrow x = 1 + \left( {n - 1} \right)5 \\
   \Rightarrow x = 1 + 5n - 5 \\
   \Rightarrow x = 5n - 4..........\left( 1 \right) \\
 $
Given, sum of n terms of an A.P ${S_n} = 148$ . So, we use the formula of sum of n terms of an A.P.
$
  {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) \\
   \Rightarrow 148 = \dfrac{n}{2}\left( {2 \times 1 + \left( {n - 1} \right) \times 5} \right) \\
   \Rightarrow 296 = n\left( {5n - 3} \right) \\
   \Rightarrow 5{n^2} - 3n - 296 = 0 \\
 $
Now, factories the quadratic equation .
$
   \Rightarrow \left( {n - 8} \right)\left( {5n + 37} \right) = 0 \\
   \Rightarrow n = 8,\dfrac{{ - 37}}{5} \\
 $
We know the number of terms cannot be negative so we eliminate the negative value.
So, $n = 8$
Now, put the value of n in (1) equation.
$
   \Rightarrow x = 5 \times 8 - 4 \\
   \Rightarrow x = 40 - 4 \\
   \Rightarrow x = 36 \\
$
So, the correct option is (a).
Note: Whenever we face such types of problems we use some important points. First we check which type of series formed then we apply the formula of nth term and sum of n terms then after some calculation we can get the required answer.
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