Answer

Verified

450.3k+ views

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.

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE