Question

Write the 6th term of arithmetic sequence 1,25,49,73,97....write the algebric form?

Answer 1

$\large\text{\underline{Let's begin:-}}$

The general term of an arithmetic sequence $\{a_{n}\}$ is,

$\hookrightarrow\boxed{ a_{n}=a+(n-1)d}$

$a$ represents the first term, and $d$ is for common difference, and $n$ is the term number. This way of showing the arithmetic sequence is a general form.

The algebraic form of an arithmetic sequence is different from above. It is a simplified form of a general term.

$\large\text{\underline{Solution:-}}$

Since $a=1,d=24$, the general form is,

$\hookrightarrow a_{n}=1+24(n-1)\ \text{[General Form]}$

Simplifying further,

$\hookrightarrow a_{n}=24n-23\ \text{[Algebraic Form]}$

We want the 6th term of $a_{n}$. So, we substitute $n=6$ into the algebraic form to get,

$\hookrightarrow a_{6}=144-23=121$

$\large\text{\underline{Conclusion:-}}$

The required answer is 121.

Answer 2

Given :-

An Arithmetic sequence 1 , 25 , 49 , 73 , 97 . . . .

To Find :-

The 6th term and the Algebraic form of the sequence

Solution :-

At first , What is the algebraic form of an A.P ;

• It is the representation of an A.P , such that one/two/all of the components of the A.P ( a , d , n ) are unknown and thus we write it's components as variable or if you have to find ( a , d or n only one of them and all others are given ) so just put all known values and the unknown in its variable form !

Consider the Sequence ;

1 , 25 , 49 , 73 , 97 , . . . . .

Here ,

• a = First term = 1
• common difference = d = 25 - 1 = 24

To Find :-

• n = no. of term = 6

We knows that

$\quad \qquad { \bigstar { \underline { \boxed { \tt { a_{n} = a + ( n - 1 ) d } } } } { \bigstar } }$

Putting all the known values we have ;

$\quad { : \longmapsto { \tt { a_{6} = 1 + ( 6 - 1 ) × 24 } } }$

$\quad { : \longmapsto { \tt { a_{6} = 1 + 5 × 24 } } }$

$\quad { : \longmapsto { \tt { a_{6} = 1 + 120 } } }$

$\quad { : \longmapsto { \tt { a_{6} = 121} } }$

Now , As here we have to find the " n " i.e 6 . So , Put all known values except of " n " we get the algebraic form ;

$\quad \qquad { \tt { a_{n} = 1 + ( n - 1 ) × 24 } }$

$\quad { : \longmapsto { \tt { a_{n} = 1 + 24n - 24 } } }$

$\quad { : \longmapsto { \tt { a_{n} = 24n - 23 } } }$

Henceforth , The Required Algebraic form is 24n - 23 and the 6th term is " 121 " .