Question

If the nth term of an AP is $\left( 5n-2 \right)$ . find its first term and common difference. Also finds its $19th$ term.

Answer 1

Hint: from the question given we have to find the first term, common difference and its $19th$ term, of the series of an AP whose nth term is $\left( 5n-2 \right)$. As we know that the if “a” is first term of an arithmetic progression series and “d” is common difference then the nth term of that series will be equal to ${{t}_{n}}=a+\left( n-1 \right)d$.

Complete step by step solution:
From the question given that the nth term of the arithmetic progression is
$\Rightarrow {{t}_{n}}=\left( 5n-2 \right)$
As we know that the if “a” is first term of an arithmetic progression series and “d” is common difference then the nth term of that series will be equal to
$\Rightarrow {{t}_{n}}=a+\left( n-1 \right)d$
Now we will rearrange the general nth term of an arithmetic progression as
$\Rightarrow {{t}_{n}}=a+\left( n-1 \right)d=a+nd-d=nd+\left( a-d \right)$
Now we will compare the general nth term of an arithmetic progression with the given nth term,
$\Rightarrow {{t}_{n}}=\left( 5n-2 \right)=nd+\left( a-d \right)$
By comparing we will get the,
$\Rightarrow d=5\,and\ a-d=-2$
From this we got the value of d, now the value of ‘a’ will be equal to,
$\Rightarrow \ a-d=-2$
As we know that the value of d, we will substitute in this
$\Rightarrow \ a-5=-2$
$\Rightarrow a=5-2=3$
Therefore, the first term of an arithmetic progression is $3$and the common difference is $5$
Now, the $19th$ term will be,
$\Rightarrow {{t}_{n}}=a+\left( n-1 \right)d$
$\Rightarrow {{t}_{19}}=3+\left( 19-1 \right)5$
By further simplification we will get,
$\Rightarrow {{t}_{19}}=3+\left( 18 \right)5$
$\Rightarrow {{t}_{19}}=3+90$
$\Rightarrow {{t}_{19}}=93$
Therefore, the $19th$ term is $93$.

Note: Students should know the all the formulas of arithmetic progression along with these students should also know about geometric progression and harmonic progression, students should also know the sum of n terms in an arithmetic progression is ${{s}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.