Question

8th term of an ap is 10 and common difference is 5 then find it's 19th term
please don't give me wrong answer then i will mark you​

Answer 1

Given :-

8th term of an AP is 10

common difference = 5

Required to find :-

• 19th term of AP

Formula used :-

$\huge{\dagger{\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}}}$

Solution :-

Given that :-

8th term of an AP = 10

Common difference ( d ) = 5

We need to find the 19th term of the arithmetic progession .

So,

In order to find the 19th term of the arithmetic progession we should find the first term of the sequence .

Hence,

8th term of the arithmetic progession can be written as ;

a + 7d = 10

However,

Common difference ( d ) = 5

So,

➦ a + 7d = 10

➦ a + 7 ( 5 ) = 10

➦ a + 35 = 10

➦ a = 10 - 35

➦ a = - 25

Hence,

The first term of the A.P is - 25

Similarly,

The values which we have are ;

First term ( a ) = - 25

Common difference (d ) = 5

Using the formula ,

$\huge{\dagger{\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}}}$

here,

a = first term

d = common difference

n = the term number which you want to find

$\rightarrow{\tt{ {a}_{nth} = {a}_{19} }}$

By substituting the values

$\rightarrow{\tt{ {a}_{19} = - 25 + ( 19 - 1 ) 5 }}$

$\rightarrow{\tt{ {a}_{19} = - 25 + ( 18 ) 5 }}$

$\rightarrow{\tt{ {a}_{19} = - 25 + 90 }}$

$\rightarrow{\tt{ {a}_{19} = 65 }}$

$\huge{\dagger{\boxed{\sf{\therefore{ 19th \; term \; = \; 65 }}}}}$

Additional Information :-

1. The arithmetic sequence of the given one till 19th term is represented as ,

a = -25 , -20 , -15 , - - - - - - , 65

2. Formula for finding the sum of nth terms is

$\large{\dagger{\boxed{\sf{ {s}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}}}$

Answer 2

Given ,

The 8th term of an AP is 10 and the common difference is 5

We know that , the general formula of an AP is given by

$\star \: \: \mathtt{ \fbox{ a_{n} = a + (n - 1)d}}$

Thus ,

$\Rightarrow \sf 10 = a + (8 - 1)5 \\ \\ \Rightarrow \sf 10 = a + 35 \\ \\ \Rightarrow \sf a = - 25$

Now , we have to find the 19th term

Thus ,

$\Rightarrow \sf a_{19} = - 25 + (19 - 1)5 \\ \\ \Rightarrow \sf a_{19} = - 25 + 90 \\ \\ \Rightarrow \sf a_{19} = 65$

$\therefore \sf \bold{ \underline{The \: 19th \: term \: is \: 65 }}$