Question

find out 50th term
4th term 12
10th term 30
​

Answer 1

Question :-

Find out the 50th term of an AP . Whose 4th term is 12 and 10th term is 30 .

Answer :-

Given :-

4th term = 12

10th term = 30

Required to find :-

• 50 term of the AP ?

Formula used :-

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

Solution :-

Given data :-

4th term = 12

10th term = 30

we need to find the 50th term of the AP .

So,

Let's consider the given information ;

4th term = 12

10th term = 30

However,

The 4th term can also be represented as " a + 3d "

=> a + 3d = 12 $\longrightarrow{\tt{Equation - 1 }}$

Consider this as equation - 1

Similarly,

The 10th term can be represented as " a + 9d "

=> a + 9d = 30 $\longrightarrow{\tt{ Equation - 2 }}$

consider this as equation - 2

Now,

Let's solve these 2 equations simultaneously using the elimination method .

$\tt{a + 9d = 30} \\ \tt{a + 3d = 12} \\ \underline{( - )( - ) \: ( - ) \: } \\ \underline{\tt{ \: \: \: \: \: \: \: 6d = 18}} \\ \\ \implies \tt{6d = 18} \\ \implies \tt{d = \frac{18}{6} } \\ \implies \tt{d = 3}$

Hence,

• Common difference ( d ) = 3

Substitute the value of d in equation 1

a + 3d = 12

a + 3 ( 3 ) = 12

a + 6 = 12

a = 12 - 6

a = 6

Hence,

• First term ( a ) = 6

Using the formula ;

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

Let's find the 50th term ;

This implies ;

$\rm{ {a}_{nth} = {a}_{50} }$

$\rm{ {a}_{50} = 6 + ( 50 - 1 ) 3 }$

$\rm{ {a}_{50} = 6 + ( 49 ) 3 }$

$\rm{ {a}_{50} = 6 + 147 }$

$\rm{ {a}_{50} = 153 }$

Therefore ,

50th term = 153

Answer 2

$\LARGE\textbf{\mathrm{ANSWER}}$

$\Large\mathrm{ 4^{th} term = 12}$

$\rm 10^{th} term = 30$

$\sf Common \:difference = {\frac{Term \: difference}{Position \: difference}$

$\sf Common \:difference = \frac{30-12}{10-4}$

                          $\sf = \frac{18}{6}$

                          $\sf={ 3}$

$\sf First \: term = 4^{th} \: term -3d$

             $\sf = 12-3\times3$

             $\sf =12-9$

             $\sf = 3$

$\sf 50^{th} \: term = 1^{st} term +49d$

             $\sf = 3 + 49\times 3$

             $\sf = 3+147$

             $\sf =150$

HOPE IT HELPS U ......