Question

for an ap 1,7,13,19....find t18,t31,t56,t101,t202​

Answer 1

Answer :-

Given :-

Arthimetic progession : 1 , 7 , 13 , 19 , - - - - - -

Required to find :-

• 18th term of the sequence ?

• 31st term of the sequence ?

• 56th term of the sequence ?

• 101th term of the sequence ?

• 202th term of the sequence ?

Formula used :-

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

Solution :-

Given arithmetic progession :-

a.p = 1 , 7 , 13 , 19 - - - - - -

He asked us to find the some of the terms in the sequence .

So,

Using the formula ,

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

Here,

a = first term

d = common difference

nth = the term which you want to find

Hence,

Consider the given arithmetic progession

a.p = 1 , 7 , 13 , 19 - - - - - -

Here,

The first term is 1 .

So, a = 1

Similarly,

Common difference = ( 2nd term - 1st term ) = ( 3rd term - 2nd term )

=> ( 7 - 1 ) = ( 13 - 7 )

=> (6) = (6)

Hence,

Common difference = 6

So, d = 6

Now , we can substitute these values in the formula .

Hence,

Let's find the 18th term . ( using the formula )

$\longrightarrow{\tt{ {a}_{nth} = {a}_{18}}}$

So,

$\longrightarrow{\tt{ {a}_{18} = 1 + ( 18 - 1 ) 6 }}$

$\longrightarrow{\tt{ {a}_{18} = 1 + (17)6 }}$

$\longrightarrow{\tt{ {a}_{18} = 1 + 102 }}$

$\longrightarrow{\tt{ {a}_{18} = 103 }}$

$\large{\leadsto{\boxed{\rm{18th \; term = 103 }}}}$

Similarly, let's find the 31st term

$\longrightarrow{\tt{ {a}_{nth} = {a}_{31}}}$

So,

$\longrightarrow{\tt{ {a}_{31} = 1 + (31 - 1 )6 }}$

$\longrightarrow{\tt{ {a}_{31} = 1 + (30)6 }}$

$\longrightarrow{\tt{ {a}_{31} = 1 + 180 }}$

$\longrightarrow{\tt{ {a}_{31} = 181 }}$

$\large{\leadsto{\boxed{\rm{31st \; term = 181 }}}}$

Let's find the 56th term

$\longrightarrow{\tt{ {a}_{nth} = {a}_{56}}}$

So,

$\longrightarrow{\tt{ {a}_{56} = 1 + ( 56 - 1 )6 }}$

$\longrightarrow{\tt{ {a}_{56} = 1 + ( 55 )6 }}$

$\longrightarrow{\tt{ {a}_{56} = 1 + 330 }}$

$\longrightarrow{\tt{ {a}_{56} = 331 }}$

$\large{\leadsto{\boxed{\rm{56th \; term = 331}}}}$

Let's find the 101th term

$\longrightarrow{\tt{ {a}_{nth} = {a}_{101}}}$

so,

$\longrightarrow{\tt{ {a}_{101} = 1 + ( 101 - 1 )6 }}$

$\longrightarrow{\tt{ {a}_{101} = 1 + ( 100 )6 }}$

$\longrightarrow{\tt{ {a}_{101} = 1 + 600 }}$

$\longrightarrow{\tt{ {a}_{101} = 601 }}$

$\large{\leadsto{\boxed{\rm{101th \; term = 601 }}}}$

Let's find the 202th term

$\longrightarrow{\tt{ {a}_{nth} = {a}_{202}}}$

So,

$\longrightarrow{\tt{ {a}_{202} = 1 + ( 202 - 1 )6 }}$

$\longrightarrow{\tt{ {a}_{202} = 1 + ( 201 )6 }}$

$\longrightarrow{\tt{ {a}_{202} = 1 + 1206 }}$

$\longrightarrow{\tt{ {a}_{202} = 1207 }}$

$\large{\leadsto{\boxed{\rm{202th \; term = 1207 }}}}$