Question

the 11th term & 21th terms of an A.P are 16 and 29 respectively, then find the 37th term . also find 'n' if nth term is 55​

Answer 1

Answer:

Step-by-step explanation:

We know the general term of an AP is $a_{n} =a+(n-1)d$.

Given $11^{th}$ term   $a_{11} =a+10d=16$ . . . . (1)

and $21^{st}$ term      $a_{21} =a+20d=29$ . . . . (2)

We are asked to find the $37^{th}$ term. So, equation (2)-(1) will give the value of $d$,  substitute it in to any of these two equations, we get the value of $a$.

and hence find the $37^{th}$ term.  So,

(2)-(1) ⇒

                 $a+20d=29\ -\\\\a+10d=16$

        ⇒      

                      $10d=13\\\\d=13/10$

applying  the value of  $d$ in (1),

                 $a+10(\frac{13}{10} )=16\\\\a+13=16\\\\a=16-13=3$

Therefore  $37^{th}$ term  $a_{37}=a+36d=3+36(\frac{13}{10} )$

                                         $=3+36(1.3)=49.8$

Hence the answer.

Also we want to find $n$ such that   $a_{n} =55$

That is     $a+(n-1)d=3+(n-1)\frac{13}{10} =55$

                                    $=3+\frac{13n}{10} -\frac{13}{10}$  $=55$

                            ⇒

                                       $\frac{13n}{10}- \frac{13}{10}=55-3=52\\\\13n-13=52*10=520$

                                       $13n=520+13=533\\ \\n=533/13=41$

Therefore it is the $41^{th}$ term.

Hence the answer.

thank you