Question

If Sn = 2165 ,then find the possible values of n , t1 and tn​

Answer 1

Answer:

fds

Step-by-step explanation:

Answer 2

given a sum of series Sn= 2165

Explanation:

1. let Sn be the sum of first 'n' terms of a series with its first term $t_1$, last term

  $t_n$ and common difference between the consecutive terms $d$.

2.then we have , $t_n=t_1+d(n-1)$    ----(i)

                            $s_n=n(\frac{t_n+t_1}{2})=2165$

 we get , $n(t_1+t_n) = 4330=433(10)$

3. since $433$ is a prime number we can say that ,

                        $n=10$

                        $t_1 + t_n = 433$    -----(ii)

4. from (i) we have , $t_n= t_1+ 9(d)$

   putting this in (ii) we get ,

                             $2(t_1) + 9(d)= 433$

5. solving the above equation we have one possible solution,  

                    $t_1=212\ and\ d=1$  

6. from (i) we get , $t_n= t_{10} = 221$