Find number of terms in arithmetic progression












0












$begingroup$


In a arithmetic progression sum of first four terms sum :
$$a_1+a_2+a_3+a_4=124$$



and sum of last four terms :
$$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
and sum of arithmetic progression is :
$$S_n=350$$



$$n=?$$



How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.






sequences-and-series summation arithmetic-progressions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    In a arithmetic progression sum of first four terms sum :
    $$a_1+a_2+a_3+a_4=124$$



    and sum of last four terms :
    $$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
    and sum of arithmetic progression is :
    $$S_n=350$$



    $$n=?$$



    How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.






    sequences-and-series summation arithmetic-progressions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In a arithmetic progression sum of first four terms sum :
      $$a_1+a_2+a_3+a_4=124$$



      and sum of last four terms :
      $$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
      and sum of arithmetic progression is :
      $$S_n=350$$



      $$n=?$$



      How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.






      sequences-and-series summation arithmetic-progressions







      share|cite|improve this question











      $endgroup$




      In a arithmetic progression sum of first four terms sum :
      $$a_1+a_2+a_3+a_4=124$$



      and sum of last four terms :
      $$a_n+a_{n-1}+a_{n-2}+a_{n-3}=156$$
      and sum of arithmetic progression is :
      $$S_n=350$$



      $$n=?$$



      How to find $n$? I tried using arithmetic progression sum formulas but getting negative or fractional numbers.






      sequences-and-series summation arithmetic-progressions




      sequences-and-series summation arithmetic-progressions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 16:19









      Lord Shark the Unknown

      105k1160133




      105k1160133










      asked Dec 14 '18 at 16:15









      Serif YaohimSerif Yaohim

      32




      32






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
          eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
          of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
          off $n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            hohohoh man thank you a lot. God bless you
            $endgroup$
            – Serif Yaohim
            Dec 14 '18 at 16:25



















          1












          $begingroup$

          Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $$350=dfrac{n(a_1+a_n)}2$$



            Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$






            share|cite|improve this answer











            $endgroup$





















              0












              $begingroup$

              Use standard formula of A.P.
              which is



              $a_n = a + (n-1)d$



              and a simple difference formula



              $a_n - a_{n-1} = d$






              share|cite|improve this answer











              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
                eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
                of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
                off $n$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  hohohoh man thank you a lot. God bless you
                  $endgroup$
                  – Serif Yaohim
                  Dec 14 '18 at 16:25
















                1












                $begingroup$

                One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
                eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
                of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
                off $n$.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  hohohoh man thank you a lot. God bless you
                  $endgroup$
                  – Serif Yaohim
                  Dec 14 '18 at 16:25














                1












                1








                1





                $begingroup$

                One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
                eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
                of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
                off $n$.






                share|cite|improve this answer









                $endgroup$



                One has $a_1+a_2+a_3+a_4+a_{n-3}+a_{n-2}+a_{n-1}+a_n=280$. The mean of those
                eight terms is $35$, so the mean of $a_1,ldots,a_n$ is also $35$. The sum
                of those $n$ terms is $n$ times their mean, and is $350$. So now you can read
                off $n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 16:23









                Lord Shark the UnknownLord Shark the Unknown

                105k1160133




                105k1160133












                • $begingroup$
                  hohohoh man thank you a lot. God bless you
                  $endgroup$
                  – Serif Yaohim
                  Dec 14 '18 at 16:25


















                • $begingroup$
                  hohohoh man thank you a lot. God bless you
                  $endgroup$
                  – Serif Yaohim
                  Dec 14 '18 at 16:25
















                $begingroup$
                hohohoh man thank you a lot. God bless you
                $endgroup$
                – Serif Yaohim
                Dec 14 '18 at 16:25




                $begingroup$
                hohohoh man thank you a lot. God bless you
                $endgroup$
                – Serif Yaohim
                Dec 14 '18 at 16:25











                1












                $begingroup$

                Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$






                    share|cite|improve this answer









                    $endgroup$



                    Use the formula $$a_n=a_1+(n-1)d$$ where $$a_{n+1}-a_{n}=d$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 14 '18 at 16:17









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    76.8k42866




                    76.8k42866























                        0












                        $begingroup$

                        $$350=dfrac{n(a_1+a_n)}2$$



                        Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          $$350=dfrac{n(a_1+a_n)}2$$



                          Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$350=dfrac{n(a_1+a_n)}2$$



                            Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$






                            share|cite|improve this answer











                            $endgroup$



                            $$350=dfrac{n(a_1+a_n)}2$$



                            Now $a_1+a_n=a_2+a_{n-1}=cdots=dfrac{124+156}4$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 14 '18 at 16:54

























                            answered Dec 14 '18 at 16:28









                            lab bhattacharjeelab bhattacharjee

                            226k15157275




                            226k15157275























                                0












                                $begingroup$

                                Use standard formula of A.P.
                                which is



                                $a_n = a + (n-1)d$



                                and a simple difference formula



                                $a_n - a_{n-1} = d$






                                share|cite|improve this answer











                                $endgroup$


















                                  0












                                  $begingroup$

                                  Use standard formula of A.P.
                                  which is



                                  $a_n = a + (n-1)d$



                                  and a simple difference formula



                                  $a_n - a_{n-1} = d$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Use standard formula of A.P.
                                    which is



                                    $a_n = a + (n-1)d$



                                    and a simple difference formula



                                    $a_n - a_{n-1} = d$






                                    share|cite|improve this answer











                                    $endgroup$



                                    Use standard formula of A.P.
                                    which is



                                    $a_n = a + (n-1)d$



                                    and a simple difference formula



                                    $a_n - a_{n-1} = d$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 18 '18 at 19:46









                                    Prakhar Nagpal

                                    752318




                                    752318










                                    answered Dec 18 '18 at 19:32







                                    user578581





































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