The sum of first 6 terms of an arithmetic sequence is 99.its 6th term is 39.what is the sum of its 3rd and 4th terms?

Question

10th  class, scert, first terminal examination 2018-2019

Answer 1

The sum of the first N terms of the arithmetic sequence

is

 

 N*(A1 + An)/2 = SUM

 

 6*(A1 + A6)/2 = 99

 

6*(A1 + 39)/2 = 99

 

3(a1+39) = 99

 

a1 + 39 = 33

 

a1 = -6

 

 

Next for divided difference D, An = a1 + (n-1)*D

 

                                           39 = -6 + (6-1)*D

 

                                          39 = -6 + 5*D

                                         

                                            45 = 5D

                                               D = 9

 

The sequence is -6, 3, 12, 21, 30, 39

 

12+21 = 33