Question

The third term of an arithmetic progression is 8 and its ninth term exceeds three
times the third term by 2 find the sum of the first 19 terms.​

Answer 1

Answer:

551

Step-by-step explanation:

given, 3rd term of AP = 8

also, given that 9th term =  3(8) + 2

=> 9th term = 24+2 = 26

so, 3rd term = a + 2d = 8

    9th term =  a + 8d = 26  

                            -6d = -18

=> d = 18/6 = 3

we have a+2d = 8

=> a+ 2(3) = 8

=> a = 8-6

=> a = 2

so, sum up to 19terms = $\frac{n}{2}$[2a+(n-1)d]

                                     = $\frac{19}{2}$[2(2)+(19-1)3]

                                     = $\frac{19}{2}$[4+(18)3]

                                     = $\frac{19}{2}$[4+54] = $\frac{19}{2}$*58

                                     = 19*29

                                     = 551