Question

the third term of an ap is 6 and the ninth term of the ap exceeds three times the 3rd time by 2 find the sum of its first 19 terms

Answer 1

a3 = 6

a + 2d = 6  …(1)

a9 = 3(6)+2

    = 20

a + 8d = 20  …(2)

(2)-(1):

6d = 14

d = 7/3

a = 6-14/3

  = 4/3

Sn = n/2[2a+(n-1)d]

    = 19/2(8/3+126/3)

    =19/2*134/3

Therefore S19 = 1273/3

Answer 2

Hey mate, the answer is in attachment

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