Question

13. The 19 term of an AP is equal to 3 times its 6" term. If its 9h term is 19, find the AP​

Answer 1

Step-by-step explanation:

a + 18d = 3(a+5d)

a+18d=3a+15d

2a+3d=0 ----(1)

it is given that a+8d=19-----(2)

multiplying sec eqn by 2

we get

2a + 16d=38----(3)

subtracting eqn 1 and 3

13d=38

d=38/13

Answer 2

Answer:

Step-by-step explanation: Let in an AP first term = a,

Common difference =d

9th term = 19

a+8d=19---------------------------------(1)

19th term =3*(6th term)

a+18d=3(a+5d)

a+18d=3a+15d

0=3a+15d-a-18d

2a-3d=0---------------------------------------------------------(2)

subtract (2) from (1)

-19d=-38

d=2

substitute d=2 in--------------------------------(1)

a=3

therefore

a=3,d=2

required AP is

3,5,7,9,....

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