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The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.

From the question, it is given that,

a_{19}=19^{\text {th }}a

19

=19

th

term of an A.P. is equal to three times its 6^{\text {th }} \text { term }=3 a_{6}6

th

term =3a

6

a_{9}=19a

9

=19

As we know, a_{n}a

n

= a + (n - 1)d

a_{9}a

9

= a + 8d = 19 … [equation (i)]

Then, a_{19}a

19

= 3(a + 5d)

a + 18d = 3a + 15d

3a – a = 18d – 15d

2a = 3d

a = (3/2)d

Now substitute the value of a in equation (i) we get,

(3/2)d + 8d = 19

(3d + 16d)/2 = 19

(19/2)d = 19

d = (19 × 2)/19

d = 2

To find out the value of a substitute the value of d in equation (i)

a + 8d = 19

a + (8 × 2) = 19

a + 16 = 19

a = 19 – 16

a = 3

Therefore, A.P. is 3, 5, 7, 9, …