Question

the 13th term of an ap is 2 times of its 10th term and third term is 6 more than the two times its 6th term then. find the first 3 term​

Answer 1

18 , 15 , 12 are first three terms of AP

Step-by-step explanation:

Let say First term = a

and common difference = d

first three terms

a  , a + d , a + 2d

13th term = a + 12d

10th term = a + 9d

a + 12d  = 2(a + 9d)

=> a= - 6d

3rd term = a + 2d  = -6d + 2d = -4d

6th  term =  a + 5d  = -6d + 5d  = -d

-4d  = 6 + 2(-d)

=> -4d = 6 - 2d

=> -2d  = 6

=> d = -3

a = -6d = - 6(-3) = 18

AP

a  , a + d , a + 2d

18 , 15 , 12

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Answer 2

The first 3 term​s are 18, 15 and 12.

Step-by-step explanation:

The 13th term of an AP is 2 times of its 10th term:

a + 12d = 2(a + 9d)

a + 12d = 2a + 18d

2a - a = 12d - 18d

a = - 6d

The third term is 6 more than the two times its 6th term:

a + 2d = 6 + 2(a + 5d)

a + 2d = 6 + 2a + 10d

2a - a = 2d - 10d - 6

a = -8d - 6

Now, on substituting the value of a, we get,

-6d = -8d - 6

8d -6d = -6

2d = -6

∴ d = -3

⇒ a = - 6(- 3)

∴ a = 18

Now, the first three terms are:

a, a + d, a + 2d = 18, 15, 12