For an AP with first term a and the common difference d, the sum of the first n terms is given as (2a + (n-1)d)(n/2).

Now for the AP given the sum of the first 15 terms is 105

=> (2a + 14d)(15/2) = 105...(1)

The sum of the next...

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For an AP with first term a and the common difference d, the sum of the first n terms is given as (2a + (n-1)d)(n/2).

Now for the AP given the sum of the first 15 terms is 105

=> (2a + 14d)(15/2) = 105...(1)

The sum of the next 15 terms is the sum of the first 15 terms subtracted from the sum of the first 30 terms.

=> (2a + 29d)(30/2) - 105 = 780

=> (2a + 29d) = 885/15

=> (2a + 29d) = 59

(1)

=> (2a + 14d)(15/2) = 105

=> 2a + 14d = 14

2a + 29d - 2a - 14d = 59 - 14

=> 15d = 45

=> d = 3

2a + 14d = 14

=> 2a = 14 - 42

=> 2a = -28

=> a = -14

**Therefore the AP has the first term as -14 and the common difference is 3.**