# What is the sum of the 11th to the 15th terms if the first term is 2?The sum of the 6th to the 10th terms of an AP is double that of the first 5 terms.

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First let us denote the first term of the AP by a and the common difference by d. Now the sum of the first n terms is (n/2)*(2a + (n-1)d).

So the sum of the first 5 terms is (5/2)*(2a + 4d). The sum of the first 10 terms is 5*(2a + 9d).

Now, we are given that 5*(2a + 9d) - (5/2)*(2a + 4d) = 2*(5/2)*(2a + 4d)

=> 5*(2a + 9d) = 2*(5/2)*(2a + 4d) + (5/2)*(2a + 4d)

=> (2a + 9d) = (2a + 4d) + (1/2)*(2a + 4d)

=> 2a + 9d = (3/2)(2a + 4d)

=> 2a + 9d = 3a + 6d

=> 3d = a

As we are given that a is 2

=> d = 2/3

The sum of the 10th to the 15th terms is given by (15/2)*(2a + 14d) - (10/2)*(2a + 9d)

=> (15/2)*(2*2 + 14*2/3) - (10/2)*(2*2 + 9*2/3)

=> 50

**The required sum is equal to 50.**