Concept:
In an AP, if the first term is a and the common difference is d then the nth term of the sequence is given by: an = a + (n - 1)d.
In an AP, if the first term is a and the common difference is d then the sum of first n terms of the sequence is given by: Sn=2n×[2a+(n−1)d]
Calculation:
Given: The 3rd term of an AP is -13 and the 6th term is - 4.
Here, we have to find the sum of first 12 terms of the AP.
Let the first term be a and the difference be d.
∵ 3rd term of an AP is -13
⇒ a + (3 - 1)d = -13
⇒ a + 2d = -13 ----(i)
Similarly, the 6th term is - 4
⇒ a + 5d = -4 ----(ii)
Subtracting (i) from (ii)
⇒ 3d = -4 - (-13) = 9
⇒ d = 3
Putting d = 3 in (i), we get
⇒ a + 6 = -13
⇒ a = -19
As we know that, sum of first n term is : Sn=2n×[2a+(n−1)d]
Here, n = 12
s12=212⋅[2×(−;19)+11×3]=;−30