T3_{3}3 = 6
⇒ d−6−2\frac{-6}{-2}−2−6 = 3
Substitute 3 for d in equation (i)
Sn_{n}n = n(2a+(n−1)d)2\frac{n(2a + (n - 1)d)}{2}2n(2a+(n−1)d)
⇒ S12_{12}12 = 12(2×0+(12−1)3)2\frac{12(2 \times 0 + (12 - 1)3)}{2}212(2×0+(12−1)3)
⇒ S12_{12}12 = 6(0 + 11 x 3)
