Sum of first 4 terms of an AP is 6 and sum of first 6 terms is 4. Find sum of first 12 terms.

Solution:

Formula:
Sn = n/2 [2a + (n−1)d]

Given:

S4 = 6
2[2a + 3d] = 6
2a + 3d = 3

S6 = 4
3[2a + 5d] = 4
2a + 5d = 4/3

Subtracting:

2d = 4/3 − 3
2d = −5/3
d = −5/6

Now,

2a + 3(−5/6) = 3
2a = 11/2
a = 11/4

Now find S12:

S12 = 12/2 [2a + 11d]

= 6[(11/2) + 11(−5/6)]

= 6[(33 − 55)/6]

= 6(−22/6)

= −22

Answer: −22