Question

in an ap if the 12th term is -13 and the sum of the first four terms is 24 what is the sum of the first 10 terms

Answer 1

a 12 = -13

since, a nth term = a+(n-1)d

a 12 = a+11d = -13

a = -13-11d 》eq 1

S4 = 24

since, s nth term = n/2 (2a + (n-1)d)

S4 = 4/2 (2a + 3d) = 24

2 (2a + 3d) = 24

2a + 3d = 12 》eq 2

from eq 1

a = -13-11d 》eq 3

from eq 2

a = 12-3d/2 》eq 4

by equating 3 & 4

-13-11d = 12-3d/2

(-13-11d)×2 = 12-3d

-26-22d = 12-3d

-38 = 19d

d = -2

substituting 'd' in eq 3

a = -13+22

a = 9

sum of first 10 terms = 10/2 (2×9 + (10-1)×-2)

S10 = 5×(18-18)

S10 = 5×(0)

therefore, S10 = 0

Answer 2

Given,t₁₂=-13
a+11d=-13----------(1)
S₄=24
2(2a+(4-1)d)=24
2a+3d=12-------(2)
Eq(1)×2,
2a+22d=-26-------(3)
Eq(3)-eq(2),
19d=-38
d=-2
Substituting d=-2 in eq(1),
a+11(-2)=-13
a=(-13)+22
a=9
S₁₀=5(18+(10-1)(-2))
S₁₀=5(18-18)
S₁₀=0.
Therefore,sum of first 10 terms is 0.