Question

In an ap 12th term is minus 13 and sum of the first 4 terms is 24 what is the sum of the first 10 terms

Answer 1

T12 = - 13
=> a +11d = 13
=> 2a +22d = - 26 -------(1)

T1 + T2+T3 + T4 = 24
=> a+a+d+a+2d+a+3d = 24
=> 4a +6d = 24
=> 2a +3d = 12 -------(2)

On subtracting equation 2 from 1,we get
19d = - 26 - 12
=> 19d = - 38
=> d = - 2

a = 9

S10 = 10[2*9 + (10-1)(-2)] / 2
= 5 (18 - 18)
= 5*0
= 0

S10 = 0

Answer 2

a12 =-13 given
s4= 24
we can write a12 =a +11d mark this as equation no 1.
s4=n÷2(2a+(n-1)d
putting values
4÷2(2a+(4-1)d
2(2a+3d)=24
4a+6d=24 mark this as equation no.2
now adding eq
1 and 2
a+11d=-13 ×4
4a+6d=24
4a+44d=-52
4a+6d=24
by solving this 38d =-28
d= -28÷38 = .... i think now u can solve this