Question

If 12th term of an A.P is -13&the sum of its first 4 terms is 24 what is the sum of its first 10 terms.

Answer 1

a12=-13
s4=24
s10=?
an=a+(n-1)d
-13=a+(12-1)d
-13=a+11d-------------1st eqn

sn= n/2[2a+(n-1)d]
s4=4/2(2a+(4-1)d)
24=2(2a+3d)-------------2nd eqn

eliminate the eqn

2(-13=a+11d)
-26=2a+22d----------3rd eqn

subtract eqn 3rd from 2nd

12=2a+3d
-26=2a+22d
+ - -
___________
38= -19d
38/-19 =d
d= -2
put the value of d in eqn 1st
-13=a+11d
-13=a+11(-2)
-13=a-22
a=-13+22
a=9


we know s=n/2(2a+(n-1)d)
s10= 10/2[2(9)+(10-1)(-2)]
s10= 5[18+9(-2)]
s10=5[18-18]
s10=5(0)
s10=0 Answer