If the 12 th term of an ap is -13 and the sum of first 4 terms is 24, what is the sum of first 10 terms?
Answers
Answered by
6
a12=-13
s4=24
s10=?
a+11d=-13
therefore, a=-13-11d
s4=4/2×(2a+3d)
s4=4a+6d
s4=4(-13-11d)+6d
s4=-52-44d+6d
s4=-38d-52=24
-38d=76
d=-2
apply 'd'value in a=-13-11d
a=-13-11(-2)
a=-13+22
a=9
s10=10/2×(2a+9d)
s10=5(2(9)+9(-2))
s10=5(18-18)
s10=0
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thank u
s4=24
s10=?
a+11d=-13
therefore, a=-13-11d
s4=4/2×(2a+3d)
s4=4a+6d
s4=4(-13-11d)+6d
s4=-52-44d+6d
s4=-38d-52=24
-38d=76
d=-2
apply 'd'value in a=-13-11d
a=-13-11(-2)
a=-13+22
a=9
s10=10/2×(2a+9d)
s10=5(2(9)+9(-2))
s10=5(18-18)
s10=0
..
..
..
..
thank u
Anonymous:
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Answered by
5
if the 12th term is -13 than
a+11d=-13 and sum of first 4 terms is 24 than
a+(a+d)+(a+2d)+(a+2d)=24 on solving this we get 4a+6d=24
solving both equ we get d=-2 and a=9 therefore sum of first 10 terms is
n/2(a+l)=10/2(9-9)which is equal to 0 so the answer is 0
a+11d=-13 and sum of first 4 terms is 24 than
a+(a+d)+(a+2d)+(a+2d)=24 on solving this we get 4a+6d=24
solving both equ we get d=-2 and a=9 therefore sum of first 10 terms is
n/2(a+l)=10/2(9-9)which is equal to 0 so the answer is 0
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