the 12th term of an ap is minus 13 and the sum of its first 4 terms is 24 find the sum of the first 10 terms
neha374:
plzz solve my sum
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Given T12 = -13
a+11d = -13____(1)
Sum of 1st four term is 24
Sn = n/2(2a+(n-1)d
Sn= a+3d= 12_____(2)
Multiplying eqn (1) by 2 and subtracting it from (2) we get
-19d=38
d= -2
So a -22= -13
a= 9
So again a=9,d=-2,n=10
S10=10/2[2*9(10-1)(-2)]
=5*0=0
Hence, sum of the first 10 terms is 0.
a+11d = -13____(1)
Sum of 1st four term is 24
Sn = n/2(2a+(n-1)d
Sn= a+3d= 12_____(2)
Multiplying eqn (1) by 2 and subtracting it from (2) we get
-19d=38
d= -2
So a -22= -13
a= 9
So again a=9,d=-2,n=10
S10=10/2[2*9(10-1)(-2)]
=5*0=0
Hence, sum of the first 10 terms is 0.
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