find the sum of the ap whose 7th term is 24 less than the 11 th term first term being 12?
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a= 12
ATQ,
a7= a11-24
a7= a+6d
a11= a+10d
a+6d= a+10d -24
12+ 6d= 12 + 10d -24
12+6d-12-10d= - 24
-4d = - 24
d=24/4=6
Sn = n/2 [2a+(n-1)d]
Sn = n/2 [2(12)+(n-1)6]
Sn = n/2 [ 24+ 6n-6]
Sn = n/2 [ 18-6n]
taking 2 common
Sn = n/2 ×2 [ 9-3n]
Sn = 9n-3n^2
here is your answer mate.
hope it helps you.
please mark it as brainliest.
ATQ,
a7= a11-24
a7= a+6d
a11= a+10d
a+6d= a+10d -24
12+ 6d= 12 + 10d -24
12+6d-12-10d= - 24
-4d = - 24
d=24/4=6
Sn = n/2 [2a+(n-1)d]
Sn = n/2 [2(12)+(n-1)6]
Sn = n/2 [ 24+ 6n-6]
Sn = n/2 [ 18-6n]
taking 2 common
Sn = n/2 ×2 [ 9-3n]
Sn = 9n-3n^2
here is your answer mate.
hope it helps you.
please mark it as brainliest.
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