sn of first n term of ap is n÷2(3n+13) find 30term
ram284:
very bad different answer
very
did you do it
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Just putting the value of n=30
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Sn = n(3n+13)/2
=> n[2a+(n-1)d] / 2= n(3n+13)/2
=> 2a + (n-1)d = 3n +13
=> 2a +(n-1)d = 16+ 3n - 3
=> 2a +(n-1)d = 16 + 3(n-1)
Comparing both sides, we get
2a = 16
a = 8
d= 3
T30 = a+29d
= 8+29*3
= 8 + 87
= 95
=> n[2a+(n-1)d] / 2= n(3n+13)/2
=> 2a + (n-1)d = 3n +13
=> 2a +(n-1)d = 16+ 3n - 3
=> 2a +(n-1)d = 16 + 3(n-1)
Comparing both sides, we get
2a = 16
a = 8
d= 3
T30 = a+29d
= 8+29*3
= 8 + 87
= 95
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