In an AP if 12th term is –13 amd sum of its first four terms is 24 find sum of its first 10 terms
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12th term=a+(n-1)d
-13=a+(12–1)d
-13=a+11d…………………(1
Sum of first four term=(4/2)[2a+(4–1)d]
24. = 2(2a+3d)……………(2
From equation (1) and (2) we get
a=9 and d=-2
Sum of first 10 term
S₁₀=(10/2)[2*9+9(-2)]
S₁₀=5(18–18)=0
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