In an ap an = 10-3n find s40
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1
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I too don't know but thanks for free points
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Step-by-step explanation:
ANSWER
tn=10-3n
T1 =10-3[1]
= 10-3
=7
T2 =10-3[2]
=10-6
= 4
d = a2 -a1
= 4-7
= -3
Sn = n/2 [ 2a + [n-1] d
S40= 40/2 [ 2 {7} + {40-1} {-3} ]
= 20[14+ + 39{-3}]
= 20 [ 14 + {-117} ]
= 20 [ -103 ]
=-2060
hence answer is -2060
pls mark me as brilliant
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