In A.P 6, 12, 18 find S40
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Answered by
2
Given sequence of AP is
6, 12, 18, 24, ....................
Here,
first term, a = 6
Common difference, d = 12 - 6 = 6
So, S(40) = sum upto 40 terms
>> Sn = n/2[2a + (n - 1)d]
Where n = no. Of terms
Now, S(40) = 40/2[2*6 + (40 - 1)6]
→ S(40) = 20[12 + 39*6]
→ S(40) = 20(12 + 234)
→ S(40) = 20(246)
→ S(40) = 4920 ______________(Ans.)
Answered by
2
Answer:
4920
Step-by-step explanation:
S40 = 20[12+39(6)]
= 20[246]
= 4920
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