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17 votes
32+40+…+120=? Someone help PLEASE

User Stoatman
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2 Answers

19 votes
19 votes
The answers u are looking for us 190
User Hrunting
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8 votes
8 votes

Answer:

912

Explanation:

the assumption is that this is an arithmetic progression

the nth term of an arithmetic progression is


a_(n) = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

use this to find which term 120 is in the sequence

with a₁ = 32 and d = a₂ - a₁ = 40 - 32 = 8 , then

32 + 8(n - 1) = 120 ( subtract 32 from both sides )

8(n - 1) = 88 ( divide both sides by 8 )

n - 1 = 11 ( add 1 to both sides )

n = 12

given the first and last terms in the sequence then sum is


S_(n) =
(n)/(2) ( first + last)

S₁₂ =
(12)/(2) (32 + 120) = 6 × 152 = 912

User Jim Harte
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