In a certain sculpture, coils of wire are arranged in rows.

Question

In a certain sculpture, coils of wire are arranged in rows. The second row has two more coils than the first, the third two more than the second, and so on, to the tenth and final row. If there is a total of 120 coils of wire in the sculpture, how many coils are in the final row?

a) 22

b) 21

c) 20

d) 19

e) 18

prov · Answer

let first row = x
second row = x+2
third row = x+4
...
tenth row = x+18

total = 10x + 2(1+2+3+...+9)
total = 10x + 2*45 = 120
x = 3
thus, tenth row = x+18 = 21

ans B

Juaz · Answer

B i.e. 21 

If x= coils in first row then
x+(x+2)+(x+4)+....(x+18) = 120
10x + 90 = 120
=> x = 3
x+18 => 21

Whatever · Answer

Ans. B,

the same method as Juaz

alpha_plus_gamma · Answer

Every row has 2 elements more than previous row i.e. arithmetic progression

sum = n  d ] / 2

120 = 10  2]/2

x1= 3

nth term of AP is An= A1 +  d

A10 = 3 +  

A10 = 21

bumpbot · Answer

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In a certain sculpture, coils of wire are arranged in rows.

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