Question
The sum of four consecutive numbers in an A.P. with common difference d>0 is 20. If the sum of their squares is 120, then the middle terms are __ and __.
Answer: Option B
:
B
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B
Let the numbers be a−3d,a−d,a+d and a+3d. These numbers form an AP with first term a−3d and common difference 2d>0.
Also, given that (a−3d)+(a−d)+(a+d)+(a+3d)=20.
⇒4a=20
⇒a=5
Also, (a−3d)2+(a−d)2+(a+d)2+(a+3d)2=120
⇒4a2+20d2=120
⇒4(5)2+20d2=120
⇒d2=1
⇒d=±1
SIince d>0, we must have d=1.
Hence, the numbers are 2, 4, 6, and 8.
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