The sum of four terms in AP is 32 and the sum of their squares is 336. Find the numbers
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2
Answer:
a+a+d+a+2d+a+3d=32
4a+6d=32
2a+3d=16
squaring on both sides 4a^2+9d^2+12ad=256. 1st en
sum of their squares is 336
a^2+(a+d)^2+(a+2d)^2+(a+3d)^2=336
4a^2+14d^2+12ad=336 2nd eqn
solve 1 and 2
we will get d=4
a=2
the numbers are 2,6, 10,14
Answered by
2
Let the 4 terms be ,
( a - 3d ) , (a - d ) , ( a + d ) & ( a + 3d )
THEIR SUM IS 32 SO....
=>(a-d) + (a-3d) + (a+d) + (a+3d) = 32
=> 4a = 32
=> a = 8
SUM OF THEIR SQUARES IS 336 SO...
=> (a-d)² + (a-3d)² + (a+d)² + (a+3d)² = 336
=> 4a² + 20d² = 336
=> a² + 5d² = 84
=> 8² + 5d² = 84
=> 5d² = 84 - 64
=> 5d² = 20
=> d² = 20/5
=> d² = 4
=> d = 2
THUS,
a = 8
d = 2
AP = 8 , 10 , 12 , 14.....
HOPE IT HELPS !!
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