Sum of first three terms in anathematic progression is 24 and the sum of square is 224 find first three terms of the arthamatic progression
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Sum of first three terms in anathematic progression is 24 and the sum of square is 224 find first three terms of the arthamatic progression.
Answer-
Let the three terms of an A.P are
(a+d), a, (a-d).
a+d+a+a-d = 24
3a = 24
a = 24/3
a = 8............. (1 )
And,
(a+d)^2+a^2+(a-d)^2 = 224
a^2+d^2+2ad+a^2+a^2+d^2-2ad=224
3a^2+2b^2 = 224............( 2 )
from equation 1 put the value of a in equation 2 ,we get
3 × 8^2 +2d^2 = 224
3 × 64+2d^2 = 224
192 + 2d^2 = 224
2d^2 = 224-192
2d^2= 32
d^2 = 32/2
ď^2 = 16
d = +_4
First three terms of an A. P are
a+d = 8+4 = 12
a = 8
a-d = 8-4 = 4
Or
a+d = 8-4 = 4
a = 8
a-d =8-(-4)= 12
Answer-
Let the three terms of an A.P are
(a+d), a, (a-d).
a+d+a+a-d = 24
3a = 24
a = 24/3
a = 8............. (1 )
And,
(a+d)^2+a^2+(a-d)^2 = 224
a^2+d^2+2ad+a^2+a^2+d^2-2ad=224
3a^2+2b^2 = 224............( 2 )
from equation 1 put the value of a in equation 2 ,we get
3 × 8^2 +2d^2 = 224
3 × 64+2d^2 = 224
192 + 2d^2 = 224
2d^2 = 224-192
2d^2= 32
d^2 = 32/2
ď^2 = 16
d = +_4
First three terms of an A. P are
a+d = 8+4 = 12
a = 8
a-d = 8-4 = 4
Or
a+d = 8-4 = 4
a = 8
a-d =8-(-4)= 12
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