ap consists of three terms their sum is 15 and sum of squares of their extremes is 58 find three terms
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Let the three no of Ap be , a, a-d, a+d,
Then a/c to your question
a+a-d+a+d=15
3a=15a
a=5
Now, again
=a^2+(a-d) ^2+(a+d)^2
=a^2+a^2+d^2-2ad+a^2+d^2+2ad
=3a^2+2d^2=58
=3*5^2+2d^2=58
=2d^2=58-75
=2d^2=-17
d=√-17/2
Now l hope you can find on( a-d), a, ( a+d) equestion, here Ap can you find on d= -√17/2.put on those equation, i think you can do it
Then a/c to your question
a+a-d+a+d=15
3a=15a
a=5
Now, again
=a^2+(a-d) ^2+(a+d)^2
=a^2+a^2+d^2-2ad+a^2+d^2+2ad
=3a^2+2d^2=58
=3*5^2+2d^2=58
=2d^2=58-75
=2d^2=-17
d=√-17/2
Now l hope you can find on( a-d), a, ( a+d) equestion, here Ap can you find on d= -√17/2.put on those equation, i think you can do it
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Answered by
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Let the AP be a,a-d,a+d
So sum of them=a+a-d+a+d
15=3a
a=5
So terms became 5,5-d,5+d
Sum of squares of extremes=(a-d)*2+(a+d)*2
58=(5-d)*2+(5+d)*2
58=25+d*2-10d+10d+25+d*2
58=50+2d*2
2d*2=8
d*2=4
d=2 or -2
So terms are
Case (1) if d is 2
a=5,(5-2)=3,5+2=7
Case 2 if d is -
a=5,(5-(-2)=7,5+(-2)=3
So sum of them=a+a-d+a+d
15=3a
a=5
So terms became 5,5-d,5+d
Sum of squares of extremes=(a-d)*2+(a+d)*2
58=(5-d)*2+(5+d)*2
58=25+d*2-10d+10d+25+d*2
58=50+2d*2
2d*2=8
d*2=4
d=2 or -2
So terms are
Case (1) if d is 2
a=5,(5-2)=3,5+2=7
Case 2 if d is -
a=5,(5-(-2)=7,5+(-2)=3
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