the sum of the first three terms of an A.P. is 48 . if the product of the first and second terms exceeds four times the third term by 12 ., find the A.P
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Let the three terms be a-d,a,a+d
Sum=a-d+a+a+d=48
Solving this, we get
3a=48
a=16
a(a-d)-4(a+d)=12
Keeping the value of a in above equation
16(16-d)-4(16+d)=12
Solving this, we get
d=9
Therefore terms are 7,16,25 (ans)
Sum=a-d+a+a+d=48
Solving this, we get
3a=48
a=16
a(a-d)-4(a+d)=12
Keeping the value of a in above equation
16(16-d)-4(16+d)=12
Solving this, we get
d=9
Therefore terms are 7,16,25 (ans)
zohanmichey:
thank u soo much
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