The sum of first three numbers in arithmetic progression is 18 if the product of the first and the third term is 5 times the common difference find the three numbers
Answers
Answered by
2
Let the first three terms of the series be.
a−d,a,a+d
Where d is the common difference.
Now as per the problem.
[math]a-d + a + a+d =48
3a=48
a=16[/math]
Product of the first and second term exceeds the 4 times the third term by 12 ie
[math](a-d)*a=4*(a+d)+12
Substitute a=16 in above equation
(16-d)*16=4*(16+d) + 12
256–16d=76 +4d
20d=256–76
d=180/20
d=9[/math]
Thus our required series is
16–9,16,16+9
Which is 7,16,25
a−d,a,a+d
Where d is the common difference.
Now as per the problem.
[math]a-d + a + a+d =48
3a=48
a=16[/math]
Product of the first and second term exceeds the 4 times the third term by 12 ie
[math](a-d)*a=4*(a+d)+12
Substitute a=16 in above equation
(16-d)*16=4*(16+d) + 12
256–16d=76 +4d
20d=256–76
d=180/20
d=9[/math]
Thus our required series is
16–9,16,16+9
Which is 7,16,25
Attachments:
bharatishita2:
it's wrong
Similar questions