Question

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers

Question # 24 in today's board paper, Set 1. Have to check if my answer is correct :)

Answer 1

Let the first three terms of the series be.

a−d,a,a+da−d,a,a+d

Where dd 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+916–9,16,16+9

Which is 7,16,25

Answer 2

refer to the attachment ............................