Question

the sum of the first three no. in an A.P is 18. if the product of the first and third term is 5 times the common difference , find the three numbers.

Answer 1

The sum of first 3 terms of an AP is 48 if the product of the first and second term exceed 4 times the third term by 12, what is AP?

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The sum of 3 terms of an AP is 30. What is the first term?

The sum of the first 25 terms of an AP is 0. Which term of this AP will be 0?

The ninth term of AP is 5 and the fourth term is 26, what is the sum of the first 7 terms of the AP?

If the nth term of an AP is (2n+1), what is the last first 8 term of AP?

Find the product of the first three terms of GP whose second term is 5?

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

Answer 2

The three numbers are "2, 6 and 10 or 10, 6, 2".

Step-by-step explanation:

Let a - d, a and a + d are the three terms of an AP.

To find, the three numbers = ?

According to question,

$a - d+a+a+d=18$

⇒ $3a=18$

⇒ a = 6

Also,

$(a - d)(a+d)=5d$

⇒ $a^{2}-d^{2}  =5d$

⇒ $6^{2}-d^{2}  =5d$

⇒$d^{2} +5d-36=0$

⇒$d^{2} +9d-4d-36=0$

⇒$(d-4)(d+9)=0$

∴ d = 4 or, - 9

The three numbers are:

6 - 4, 6 and 6 + 4  or 2, 6 and 10 or 10, 6, 2.

Hence, the three numbers are 2, 6 and 10 or 10, 6, 2.