Question

The sum of the three number in AP is 9. if 4 is added to the
third term then the resulting numbers are in Gp. Find the
numbers
​

Answer 1

Answer:

Let the numbers a-d , a and a+d

Given

$a - d + a + a + d = 9 \\ 3a = 9 \\ a = 3$

Add 4 in third term

$a + d + 4 = 3 + d + 4 = 7 + d$

Condition that

3- d , 3 and 7+d are in gp

${3}^{2} = (3- d)(7 + d) \\ \\ 9 = 21 + 3d - 7d - {d}^{2} \\ \\ {d}^{2} + 4d - 12 = 0 \\ \\ (d + 6)(d - 2) = 0 \\ \\ d = - 6 \: \: or \: \: 2$

So numbers are

a- d , a and a+d

a = 3 and d = -6 or 2

which means

9 , 3 and -3

or

1 , 3 and 5

Answer 2

The numbers are 1,3,5 or 9,3,-3.

• Let the three numbers in AP be a-d,a,a+d.(where d is the common difference)
• Sum of the numbers in AP is 9

a - d + a + a + d =9

a=3

• The resulting AP becomes 3-d,3,3+d
• Now when 4 is added to the third term that is 'a+d' the resulting sequence becomes GP.
• Let us assume the three numbers in GP to be $\frac{a}{r}$ , a , ar (where r is the common ratio)
• now that we know one number from the AP is 3.
• We equate the first and the third term of the AP and the GP

First term :

3-d = $\frac{3}{r}$

d = 3 - $\frac{3}{r}$

Third term :

Third term of AP + 4= Third term of GP

(3 + d) + 4= 3r

• Now substituting the value of d in this equation we get

7 + 3 - $\frac{3}{r}$ = 3r

3r² - 10r + 3 = 0

• On solving this quadratic equation we get , r=3 and r = $\frac{1}{3}$
• This shows that there can be 2 possible set of numbers that satisfy the given condition.
• Substituting r=3 we get the numbers in GP as 1,3,9.
• The original numbers that are in AP are 1,3,5.
• Now substituting the value of r=$\frac{1}{3}$ we get the numbers in GP as 9,3,1.
• The original numbers in AP are 9,3,-3.