Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.

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Three numbers are in A.P., and their sum is 15. If 1, 3, 9 be added to them respectively, they from a G.P. find the numbers.

Solution:

Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’.

We have, a₁ + a₂ + a₃ = 15

Where the three numbers are as follows:

a, a + d, and a + 2d

So, we can write:

$a\text{ }+\text{ }a\text{ }+\text{ }d\text{ }+\text{ }a\text{ }+\text{ }2d\text{ }=\text{ }15$

$3a\text{ }+\text{ }3d\text{ }=\text{ }15\text{ }$

$or\text{ }a\text{ }+\text{ }d\text{ }=\text{ }5$

$d\text{ }=\text{ }5-a\text{ }\ldots \text{ }\left( i \right)$

Now, according to the question:

a + 1, a + d + 3, and a + 2d + 9

they are in GP, that is:

$\left( a+d+3 \right)/\left( a+1 \right)\text{ }=\text{ }\left( a+2d+9 \right)/\left( a+d+3 \right)$

${{\left( a\text{ }+\text{ }d\text{ }+\text{ }3 \right)}^{2}}~=~\left( a\text{ }+\text{ }2d\text{ }+\text{ }9 \right)\text{ }\left( a\text{ }+\text{ }1 \right)$

${{a}^{2}}~+\text{ }{{d}^{2}}~+\text{ }9\text{ }+\text{ }2ad\text{ }+\text{ }6d\text{ }+\text{ }6a\text{ }=\text{ }{{a}^{2}}~+\text{ }a\text{ }+\text{ }2da\text{ }+\text{ }2d\text{ }+\text{ }9$