The sum of three numbers in A.P. is 36. When the numbers are increased by 1, 4, 43 respectively, the resulting numbers are in G.P. Find the numbers.

Let the three numbers in Arithmetic Progression (A.P.) be: $$a-d,a,a+d$$ where:

• $a$ is the middle term,
• $d$ is the common difference.

Step 1: Use the sum condition

The sum of the three numbers is 36: $$(a-d)+a+(a+d)=36$$ Simplify: $$3a=36⟹a=12$$ So, the three numbers are: $$12-d,12,12+d$$

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Step 2: Apply the Geometric Progression (G.P.) condition

When the numbers are increased by 1, 4, and 43 respectively, they form a G.P.: $$(12-d)+1,12+4,(12+d)+43$$ Simplify: $$13-d,16,55+d$$ Since these form a G.P., the square of the middle term equals the product of the first and third terms: $$(16{)}^2=(13-d)(55+d)$$ Calculate: $$256=(13-d)(55+d)$$ Expand the right-hand side: $$256=13\times 55+13d-55d-d^2$$ $$256=715-42d-d^2$$ Rearrange to form a quadratic equation: $$d^2+42d-459=0$$

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Step 3: Solve the quadratic equation

The quadratic equation is: $$d^2+42d-459=0$$ Use the quadratic formula: $$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ where $a=1$, $b=42$, and $c=-459$: $$d=\frac{-42\pm \sqrt{(42{)}^2-4\times 1\times (-459)}}{2\times 1}$$ Calculate the discriminant: $$\sqrt{1764+1836}=\sqrt{3600}=60$$ Thus: $$d=\frac{-42\pm 60}{2}$$ This gives two solutions:

1. $d=\frac{-42+60}{2}=\frac{18}{2}=9$
2. $d=\frac{-42-60}{2}=\frac{-102}{2}=-51$

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Step 4: Find the three numbers for each case

Case 1: $d=9$

The three numbers are: $$12-9=3,12,12+9=21$$ Verify the G.P. condition: $$3+1=4,12+4=16,21+43=64$$ Check if $4,16,64$ form a G.P.: $$\frac{16}{4}=4,\frac{64}{16}=4$$ The common ratio is consistent, so this is valid.

Case 2: $d=-51$

The three numbers are: $$12-(-51)=63,12,12+(-51)=-39$$ Verify the G.P. condition: $$63+1=64,12+4=16,-39+43=4$$ Check if $64,16,4$ form a G.P.: $$\frac{16}{64}=\frac{1}{4},\frac{4}{16}=\frac{1}{4}$$ The common ratio is consistent, so this is also valid.

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Final Answer

The three numbers are: 3, 12, and 21 or 63, 12, and -39.