Question

Three numbers whose sum is 18 are in A.P. If the first is increased by 1, the resulting numbers are in G.P. Find the numbers in A.P.​

Answer 1

Three numbers in A.P can be written as $a-d,a,a+d$.

As it is given that the three numbers sum to 18,

$\hookrightarrow(a-d)+a+(a+d)=18$

$\hookrightarrow 3a=18$

$\large\hookrightarrow\red{\boxed{\red{\bold{\red{a=6}}}}}$

The definition of the geometric progression is that any three consecutive terms a, b and c will satisfy the following equation,

$\hookrightarrow\large\red{\boxed{\bold{\red{b^{2}=ac}}}}$

As it is given that three numbers form a G.P if the first term increases by 1, according to the question we get,

$\hookrightarrow\red{\boxed{\bold{\red{(7-d),6,(6+d)\ (G.P)}}}}$

So we will get,

$\hookrightarrow 36=(7-d)(6+d)$

$\hookrightarrow 36=42+7d-6d-d^{2}$

$\hookrightarrow d^{2}-d-6=0$

$\hookrightarrow(d+2)(d-3)=0$

$\large\hookrightarrow\red{\boxed{\bold{d=-2\ or\ d=3}}}$

So, possible A.P are

$\large\hookrightarrow\bold{(a=6,d=-2)\iff\red{\boxed{\red{8,6,4\ (A.P)}}}}$

$\large\hookrightarrow\bold{(a=6,d=3)\iff\red{\boxed{\red{3,6,9\ (A.P)}}}}$

For verification,

$\large\red{\boxed{\red{8,6,4\ (A.P)}}}\iff\red{\boxed{\red{9,6,4\ (G.P)}}}$

It is a G.P with the first term of 9 and a common ratio of $\dfrac{2}{3}$.

$\large\red{\boxed{\red{3,6,9\ (A.P)}}}\iff\red{\boxed{\red{4,6,9\ (G.P)}}}$

It is a G.P with the first term of 4 and a common ratio of $\dfrac{3}{2}$.