Question

Three non-zero numbers A, B and C are in A P. Increasing the number A by 1 or increasing the number C by 2, the numbers are in GP then, the values of A, B and C are?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Three non - zero numbers, A, B, C are in AP.

We know, 3 numbers are in AP iff common difference between the consecutive terms is same.

$\rm :\longmapsto\:B - A = C - B$

$\purple{\rm\implies \:2B = A + C - - - - (1)}$

Now, According to statement, A + 1, B, C are in GP

We know, three numbers are in GP iff common ratio between the consecutive terms is same

$\rm :\longmapsto\:\dfrac{B}{A + 1} = \dfrac{C}{B}$

$\purple{\rm\implies \: {B}^{2} = (A + 1)C - - - - (2)}$

Again, According to statement, A, B, C + 2 are in GP

$\rm :\longmapsto\:\dfrac{B}{A} = \dfrac{C + 2}{B}$

$\purple{\rm\implies \: {B}^{2} = A(C + 2) - - - - (3)}$

Now, from equation (2) and (3), we get

$\rm :\longmapsto\:(A + 1)C = A(C + 2)$

$\rm :\longmapsto\:AC + C = AC + 2A$

$\purple{\rm\implies \:C = 2A - - - (4)}$

On substituting the value of C in equation (1), we get

$\rm :\longmapsto\:2B = A + 2A$

$\rm :\longmapsto\:2B = 3A$

$\purple{\rm\implies \:B = \dfrac{3A}{2} - - - (5)}$

On substituting equation (4) and (5) in equation (2), we get

$\rm :\longmapsto\:\dfrac{9 {A}^{2} }{4} = (A + 1)2A$

$\rm :\longmapsto\:\dfrac{9 {A}}{4} = (A + 1)2$

$\rm :\longmapsto\:\dfrac{9 {A}}{4} = 2A + 2$

$\rm :\longmapsto\:9A = 8A + 8$

$\red{\rm\implies \:\boxed{\sf{ \: A \: = \: 8 \: }}}$

On substituting A = 8 in equation (5), we get

$\red{\rm\implies \:\boxed{\sf{ \: B \: = \: 12 \: }}}$

On substituting A = 8 in equation (4), we get

$\red{\rm\implies \:\boxed{\sf{ \: C \: = \: 16\: }}}$

Hence,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{A = 8} \\ \\ &\sf{B = 12}\\ \\ &\sf{C = 16} \end{cases}\end{gathered}\end{gathered}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Three non - zero numbers, A, B, C are in AP.

We know, 3 numbers are in AP iff common difference between the consecutive terms is same.

$\rm :\longmapsto\:B - A = C - B$

$\purple{\rm\implies \:2B = A + C - - - - (1)}$

Now, According to statement, A + 1, B, C are in GP

We know, three numbers are in GP iff common ratio between the consecutive terms is same

$\rm :\longmapsto\:\dfrac{B}{A + 1} = \dfrac{C}{B}$

$\purple{\rm\implies \: {B}^{2} = (A + 1)C - - - - (2)}$

Again, According to statement, A, B, C + 2 are in GP

$\rm :\longmapsto\:\dfrac{B}{A} = \dfrac{C + 2}{B}$

$\purple{\rm\implies \: {B}^{2} = A(C + 2) - - - - (3)}$

Now, from equation (2) and (3), we get

$\rm :\longmapsto\:(A + 1)C = A(C + 2)$

$\rm :\longmapsto\:AC + C = AC + 2A$

$\purple{\rm\implies \:C = 2A - - - (4)}$

On substituting the value of C in equation (1), we get

$\rm :\longmapsto\:2B = A + 2A$

$\rm :\longmapsto\:2B = 3A$

$\purple{\rm\implies \:B = \dfrac{3A}{2} - - - (5)}$

On substituting equation (4) and (5) in equation (2), we get

$\rm :\longmapsto\:\dfrac{9 {A}^{2} }{4} = (A + 1)2A$

$\rm :\longmapsto\:\dfrac{9 {A}}{4} = (A + 1)2$

$\rm :\longmapsto\:\dfrac{9 {A}}{4} = 2A + 2$

$\rm :\longmapsto\:9A = 8A + 8$

$\red{\rm\implies \:\boxed{\sf{ \: A \: = \: 8 \: }}}$

On substituting A = 8 in equation (5), we get

$\red{\rm\implies \:\boxed{\sf{ \: B \: = \: 12 \: }}}$

On substituting A = 8 in equation (4), we get

$\red{\rm\implies \:\boxed{\sf{ \: C \: = \: 16\: }}}$

Hence,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{A = 8} \\ \\ &\sf{B = 12}\\ \\ &\sf{C = 16} \end{cases}\end{gathered}\end{gathered}}$