Question

the sum of three numbers which are consecutive terms of an A. P. is 21. if the second number is reduced by 1 while the third is increased by 1, three consecutive terms of a G. P. result. find these numbers. ​

Answer 1

Given :-

• The sum of three numbers which are consecutive terms of an A. P. is 21.
• If the second number is reduced by 1 while the third is increased by 1, three consecutive terms are in G. P.

To Find :-

• Three numbers in AP.

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let \: three \: numbers \: be   \: \begin{cases} &\sf{a - d} \\ &\sf{a}\\ &\sf{a + d} \end{cases}\end{gathered}\end{gathered}$

☆ According to statement,

☆ Sum of three numbers of AP is 21.

$\tt\implies \:a - d + a + a + d = 21$

$\tt\implies \:3a = 21$

$\tt\implies \:a = 7$

☆ According to the statement,

If the second number is reduced by 1 while the third is increased by 1, three consecutive terms are in G. P.

$\begin{gathered}\begin{gathered}\bf So, \: terms \: are \: -  \begin{cases} &\sf{a - d = 7 - d} \\ &\sf{a - 1 = 7 - 1 = 6}\\ &\sf{a + d + 1 = 7 + d + 1 = 8 + d} \end{cases}\end{gathered}\end{gathered}$

$\tt\implies \:7-d, \: 6, \: 8+d \: are \: in \: GP$

$\tt\implies \:\dfrac{6}{7 - d} = \dfrac{8 + d}{6}$

$\tt\implies \:36 = (7 - d)(8 + d)$

$\tt\implies \:36 = 56 + 7d - 8d - {d}^{2}$

$\tt\implies \:36 = 56 - d - {d}^{2}$

$\tt\implies \: {d}^{2} + d - 20 = 0$

$\tt\implies \: {d}^{2} + 5d - 4d - 20 = 0$

$\tt\implies \:d(d + 5) - 4(d + 5) = 0$

$\tt\implies \:(d + 5)(d - 4) = 0$

$\tt\implies \: \boxed{ \purple{ \bf \: d \: = - 5 \: or \: d \: = 4}}$

Now, two cases arises,

☆ When d = 4

$\begin{gathered}\begin{gathered}\bf So, \: terms \: are \: -  \begin{cases} &\sf{a - d = 7 - 4 = 3} \\ &\sf{a = 7}\\ &\sf{a + d = 7 + 4 = 11} \end{cases}\end{gathered}\end{gathered}$

☆ When d = - 5,

$\begin{gathered}\begin{gathered}\bf So, \: terms \: are \: -  \begin{cases} &\sf{a - d = 7 + 5 = 12} \\ &\sf{a = 7}\\ &\sf{a + d = 7 - 5 = 2} \end{cases}\end{gathered}\end{gathered}$