Question

Please solve this above questions. ​

Answer 1

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

Since 3 numbers are in AP.

So, let we assume that three numbers be a - d, a, a + d.

According to statement,

Sum of three numbers is 15.

$\rm :\longmapsto\:a - d + a + a + d = 15$

$\rm :\longmapsto\:3a = 15$

$\rm :\implies\:a = 5$

According to statement,

When 1, 4, 19 are added to a - d, a, a + d, the numbers are in GP.

$\rm :\longmapsto\:a - d + 1, \: a + 4, \: a + d + 19 \: are \: in \: GP$

We know that,

If a, b, c are in GP, then b² = ac

So, using this identity, we have

$\rm :\longmapsto\: {(a + 4)}^{2} = (a - d + 1)(a + d + 19)$

On substituting the value of a, we get

$\rm :\longmapsto\: {(5 + 4)}^{2} = (5 - d + 1)(5 + d + 19)$

$\rm :\longmapsto\: {(9)}^{2} = (6 - d)(24 + d)$

$\rm :\longmapsto\: 81= 144 + 6d - 24d - {d}^{2}$

$\rm :\longmapsto\: = 144 - 18d - {d}^{2} - 81 = 0$

$\rm :\longmapsto\: = 63 - 18d - {d}^{2} = 0$

$\rm :\longmapsto\: {d}^{2} + 18d - 63 = 0$

$\rm :\longmapsto\: {d}^{2} + 21d - 3d - 63 = 0$

$\rm :\longmapsto\:d(d + 21) - 3(d + 21) = 0$

$\rm :\longmapsto\:(d + 21)(d - 3) = 0$

$\bf\implies \:d = - 21 \: \: \: or \: \: \: d = 3$

So,

Two cases arises

Case :- 1 When a = 5 and d = 3

Numbers are 2, 5, 8

Case :- 2 When a = 5 and d = - 21

Numbers are 26, 5, - 16

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

Given that,

a, b, c are in AP

It means, b - a = c - b

⇛ 2b = a + c --------[ 1 ]

Also, given that,

x, y, z are in GP

It means y² = xz --------- [ 2 ]

Consider,

$\rm :\longmapsto\: {x}^{b - c} {y}^{c - a} {z}^{a - b}$

$\rm \: = \: {x}^{b - c} \: { ( \: \sqrt{xz} \: )}^{c - a} \: {z}^{a - b}$

$\: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {y}^{2} = xz \: \implies \: y = \sqrt{xz} \bigg \}}$

$\rm \: = \: {x}^{b - c} \: {\bigg(x \bigg) }^{\dfrac{c - a}{2} } \: {\bigg(z \bigg) }^{\dfrac{c - a}{2} } \: {z}^{a - b}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {(xy)}^{z} = {x}^{z} \times {y}^{z} \bigg \}}$

$\rm \: = \: {\bigg(x \bigg) }^{b - c \: + \: \dfrac{c - a}{2} } \: {\bigg(z \bigg) }^{\dfrac{c - a}{2} \: + \: a - b} \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{m} \times {x}^{n} = {x}^{m + n} \bigg \}}$

$\rm \: = \: {\bigg(x \bigg) }^{ \: \dfrac{2b - 2c + c - a}{2} } \: {\bigg(z \bigg) }^{\dfrac{c - a + 2a - 2b}{2} } \:$

$\rm \: = \: {\bigg(x \bigg) }^{ \: \dfrac{2b - c - a}{2} } \: {\bigg(z \bigg) }^{\dfrac{c + a - 2b}{2} } \:$

$\rm \: = \: {\bigg(x \bigg) }^{ \: \dfrac{a + c - c - a}{2} } \: {\bigg(z \bigg) }^{\dfrac{c + a - a - c}{2} } \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: 2b = a + c\bigg \}}$

$\rm \: = \: {x}^{0} \: {z}^{0}$

$\rm \: = \: \:1 \times 1$

$\rm \: = \: \:1$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$