Question

How to do this type of questions

Answer 1

Given that,

a, b and c are in AP and 1/a², 1/b² and 1/c² also in AP.

Therefore common difference will be same between two consecutive terms.

1/c² - 1/b² = 1/b² - 1/a²

1/c² + 1/a² = 1/b² + 1/b²

1/c² + 1/a² = 2/b²

Hence option (1) is correct.

Answer 2

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

Given that

$\red{\rm :\longmapsto\:a,b,c \: are \: in \: AP \: }$

$\red{\rm \implies\:b - a = c - b - - - - (1)}$

Also, given that,

$\rm :\longmapsto\:\dfrac{1}{ {a}^{2} }, \: \dfrac{1}{ {b}^{2} }, \: \dfrac{1}{ {c}^{2} } \: are \: in \: AP$

$\rm :\longmapsto\:\dfrac{1}{ {b}^{2} } - \dfrac{1}{ {a}^{2} } = \dfrac{1}{ {c}^{2} } - \dfrac{1}{ {b}^{2} }$

$\rm :\longmapsto\:\dfrac{ {a}^{2} - {b}^{2} }{ {b}^{2} {a}^{2} } = \dfrac{ {b}^{2} - {c}^{2} }{ {c}^{2} {b}^{2} }$

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

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

$\rm :\longmapsto\:\dfrac{a + b}{ {a}^{2} } = \dfrac{b + c}{ {c}^{2} }$

$\rm :\longmapsto\: {c}^{2}(a + b) = {a}^{2}(b + c)$

$\rm :\longmapsto\: {c}^{2}a + b {c}^{2} = {a}^{2}b + c {a}^{2}$

$\rm :\longmapsto\: {c}^{2}a + b {c}^{2} - {a}^{2}b - c {a}^{2} = 0$

$\rm :\longmapsto\:ac(c - a) + b( {c}^{2} - {a}^{2}) = 0$

$\rm :\longmapsto\:ac(c - a) + b( c + a)(c - a) = 0$

$\rm :\longmapsto\:(c - a)[ac + b( c + a)] = 0$

$\rm :\longmapsto\:(c - a)[ac + bc + ba] = 0$

$\bf\implies \:c = a - - - (2)$

or

$\rm :\longmapsto\:ac + bc + ab = 0$

Also, from equation (1)

$\rm :\longmapsto\:b - a = c - b$

$\rm :\longmapsto\:2b = a + c$

$\rm :\longmapsto\:2b = a + a$

$\rm :\longmapsto\:2b = 2a$

$\bf\implies \:b = a - - - - (3)$

From equation (2) and (3), we concluded that

$\bf\implies \:a = b = c$

• Hence, option (1) is correct.