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Answer 1

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

• Let us assume that first term and common ratio of G P. series be A and R respectively.

Since, it is given that

$\bullet \: \red{ \rm \: {p}^{th} \: term \: of \: GP \: is \: \bf \: a}$

$\rm : \implies \:a_p \: = \: a$

$\rm : \implies \: {AR}^{p - 1} = a$

Taking log on both sides, we get

$\rm : \implies \:loga \: = log(A {R}^{p - 1} )$

$\boxed{ \purple {\rm : \implies \: log(a) = log(A) + (p - 1) log(R)}}$

Similarly,

$\boxed{ \purple{ \rm : \implies \: log(b) = log(A) + (q - 1) log(R)}}$

Similarly,

$\boxed{ \purple{ \rm : \implies \: log(c) = log(A) + (r - 1) log(R) }}$

Now, Consider

$\rm : \implies \:\begin{array}{|ccc|}</p><p> log(a) & p & 1 \\</p><p> log(b) & q & 1 \\</p><p> log(c) &r & 1 \\</p><p>\end{array}$

On substituting the values of log(a), log(b) and log(c), we get

$\rm : \implies \:\begin{array}{|ccc|}</p><p> log(A) + (p - 1) log(R) & p & 1 \\</p><p> log(A) + (q - 1) log(R) & q & 1 \\</p><p> log(A) + (r - 1) log(R) &r & 1 \\</p><p>\end{array}$

Now, use splitting property of determinants, split along first Column, we get

$= \rm \begin{array}{|ccc|}</p><p> log(A) & p & 1 \\</p><p> log(A) & q & 1 \\</p><p> log(A) &r & 1 \\</p><p>\end{array} \: + \rm \begin{array}{|ccc|}</p><p> (p - 1)log(R) & p & 1 \\</p><p>(q - 1) log(R) & q & 1 \\</p><p>(r - 1) log(R) &r & 1 \\</p><p>\end{array}$

Taking log(A) common from first determinant and taking log(R) common from second determinant, So we get

$= \rm log(a) \begin{array}{|ccc|}</p><p> 1 & p & 1 \\</p><p> 1 & q & 1 \\</p><p> 1 &r & 1 \\</p><p>\end{array} \: + \: \rm \: log(R) \begin{array}{|ccc|}</p><p> (p - 1) & p & 1 \\</p><p>(q - 1) & q & 1 \\</p><p>(r - 1) &r & 1 \\</p><p>\end{array}$

Since, column 1 and column 3 are identical in first determinant, so its value is 0.

$\rm : \implies \: log(a) \times 0 \: \: + \: \rm \: log(R) \begin{array}{|ccc|}</p><p> (p - 1) & p & 1 \\</p><p>(q - 1) & q & 1 \\</p><p>(r - 1) &r & 1 \\</p><p>\end{array}$

$\rm : \implies \:OP \: C_1 \: \longrightarrow \: C_1 \: + \: C_3$

$\rm : \implies \:0 \: \: + \: \rm \: log(R) \begin{array}{|ccc|}</p><p> p & p & 1 \\</p><p>q & q & 1 \\</p><p>r &r & 1 \\</p><p>\end{array}$

Since, column 1 and column 2 are identical, so its value is 0.

$\rm : \implies \: log(R) \times 0$

$\rm : \implies \:0$

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