If pth ,qth ,rth terms of a G.Pare a, b, c then Σ(q – r) log a = A

1.	If pth ,qth ,rth terms of a G.Pare a, b, c then Σ(q – r) log a = A) 0 B) 1 C) abc D) pqr
Answer» Correct option is (A) 0 Given that \(p^{th},q^{th}\;\&\;r^{th}\) terms of a G.P. are a, b and c. \(i.e.,a_p=a,a_q=b\;\&\;a_r=c\) Let A and R be first term and common difference of given G.P. \(\therefore\) \(AR^{p-1}=a\) _______________(1) \(AR^{q-1}=b\) _______________(2) and \(AR^{r-1}=c\) _______________(3) Now, \(\sum(q-r)\,log\,a\) \(=(q-r)\,log\,a+(r-p)\,log\,b+(p-q)\,log\,c\) \(=log\,a^{q-r}+log\,b^{r-p}+log\,c^{p-q}\) \(=log\,(a^{q-r}.b^{r-p}.c^{p-q})\) \(=log\,\left((AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}.(AR^{r-1})^{p-q}\right)\) (From (1), (2) & (3)) \(=log\,(A^{(q-r)+(r-p)+(p-q)}\,R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})\) \(=log\,(A^{q-q-r+r-p+p}\,R^{(pq-pr-q+r)+(qr-qp-r+p)+(rp-rq-p+q)})\) \(=log\,(A^{0}\,R^{pq-pq-pr+pr-q+q+r-r+qr-qr+p-p})\) \(=log\,(A^{0}\,R^{0})\) \(=log\,1=0\) Hence, \(\sum(q-r)\,log\,a=0\) Correct option is A) 0

If pth ,qth ,rth terms of a G.Pare a, b, c then Σ(q – r) log a = A) 0 B) 1 C) abc D) pqr

Answer»

Correct option is (A) 0

Given that \(p^{th},q^{th}\;\&\;r^{th}\) terms of a G.P. are a, b and c.

\(i.e.,a_p=a,a_q=b\;\&\;a_r=c\)

Let A and R be first term and common difference of given G.P.

\(\therefore\) \(AR^{p-1}=a\) _______________(1)

\(AR^{q-1}=b\) _______________(2)

and \(AR^{r-1}=c\) _______________(3)

Now, \(\sum(q-r)\,log\,a\) \(=(q-r)\,log\,a+(r-p)\,log\,b+(p-q)\,log\,c\)

\(=log\,a^{q-r}+log\,b^{r-p}+log\,c^{p-q}\)

\(=log\,(a^{q-r}.b^{r-p}.c^{p-q})\)

\(=log\,\left((AR^{p-1})^{q-r}.(AR^{q-1})^{r-p}.(AR^{r-1})^{p-q}\right)\) (From (1), (2) & (3))

\(=log\,(A^{(q-r)+(r-p)+(p-q)}\,R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})\)

\(=log\,(A^{q-q-r+r-p+p}\,R^{(pq-pr-q+r)+(qr-qp-r+p)+(rp-rq-p+q)})\)

\(=log\,(A^{0}\,R^{pq-pq-pr+pr-q+q+r-r+qr-qr+p-p})\)

\(=log\,(A^{0}\,R^{0})\)

\(=log\,1=0\)

Hence, \(\sum(q-r)\,log\,a=0\)

Correct option is A) 0