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Question: If the \({{p}^{th}},{{q}^{th}}\) and \({{r}^{th}}\) terms of an A.P. are in G.P. . Then find the com...

If the pth,qth{{p}^{th}},{{q}^{th}} and rth{{r}^{th}} terms of an A.P. are in G.P. . Then find the common ratio of the GP

Explanation

Solution

We will first write the given terms in the form of aa and dd where, aa is the first term of the sequence and dd is the common difference between terms. This we will do by applying the formula for the nth{{n}^{th}} term of an arithmetic sequence that is an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d . Then we will apply the definition of geometric sequences and we will write the obtained terms as \left\\{ A,AR,A{{R}^{2}},A{{R}^{3}}..... \right\\}, finally we will get two equations , we will divide them and get our common ratio.

Complete step-by-step answer:
We know that the formula for the nth{{n}^{th}} term of an arithmetic sequence is as follows:
an=a+(n−1)d{{a}_{n}}=a+\left( n-1 \right)d
Where, an{{a}_{n}} is the nth{{n}^{th}} term, aa is the first term of the sequence and dd is the common difference between terms.
Now, we will find the pth{{p}^{th}} term , by applying the above formula. We will take n=pn=p. Therefore, ap=a+(p−1)d .........Equation 1.{{a}_{p}}=a+\left( p-1 \right)d\text{ }.........\text{Equation 1}\text{.}
Similarly we will find the qth{{q}^{th}} term and we will now take n=qn=q. So,
aq=a+(q−1)d{{a}_{q}}=a+\left( q-1 \right)d
Finally, we will find the rth{{r}^{th}} term and we will be taking n=rn=r. So,
ar=a+(r−1)d{{a}_{r}}=a+\left( r-1 \right)d

We now know that in a geometric sequence each term is found by multiplying the previous term by a constant. In general, a sequence say \left\\{ A,AR,A{{R}^{2}},A{{R}^{3}}..... \right\\} is in geometric progression, where, AA is the first term and RR is the common ratio.
Since, it is given in the question that pth,qth{{p}^{th}},{{q}^{th}} and rth{{r}^{th}} terms of the arithmetic progression are in geometric progression,
Now, the pth{{p}^{th}} term of the A.P. that is a+(p−1)da+\left( p-1 \right)d is the first term of the GP, Therefore: a+(p−1)d=A ..........Equation 1.a+\left( p-1 \right)d=A\text{ }..........\text{Equation 1}\text{.}

Similarly the qth{{q}^{th}} term that is a+(q−1)da+\left( q-1 \right)d will be the second term of the GP: a+(q−1)d=AR ........Equation 2.a+\left( q-1 \right)d=AR\text{ }........\text{Equation 2}\text{.}
And finally the rth{{r}^{th}} term which is a+(r−1)da+\left( r-1 \right)d will be the third term of the GP: a+(r−1)d=AR2 ............Equation 3.a+\left( r-1 \right)d=A{{R}^{2}}\text{ }............\text{Equation 3}\text{.}
Now we will subtract equation 1 from equation 2 , we will get: [a+(q−1)d]−[a+(p−1)d]=AR−A a+(q−1)d−a−(p−1)d=A(R−1) [(q−1)−(p−1)]d=A(R−1) ⇒(q−p)d=A(R−1) ......... Equation 4. \begin{aligned} & \left[ a+\left( q-1 \right)d \right]-\left[ a+\left( p-1 \right)d \right]=AR-A \\\ & a+\left( q-1 \right)d-a-\left( p-1 \right)d=A\left( R-1 \right) \\\ & \left[ \left( q-1 \right)-\left( p-1 \right) \right]d=A\left( R-1 \right) \\\ & \Rightarrow \left( q-p \right)d=A\left( R-1 \right)\text{ }.........\text{ Equation 4}\text{.} \\\ \end{aligned}
Similarly we will subtract equation 2 from equation 3:
[a+(r−1)d]−[a+(q−1)d]=AR2−AR a+(r−1)d−a−(q−1)d=AR(R−1) [(r−1)−(q−1)]d=AR(R−1) ⇒(r−q)d=AR(R−1) ......... Equation 5. \begin{aligned} & \left[ a+\left( r-1 \right)d \right]-\left[ a+\left( q-1 \right)d \right]=A{{R}^{2}}-AR \\\ & a+\left( r-1 \right)d-a-\left( q-1 \right)d=AR\left( R-1 \right) \\\ & \left[ \left( r-1 \right)-\left( q-1 \right) \right]d=AR\left( R-1 \right) \\\ & \Rightarrow \left( r-q \right)d=AR\left( R-1 \right)\text{ }.........\text{ Equation 5}\text{.} \\\ \end{aligned}
We will now divide equation 5 by equation 4:
⇒(r−q)d(q−p)d=AR(R−1)A(R−1)⇒(r−q)(q−p)=R ⇒R=−(q−r)−(p−q) ⇒R=q−rp−q \begin{aligned} & \Rightarrow \dfrac{\left( r-q \right)d}{\left( q-p \right)d}=\dfrac{AR\left( R-1 \right)}{A\left( R-1 \right)}\Rightarrow \dfrac{\left( r-q \right)}{\left( q-p \right)}=R \\\ & \Rightarrow R=\dfrac{-\left( q-r \right)}{-\left( p-q \right)} \\\ & \Rightarrow R=\dfrac{q-r}{p-q} \\\ \end{aligned}
Hence, the common ratio for the G.P. is q−rp−q\dfrac{q-r}{p-q} .

Note: The resultant common ratio can be applied to general terms of an arithmetic progression which are in G.P. . While solving such types of questions, remember to proceed in such a way that we eliminate the terms which we do not require or we are not given the value of it. For example, here we eliminated the variable AA .