If the pth term of a G.P. is x and the qth term is y , then find the nth term:

Option d: xn-q/p-q yn-p/q-p. Solution: Let the G.P. have first term a and common ratio r . Given: x = arp-1 --- (1) y = arq-1 --- (2) Dividing equation (1) by equation (2): x/y = rp-q r = (x/y)1/p-q The nth term tn is: tn = arn-1 = arp-1 × rn-p = x × rn-p tn = x × ((x/y)1/p-q)n-p = x × xn-p/p-q × y-n-p/p-q tn = x1 + n-p/p-q × yn-p/q-p = xn-q/p-q yn-p/q-p Hence, Option D is the correct answer.

If the $p^{th}$ term of a G.P. is $x$ and the $q^{th}$ term is $y$, then find the $n^{th}$ term:

The correct answer is Option d: $x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}$. Let the G.P. have first term $a$ and common ratio $r$. Given: $$x = ar^{p-1} \quad \text{--- (1)}$$ $$y = ar^{q-1} \quad \text{--- (2)}$$ Dividing equation (1) by equation (2): $$\frac{x}{y} = r^{p-q} \implies r = \left(\frac{x}{y}\right)^{\frac{1}{p-q}}$$ The $n^{\text{th}}$ term $t_n$ is: $$t_n = ar^{n-1} = ar^{p-1} \cdot r^{n-p} = x \cdot r^{n-p}$$ $$t_n = x \cdot \left(\left(\frac{x}{y}\right)^{\frac{1}{p-q}}\right)^{n-p} = x \cdot x^{\frac{n-p}{p-q}} \cdot y^{-\frac{n-p}{p-q}}$$ $$t_n = x^{1 + \frac{n-p}{p-q}} \cdot y^{\frac{n-p}{q-p}} = x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}$$ Hence, **Option D** is the correct answer.

Sequence and SeriesMCQMTP Apr 21Question 1846 of 212

All Questions

If the $\displaystyle p^{th}$ term of a G.P. is $\displaystyle x$ and the $\displaystyle q^{th}$ term is $\displaystyle y$ , then find the $\displaystyle n^{th}$ term:

Options

\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{n-p}{p-q}}

\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{p-n}{p-q}}

\displaystyle 1

\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}

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Correct Answer

✅ Option d — $\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}$

All Options:

A $\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{n-p}{p-q}}$
B $\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{p-n}{p-q}}$
C $\displaystyle 1$
D $\displaystyle x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}$

Detailed Solution & Explanation

Let the G.P. have first term

\displaystyle a

and common ratio

\displaystyle r

.
Given:

x = ar^{p-1} \quad \text{--- (1)}

y = ar^{q-1} \quad \text{--- (2)}

Dividing equation (1) by equation (2):

\frac{x}{y} = r^{p-q} \implies r = \left(\frac{x}{y}\right)^{\frac{1}{p-q}}

The

\displaystyle n^{\text{th}}

term

\displaystyle t_n

is:

t_n = ar^{n-1} = ar^{p-1} \cdot r^{n-p} = x \cdot r^{n-p}

t_n = x \cdot \left(\left(\frac{x}{y}\right)^{\frac{1}{p-q}}\right)^{n-p} = x \cdot x^{\frac{n-p}{p-q}} \cdot y^{-\frac{n-p}{p-q}}

t_n = x^{1 + \frac{n-p}{p-q}} \cdot y^{\frac{n-p}{q-p}} = x^{\frac{n-q}{p-q}} y^{\frac{n-p}{q-p}}

Hence, **Option D** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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If the pth\displaystyle p^{th}pth term of a G.P. is x\displaystyle xx and the qth\displaystyle q^{th}qth term is y\displaystyle yy, then find the nth\displaystyle n^{th}nth term: