The nth term of the series 9,7,5,... and 15,12,9,... are same. Find the nth term?

Option a: 7. Solution: Let's find the expression for the nth term of each series: First series: 9, 7, 5, ... First term a1 = 9 , common difference d1 = -2 tn(1) = a1 + (n-1)d1 = 9 + (n-1)(-2) = 11 - 2n Second series: 15, 12, 9, ... First term a2 = 15 , common difference d2 = -3 tn(2) = a2 + (n-1)d2 = 15 + (n-1)(-3) = 18 - 3n Since the nth terms are equal: 11 - 2n = 18 - 3n 3n - 2n = 18 - 11 n = 7 Thus, the terms are equal when n = 7 . Hence, Option A is the correct answer.

The $n^{th}$ term of the series $9,7,5,...$ and $15,12,9,...$ are same. Find the $n^{th}$ term?

The correct answer is Option a: $7$. Let's find the expression for the $n^{\text{th}}$ term of each series: First series: $9, 7, 5, \dots$ First term $a_1 = 9$, common difference $d_1 = -2$ $$t_n^{(1)} = a_1 + (n-1)d_1 = 9 + (n-1)(-2) = 11 - 2n$$ Second series: $15, 12, 9, \dots$ First term $a_2 = 15$, common difference $d_2 = -3$ $$t_n^{(2)} = a_2 + (n-1)d_2 = 15 + (n-1)(-3) = 18 - 3n$$ Since the $n^{\text{th}}$ terms are equal: $$11 - 2n = 18 - 3n$$ $$3n - 2n = 18 - 11$$ $$n = 7$$ Thus, the terms are equal when $n = 7$. Hence, **Option A** is the correct answer.

Sequence and SeriesMCQPYQ June 22Question 1764 of 212

All Questions

The $\displaystyle n^{th}$ term of the series $\displaystyle 9,7,5,...$ and $\displaystyle 15,12,9,...$ are same. Find the $\displaystyle n^{th}$ term?

Options

\displaystyle 7

\displaystyle 8

\displaystyle 9

\displaystyle 10

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Correct Answer

✅ Option a — $\displaystyle 7$

All Options:

A $\displaystyle 7$
B $\displaystyle 8$
C $\displaystyle 9$
D $\displaystyle 10$

Detailed Solution & Explanation

Let's find the expression for the

\displaystyle n^{\text{th}}

term of each series:
First series:

\displaystyle 9, 7, 5, \dots

First term

\displaystyle a_1 = 9

, common difference

\displaystyle d_1 = -2

t_n^{(1)} = a_1 + (n-1)d_1 = 9 + (n-1)(-2) = 11 - 2n

Second series:

\displaystyle 15, 12, 9, \dots

First term

\displaystyle a_2 = 15

, common difference

\displaystyle d_2 = -3

t_n^{(2)} = a_2 + (n-1)d_2 = 15 + (n-1)(-3) = 18 - 3n

Since the

\displaystyle n^{\text{th}}

terms are equal:

11 - 2n = 18 - 3n

3n - 2n = 18 - 11

n = 7

Thus, the terms are equal when

\displaystyle n = 7

.
Hence, **Option A** 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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The nth\displaystyle n^{th}nth term of the series 9,7,5,...\displaystyle 9,7,5,...9,7,5,... and 15,12,9,...\displaystyle 15,12,9,...15,12,9,... are same. Find the nth\displaystyle n^{th}nth term?