The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1 . Find the first term 'a' and common difference 'd'.

Option a: a=3, d=2. Solution: We are given: 1) 4th term is 3 times the first term ( t4 = 3a ): a + 3d = 3a 2a = 3d a = 1.5d 2) 7th term exceeds twice the 3rd term by 1 ( t7 = 2t3 + 1 ): a + 6d = 2(a + 2d) + 1 a + 6d = 2a + 4d + 1 2d = a + 1 Substitute a = 1.5d into the second relation: 2d = 1.5d + 1 0.5d = 1 d = 2 Then, find a : a = 1.5(2) = 3 So the first term is 3 and common difference is 2 . Hence, Option A is the correct answer.

The $4^{th}$ term of an A.P. is three times the first and the $7^{th}$ term exceeds twice the third term by $1$. Find the first term 'a' and common difference 'd'.

The correct answer is Option a: $a=3, d=2$. We are given: 1) $4^{\text{th}}$ term is $3$ times the first term ($t_4 = 3a$): $$a + 3d = 3a \implies 2a = 3d \implies a = 1.5d$$ 2) $7^{\text{th}}$ term exceeds twice the $3^{\text{rd}}$ term by $1$ ($t_7 = 2t_3 + 1$): $$a + 6d = 2(a + 2d) + 1$$ $$a + 6d = 2a + 4d + 1$$ $$2d = a + 1$$ Substitute $a = 1.5d$ into the second relation: $$2d = 1.5d + 1 \implies 0.5d = 1 \implies d = 2$$ Then, find $a$: $$a = 1.5(2) = 3$$ So the first term is $3$ and common difference is $2$. Hence, **Option A** is the correct answer.

Sequence and SeriesMCQMTP Dec 2023 Series IQuestion 1814 of 212

All Questions

The $\displaystyle 4^{th}$ term of an A.P. is three times the first and the $\displaystyle 7^{th}$ term exceeds twice the third term by $\displaystyle 1$ . Find the first term 'a' and common difference 'd'.

Options

\displaystyle a=3, d=2

\displaystyle a=4, d=3

\displaystyle a=5, d=4

\displaystyle a=6, d=5

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — $\displaystyle a=3, d=2$

All Options:

A $\displaystyle a=3, d=2$
B $\displaystyle a=4, d=3$
C $\displaystyle a=5, d=4$
D $\displaystyle a=6, d=5$

Detailed Solution & Explanation

We are given:
1)

\displaystyle 4^{\text{th}}

term is

\displaystyle 3

times the first term (

\displaystyle t_4 = 3a

a + 3d = 3a \implies 2a = 3d \implies a = 1.5d

\displaystyle 7^{\text{th}}

term exceeds twice the

\displaystyle 3^{\text{rd}}

term by

\displaystyle 1

(

\displaystyle t_7 = 2t_3 + 1

a + 6d = 2(a + 2d) + 1

a + 6d = 2a + 4d + 1

2d = a + 1

Substitute

\displaystyle a = 1.5d

into the second relation:

2d = 1.5d + 1 \implies 0.5d = 1 \implies d = 2

Then, find

\displaystyle a

a = 1.5(2) = 3

So the first term is

\displaystyle 3

and common difference is

\displaystyle 2

.
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.

Related Comparison Tables

Void vs Voidable Contract: CA Foundation Law Distinction AP vs GP: Sequences and Series for CA Foundation Maths

Ready to Master Sequence and Series?

Practice all 212 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

The 4th\displaystyle 4^{th}4th term of an A.P. is three times the first and the 7th\displaystyle 7^{th}7th term exceeds twice the third term by 1\displaystyle 11. Find the first term 'a' and common difference 'd'.