JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a, b$ and $c$ be in G. P. with common ratio $r$ , where $a \neq = 0$ and $0 < r \leq 1/2$ . If $3 a, 7 b$ and $15 c$ are the first three terms of an A. P., then the 4th term of this A. P. is :

Q: Do I really need to divide by $a$ in the equation $14ar = 3a + 15ar^2$? What if $a=0$?

Excellent observation! The problem states $a \neq 0$. If $a$ were zero, the entire sequence would be zero, which is trivial and usually excluded in G.P. definitions. Because $a \neq 0$, we are mathematically safe to divide.

Select Answer:

Visualized Solution

Problem Setup

Given: $a, b, c$ are in G.P. with common ratio $r$ .
Constraint: $a \neq = 0$ and $0 < r \leq \frac{1}{2}$ .
Given: $3 a, 7 b, 15 c$ are in A.P.
Goal: Find the 4th term of this A.P.

Expressing G.P. Terms

Since $a, b, c$ are in G.P. with ratio $r$ :
$b = a r$
$c = a r^{2}$

Identifying the A.P. Terms

The first three terms of the A.P. are:
$T_{1} = 3 a$
$T_{2} = 7 b = 7 a r$
$T_{3} = 15 c = 15 a r^{2}$

Applying the A.P. Condition

For any three terms $x, y, z$ in A.P.:
$2 y = x + z$
Substituting our terms:
$2 (7 a r) = 3 a + 15 a r^{2}$

Simplifying the Equation

$14 a r = 3 a + 15 a r^{2}$
Since $a \neq = 0$ , divide the entire equation by $a$ :
$14 r = 3 + 15 r^{2}$

Forming the Quadratic Equation

Rearrange terms to form a standard quadratic equation in $r$ :
$15 r^{2} - 14 r + 3 = 0$

Solving the Quadratic Equation

Factorize by splitting the middle term:
$15 r^{2} - 9 r - 5 r + 3 = 0$
$3 r (5 r - 3) - 1 (5 r - 3) = 0$
$(3 r - 1) (5 r - 3) = 0$

Finding the Values of $r$

Equating each factor to zero:
$3 r - 1 = 0 ⟹ r = \frac{1}{3}$
$5 r - 3 = 0 ⟹ r = \frac{3}{5}$

Applying the Constraint on $r$

Given constraint: $0 < r \leq \frac{1}{2}$
Check $r = \frac{3}{5}$ : $\frac{3}{5} = 0.6 > 0.5$ (Rejected)
Check $r = \frac{1}{3}$ : $\frac{1}{3} \approx 0.33 \leq 0.5$ (Accepted)
Therefore, $r = \frac{1}{3}$

Calculating the A.P. Terms

Substitute $r = \frac{1}{3}$ into the A.P. terms:
$T_{1} = 3 a$
$T_{2} = 7 a (\frac{1}{3}) = \frac{7 a}{3}$
$T_{3} = 15 a (\frac{1}{3})^{2} = \frac{15 a}{9} = \frac{5 a}{3}$

Finding the Common Difference

Common difference $d = T_{2} - T_{1}$
$d = \frac{7 a}{3} - 3 a$
$d = \frac{7 a - 9 a}{3} = - \frac{2 a}{3}$

Calculating the 4th Term

4th term $T_{4} = T_{3} + d$
$T_{4} = \frac{5 a}{3} + (- \frac{2 a}{3})$
$T_{4} = \frac{3 a}{3} = a$

Final Conclusion

The 4th term of the A.P. is $a$ .
The correct option is (2).
Key Takeaway: Always verify multiple roots against the given constraints in the problem.

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Setup

We begin with three numbers

a, b, c

that form a Geometric Progression (G.P.). By definition, we can express these terms using a common ratio

r

b = a r

c = a r^{2}

This reduction of variables allows us to manage the progression with only two unknowns,

a

and

r

The Master Equation

We are given that the sequence

3 a, 7 b, 15 c

forms an Arithmetic Progression (A.P.). The fundamental property of an A.P. is that the middle term is the arithmetic mean of its neighbors, expressed as:

2 (7 b) = 3 a + 15 c

Substituting our G.P. definitions into this equation, we obtain:

14 a r = 3 a + 15 a r^{2}

Since $a eq 0$, we can safely divide the entire equation by

a

to arrive at the quadratic equation:

15 r^{2} - 14 r + 3 = 0

Solving the Quadratic

To find the common ratio

r

, we solve the quadratic equation using the factorization method or the quadratic formula:

(3 r - 1) (5 r - 3) = 0

This yields two potential values for the ratio:

r = \frac{1}{3} or r = \frac{3}{5}

Applying Constraints and Final Calculation

We are provided with the specific constraint

0 < r \leq \frac{1}{2}

. Comparing our results, we see that

r = \frac{3}{5} = 0.6

exceeds the upper limit. Therefore, we reject

r = \frac{3}{5}

and accept $r = \frac{1}{3}$ .

Now, we determine the terms of the A.P. using

r = \frac{1}{3}

: The first term is

3 a

. The second term is

7 b = 7 (a \cdot \frac{1}{3}) = \frac{7 a}{3}

. The third term is

15 c = 15 (a \cdot \frac{1}{9}) = \frac{15 a}{9} = \frac{5 a}{3}

The common difference

d

of this A.P. is:

d = \frac{7 a}{3} - 3 a = - \frac{2 a}{3}

The fourth term of the A.P. is calculated by adding the common difference to the third term:

T_{4} = \frac{5 a}{3} + (- \frac{2 a}{3}) = \frac{3 a}{3} = a

The final result is $a$ .

Common Traps & FAQs

Why did we reject

r = 3/5

? It is a valid root of the quadratic equation, isn't it?

It is indeed a root of the quadratic, but in the world of JEE, a solution must satisfy all conditions. The problem explicitly restricts

r

0 < r \leq 1/2

. Since

3/5 = 0.6

, which is greater than

0.5

, it violates the domain. Always check your constraints!

Do I really need to divide by

a

in the equation

14 a r = 3 a + 15 a r^{2}

? What if

a = 0

Excellent observation! The problem states

a \neq = 0

. If

a

were zero, the entire sequence would be zero, which is trivial and usually excluded in G.P. definitions. Because

a \neq = 0

, we are mathematically safe to divide.

Why is the common difference

d

negative? Did I make a calculation error?

Not at all! A negative common difference simply means your Arithmetic Progression is a decreasing sequence. It is perfectly valid. Trust your algebra; if the math leads to a negative

d

, embrace it.

Animated Solution for Mathematics - Sequence and Series: Let $a, b$ and $c$ be in G. P. with common ratio $r$ , where $a \neq = 0$ and $0 < r \leq 1/2$ . If $3 a, 7 b$ and $15 c$ are the first three terms of an A. P., then the 4th term of this A. P. is :

Select Answer:

Visualized Solution

Problem Setup

Expressing G.P. Terms

Identifying the A.P. Terms

Applying the A.P. Condition

Simplifying the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Finding the Values of $r$

Applying the Constraint on $r$

Calculating the A.P. Terms

Finding the Common Difference

Calculating the 4th Term

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Master Equation

Solving the Quadratic

Applying Constraints and Final Calculation

Similar Questions

Three numbers are in an increasing geometric progression with common ratio $r$ . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$ . If the fourth term of $GP$ is $3 r^{2}$ , then $r^{2} - d$ is equal to :

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

Let $a, b$ and $c$ be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then $a / c$ is equal to:

Let the first three terms $2, p$ and $q$ , with $q \neq = 2$ , of a G.P. be respectively the $7^{th}, 8^{th}$ and $1 3^{th}$ terms of an A.P. If the $5^{th}$ term of the G.P. is the $n^{th}$ term of the A.P., then $n$ is equal to:

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

The third term of a geometric progression is $4$ . The product of the first five terms is

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is:

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-

Let $a_{1}, a_{2}, a_{3}, \dots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Select Answer:

Visualized Solution

Problem Setup

Expressing G.P. Terms

Identifying the A.P. Terms

Applying the A.P. Condition

Simplifying the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Finding the Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Applying the Constraint on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Calculating the A.P. Terms

Finding the Common Difference

Calculating the 4th Term

Final Conclusion

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Master Equation

Solving the Quadratic

Applying Constraints and Final Calculation

Similar Questions

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then a/c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The third term of a geometric progression is 4. The product of the first five terms is

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a, b, c are in A.P. and a2,b2,c2 are in G.P. such that a<b<c and a+b+c=43​, then the value of a is :-

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Finding the Values of $r$

Applying the Constraint on $r$

Let $a, b$ and $c$ be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then $a / c$ is equal to:

The third term of a geometric progression is $4$ . The product of the first five terms is

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-