JEE Main 2019 (9 January)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: 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:

Q: Is there a faster way than expanding the quadratic?

You could potentially use the property of the common ratio $r = b/a = c/b$, but that often leads to more complex fractions. Expanding the quadratic is the most robust and error-proof path. Trust the algebra—it is designed to simplify beautifully!

Select Answer:

Visualized Solution

Defining the A.P. Terms $T_{n}$

Let the first term of the A.P. be $A$ and the common difference be $d$ .
The general term of an A.P. is given by $T_{n} = A + (n - 1) d$ .
We are given three specific terms: $a = T_{7}$ , $b = T_{11}$ , and $c = T_{13}$ .

Expressing $a, b, c$

$a = A + (7 - 1) d = A + 6 d$
$b = A + (11 - 1) d = A + 10 d$
$c = A + (13 - 1) d = A + 12 d$
Note: Since the A.P. is non-constant, $d \neq = 0$ .

The G.P. Condition

Since $a, b, c$ are in G.P., they share a common ratio.
The middle term squared equals the product of the outer terms.
Condition: $b^{2} = a c$

Substituting A.P. into G.P.

Substitute the expressions for $a, b,$ and $c$ into the G.P. condition:
$(A + 10 d)^{2} = (A + 6 d) (A + 12 d)$

Expanding $(A + 10 d)^{2}$

Expand the Left Hand Side (LHS) using $(x + y)^{2} = x^{2} + 2 x y + y^{2}$ :
$(A + 10 d)^{2} = A^{2} + 2 (A) (10 d) + (10 d)^{2}$
LHS $= A^{2} + 20 A d + 100 d^{2}$

Expanding $(A + 6 d) (A + 12 d)$

Expand the Right Hand Side (RHS) by multiplying the binomials:
$(A + 6 d) (A + 12 d) = A (A + 12 d) + 6 d (A + 12 d)$
RHS $= A^{2} + 12 A d + 6 A d + 72 d^{2}$
RHS $= A^{2} + 18 A d + 72 d^{2}$

Simplifying the Equation

Equate LHS and RHS:
$A^{2} + 20 A d + 100 d^{2} = A^{2} + 18 A d + 72 d^{2}$
Subtract $A^{2}$ from both sides and rearrange:
$(20 A d - 18 A d) + (100 d^{2} - 72 d^{2}) = 0$
$2 A d + 28 d^{2} = 0$

Finding $A$ in terms of $d$

$2 A d = - 28 d^{2}$
Divide by $2 d$ (since $d \neq = 0$ ):
$A = - 14 d$

Setting up the Ratio $\frac{a}{c}$

We need to find the ratio $\frac{a}{c}$ .
Substitute $A = - 14 d$ into the expressions for $a$ and $c$ :
$\frac{a}{c} = \frac{A + 6 d}{A + 12 d}$
$\frac{a}{c} = \frac{- 14 d + 6 d}{- 14 d + 12 d}$

Final Calculation

Simplify the numerator and denominator:
Numerator: $- 14 d + 6 d = - 8 d$
Denominator: $- 14 d + 12 d = - 2 d$
$\frac{a}{c} = \frac{- 8 d}{- 2 d} = 4$

Key Takeaways

Key Concept: Using the $n$ -th term of an A.P. to satisfy G.P. conditions.
Constraint Check: The 'non-constant' condition ( $d \neq = 0$ ) was essential to simplify the ratio.
Final Result: The ratio $\frac{a}{c}$ is 4.

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

Solution Diagram

Analyzing the Setup

We begin by defining our landscape using the standard notation for an Arithmetic Progression (A.P.). Let the first term be

A

and the common difference be

d

The general term of an A.P. is given by

T_{n} = A + (n - 1) d

. Our three specific terms are defined as follows:

a = T_{7} = A + 6 d

b = T_{11} = A + 10 d

c = T_{13} = A + 12 d

The Geometric Bridge

We are given that

a

b

, and

c

are consecutive terms of a Geometric Progression (G.P.). The fundamental property of any three consecutive terms in a G.P. is that the square of the middle term equals the product of the outer terms.

This relationship is expressed as:

b^{2} = a c

Substituting our A.P. expressions into this condition, we obtain:

(A + 10 d)^{2} = (A + 6 d) (A + 12 d)

The Algebraic Dance

To solve for the relationship between

A

and

d

, we expand both sides of the equation. Expanding the left side yields:

A^{2} + 20 A d + 100 d^{2}

Expanding the right side yields:

A^{2} + 12 A d + 6 A d + 72 d^{2} = A^{2} + 18 A d + 72 d^{2}

Equating the two sides, the

A^{2}

terms cancel out, leaving us with:

20 A d + 100 d^{2} = 18 A d + 72 d^{2}

Rearranging the terms to isolate

A

and

d

, we find:

2 A d = - 28 d^{2}

Since the problem implies a non-constant sequence, we assume $d eq 0$. Dividing both sides by

2 d

, we arrive at the critical relationship:

A = - 14 d

The Final Revelation

The question asks for the ratio of the first term to the third term of the G.P., which is

\frac{a}{c}

. Substituting

A = - 14 d

into the expressions for

a

and

c

\frac{a}{c} = \frac{A + 6 d}{A + 12 d}

\frac{a}{c} = \frac{- 14 d + 6 d}{- 14 d + 12 d}

\frac{a}{c} = \frac{- 8 d}{- 2 d}

The common difference

d

cancels out, leaving us with the final result:

\frac{a}{c} = 4

The ratio of the 7th term to the 13th term is 4.

Common Traps & FAQs

Wait, why can't we just pick values for

A

and

d

to solve this?

You absolutely can! In a competitive exam like JEE, if you are stuck, picking simple values (like

A = 1, d = 1

) is a brilliant strategy. However, the algebraic method is the 'soul' of the problem—it proves that the ratio is independent of the specific values of

A

and

d

, provided the A.P. is non-constant.

Why does the problem specify a 'non-constant' A.P.?

This is a crucial constraint! If the A.P. were constant, then

d = 0

. If

d = 0

, then

a = b = c = A

. While this technically forms a G.P. (with common ratio 1), the problem asks for a ratio

a / c

. If

d = 0

, we cannot divide by

d

to find the relationship between

A

and

d

. The 'non-constant' condition ensures

d \neq = 0

, allowing us to perform the division safely.

Is there a faster way than expanding the quadratic?

You could potentially use the property of the common ratio

r = b / a = c / b

, but that often leads to more complex fractions. Expanding the quadratic is the most robust and error-proof path. Trust the algebra—it is designed to simplify beautifully!

Animated Solution for Mathematics - Sequence and Series: 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:

Select Answer:

Visualized Solution

Defining the A.P. Terms $T_{n}$

Expressing $a, b, c$

The G.P. Condition

Substituting A.P. into G.P.

Expanding $(A + 10 d)^{2}$

Expanding $(A + 6 d) (A + 12 d)$

Simplifying the Equation

Finding $A$ in terms of $d$

Setting up the Ratio $\frac{a}{c}$

Final Calculation

Key Takeaways

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

Analyzing the Setup

The Geometric Bridge

The Algebraic Dance

The Final Revelation

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:

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 :

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

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:

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

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

If $x, y$ and $z$ are $p$ th, $q$ th and $r$ th terms respectively of an A.P. and also of a G.P., then $x^{y - z} y^{z - x} z^{x - y}$ is equal to :

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 :-

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

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:

Select Answer:

Visualized Solution

Defining the A.P. Terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

The G.P. Condition

Substituting A.P. into G.P.

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A+10d)2

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A+6d)(A+12d)

Simplifying the Equation

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Setting up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ca​

Final Calculation

Key Takeaways

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

Analyzing the Setup

The Geometric Bridge

The Algebraic Dance

The Final Revelation

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 in G. P. with common ratio r, where a=0 and 0<r≤1/2. If 3a,7b and 15c are the first three terms of an A. P., then the 4th term of this A. P. is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 3,a,b,c be in A.P. and 3,a−1,b+1,c+9 be in G.P. Then, the arithmetic mean of a,b and c 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

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x,y and z are pth, qth and rth terms respectively of an A.P. and also of a G.P., then xy−zyz−xzx−y is equal to :

.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 :-

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:

Defining the A.P. Terms $T_{n}$

Expressing $a, b, c$

Expanding $(A + 10 d)^{2}$

Expanding $(A + 6 d) (A + 12 d)$

Finding $A$ in terms of $d$

Setting up the Ratio $\frac{a}{c}$

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 :

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

If $x, y$ and $z$ are $p$ th, $q$ th and $r$ th terms respectively of an A.P. and also of a G.P., then $x^{y - z} y^{z - x} z^{x - y}$ is equal to :

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 :-