Let $a_1, a_2, a_3,...$ be a G.P. such that $a_1 < 0, a_1 + a_2 = 4$ and $a_3 + a_4 = 16$. If $\sum_{i=1}^{9} a_i = 4\lambda$, then $\lambda$ is equal to : Options: (A) $-513$ (B) $\frac{511}{3}$ (C) $-171$ (D) 171

### Analyzing the Setup We are exploring a Geometric Progression (G.P.) defined by the sequence $a_1, a_2, a_3, \dots$. We are provided with two primary constraints: $a_1 + a_2 = 4$ and $a_3 + a_4 = 16$. Crucially, we must adhere to the condition that $a_1 < 0$. This constraint serves as our guide to selecting the correct common ratio. ### The Algebraic Detective Work Any term in a G.P. is expressed as $a_n = a_1 r^{n-1}$. We can rewrite our given equations as follows: $$a_1(1 + r) = 4$$ $$a_1 r^2(1 + r) = 16$$ By dividing the second equation by the first, we eliminate $a_1$ and $(1 + r)$ through cancellation: $$\frac{a_1 r^2(1 + r)}{a_1(1 + r)} = \frac{16}{4}$$ $$r^2 = 4$$ ### The Trap Door and the Truth The equation $r^2 = 4$ yields two potential values for the common ratio: $r = 2$ or $r = -2$. We must test these against the condition $a_1 < 0$. If $r = 2$, substituting into $a_1(1 + r) = 4$ gives $3a_1 = 4$, resulting in $a_1 = \frac{4}{3}$. Since this is positive, we must reject $r = 2$. If $r = -2$, substituting into $a_1(1 + r) = 4$ gives $a_1(1 - 2) = 4$, which simplifies to $-a_1 = 4$, or $a_1 = -4$. This satisfies the condition $a_1 < 0$, confirming our parameters are $a_1 = -4$ and $r = -2$. ### The Final Calculation To find the sum of the first nine terms, $S_9$, we use the standard G.P. summation formula: $$S_n = \frac{a_1(r^n - 1)}{r - 1}$$ Substituting our known values: $$S_9 = \frac{-4((-2)^9 - 1)}{-2 - 1}$$ Given that $(-2)^9 = -512$, the calculation proceeds as follows: $$S_9 = \frac{-4(-512 - 1)}{-3} = \frac{-4(-513)}{-3} = \frac{2052}{-3} = -684$$ The problem states that this sum is equal to $4\lambda$. Therefore, $4\lambda = -684$, which leads to the final result: $$\lambda = -171$$

JEE Main 2020 (7 Jan Evening)

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Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. such that $a_{1} < 0, a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$ . If $\sum_{i = 1}^{9} a_{i} = 4 λ$ , then $λ$ is equal to :

Q: Wait, why do we reject the positive root for $r$?

That is a fantastic observation! When we solve $r^2 = 4$, we get $r = 2$ and $r = -2$. If we choose $r = 2$, our first equation $a_1(1+r) = 4$ becomes $3a_1 = 4$, which gives $a_1 = \frac{4}{3}$. This violates the initial condition that $a_1 < 0$. Therefore, we must reject $r = 2$ and accept $r = -2$.

Q: Why can't we just use the sum formula $S_n = \frac{a_1(r^n - 1)}{r - 1}$ directly without finding $a_1$ and $r$?

The sum formula is a powerful tool, but it is a skeleton that needs the flesh and blood of the specific parameters $a_1$ and $r$ to function. Without knowing these, the formula is just an algebraic expression. We must first deduce the specific values of the sequence to make the formula work for us.

Q: Is there a faster way to find the sum of the first 9 terms?

While you could potentially group terms or look for patterns, finding $a_1$ and $r$ is the most robust and reliable method. In JEE Advanced, precision and a systematic approach are your best friends. Don't look for shortcuts that might lead to errors; trust the process!

Select Answer:

Visualized Solution

Introduction to the G.P.

Let the G.P. be $a_{1}, a_{2}, a_{3}, \dots$
First term = $a_{1}$
Common ratio = $r$
Given constraint: $a_{1} < 0$

Expressing Terms in $a_{1}$ and $r$

$a_{2} = a_{1} r$
$a_{3} = a_{1} r^{2}$
$a_{4} = a_{1} r^{3}$

Setting up Equation 1

Given: $a_{1} + a_{2} = 4$
Substitute $a_{2} = a_{1} r$ : $a_{1} + a_{1} r = 4$
Factor out $a_{1}$ : $a_{1} (1 + r) = 4 \dots (i)$

Setting up Equation 2

Given: $a_{3} + a_{4} = 16$
Substitute $a_{3}, a_{4}$ : $a_{1} r^{2} + a_{1} r^{3} = 16$
Factor out $a_{1} r^{2}$ : $a_{1} r^{2} (1 + r) = 16 \dots (ii)$

Dividing Equations to find $r^{2}$

Divide equation $(ii)$ by $(i)$
$\frac{a _{1} r ^{2} ( 1 + r )}{a _{1} ( 1 + r )} = \frac{16}{4}$
Simplifying gives: $r^{2} = 4$

Solving for $r$

$r^{2} = 4 \Rightarrow r = \pm 2$
Case 1: $r = 2$
Case 2: $r = - 2$

Testing Case 1: $r = 2$

If $r = 2$ , substitute into $(i)$
$a_{1} (1 + 2) = 4 \Rightarrow 3 a_{1} = 4$
Result: $a_{1} = \frac{4}{3}$
Contradiction: $a_{1} > 0$ , but we need $a_{1} < 0$ .

Testing Case 2: $r = - 2$

If $r = - 2$ , substitute into $(i)$
$a_{1} (1 - 2) = 4 \Rightarrow a_{1} (- 1) = 4$
Result: $a_{1} = - 4$
Condition Met: $a_{1} < 0$ is satisfied.

The Sum of G.P. Formula

Sum of $n$ terms $S_{n} = \frac{a _{1} ( r ^{n} - 1 )}{r - 1}$
We need $S_{9}$ where $n = 9$ , $a_{1} = - 4$ , $r = - 2$

Substituting Values into $S_{9}$

$S_{9} = \frac{- 4 (( - 2 ) ^{9} - 1 )}{- 2 - 1}$
$S_{9} = \frac{- 4 ( - 512 - 1 )}{- 3}$
$S_{9} = \frac{- 4 ( - 513 )}{- 3}$

Calculating the Value of $S_{9}$

$S_{9} = \frac{4 \times 513}{- 3}$
$S_{9} = 4 \times (- 171)$
$S_{9} = - 684$

Equating to $4 λ$ and Solving

Given: $\sum_{i = 1}^{9} a_{i} = 4 λ$
So, $4 λ = - 684$
$λ = \frac{- 684}{4}$
$λ = - 171$

00:00 / 00:00

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

Analyzing the Setup

We are exploring a Geometric Progression (G.P.) defined by the sequence

a_{1}, a_{2}, a_{3}, \dots

. We are provided with two primary constraints:

a_{1} + a_{2} = 4

and

a_{3} + a_{4} = 16

Crucially, we must adhere to the condition that

a_{1} < 0

. This constraint serves as our guide to selecting the correct common ratio.

The Algebraic Detective Work

Any term in a G.P. is expressed as

a_{n} = a_{1} r^{n - 1}

. We can rewrite our given equations as follows:

a_{1} (1 + r) = 4

a_{1} r^{2} (1 + r) = 16

By dividing the second equation by the first, we eliminate

a_{1}

and

(1 + r)

through cancellation:

\frac{a _{1} r ^{2} ( 1 + r )}{a _{1} ( 1 + r )} = \frac{16}{4}

r^{2} = 4

The Trap Door and the Truth

The equation

r^{2} = 4

yields two potential values for the common ratio:

r = 2

r = - 2

. We must test these against the condition

a_{1} < 0

r = 2

, substituting into

a_{1} (1 + r) = 4

gives

3 a_{1} = 4

, resulting in

a_{1} = \frac{4}{3}

. Since this is positive, we must reject

r = 2

r = - 2

, substituting into

a_{1} (1 + r) = 4

gives

a_{1} (1 - 2) = 4

, which simplifies to

- a_{1} = 4

, or

a_{1} = - 4

. This satisfies the condition

a_{1} < 0

, confirming our parameters are

a_{1} = - 4

and

r = - 2

The Final Calculation

To find the sum of the first nine terms,

S_{9}

, we use the standard G.P. summation formula:

S_{n} = \frac{a _{1} ( r ^{n} - 1 )}{r - 1}

Substituting our known values:

S_{9} = \frac{- 4 (( - 2 ) ^{9} - 1 )}{- 2 - 1}

Given that

(- 2)^{9} = - 512

, the calculation proceeds as follows:

S_{9} = \frac{- 4 ( - 512 - 1 )}{- 3} = \frac{- 4 ( - 513 )}{- 3} = \frac{2052}{- 3} = - 684

The problem states that this sum is equal to

4 λ

. Therefore,

4 λ = - 684

, which leads to the final result:

λ = - 171

Common Traps & FAQs

Wait, why do we reject the positive root for

r

That is a fantastic observation! When we solve

r^{2} = 4

, we get

r = 2

and

r = - 2

. If we choose

r = 2

, our first equation

a_{1} (1 + r) = 4

becomes

3 a_{1} = 4

, which gives

a_{1} = \frac{4}{3}

. This violates the initial condition that

a_{1} < 0

. Therefore, we must reject

r = 2

and accept

r = - 2

Why can't we just use the sum formula

S_{n} = \frac{a _{1} ( r ^{n} - 1 )}{r - 1}

directly without finding

a_{1}

and

r

The sum formula is a powerful tool, but it is a skeleton that needs the flesh and blood of the specific parameters

a_{1}

and

r

to function. Without knowing these, the formula is just an algebraic expression. We must first deduce the specific values of the sequence to make the formula work for us.

Is there a faster way to find the sum of the first 9 terms?

While you could potentially group terms or look for patterns, finding

a_{1}

and

r

is the most robust and reliable method. In JEE Advanced, precision and a systematic approach are your best friends. Don't look for shortcuts that might lead to errors; trust the process!

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. such that $a_{1} < 0, a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$ . If $\sum_{i = 1}^{9} a_{i} = 4 λ$ , then $λ$ is equal to :

Select Answer:

Visualized Solution

Introduction to the G.P.

Expressing Terms in $a_{1}$ and $r$

Setting up Equation 1

Setting up Equation 2

Dividing Equations to find $r^{2}$

Solving for $r$

Testing Case 1: $r = 2$

Testing Case 2: $r = - 2$

The Sum of G.P. Formula

Substituting Values into $S_{9}$

Calculating the Value of $S_{9}$

Equating to $4 λ$ and Solving

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

Analyzing the Setup

The Algebraic Detective Work

The Trap Door and the Truth

The Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Introduction to the G.P.

Expressing Terms in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Setting up Equation 1

Setting up Equation 2

Dividing Equations to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Testing Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=2

Testing Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=−2

The Sum of G.P. Formula

Substituting Values into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S9​

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S9​

Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4λ and Solving

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

Analyzing the Setup

The Algebraic Detective Work

The Trap Door and the Truth

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​(>0),a2​,a3​,a4​,a5​ are in a G.P., a2​+a4​=2a3​+1 and 3a2​+a3​=2a4​, then a2​+a4​+2a5​ is equal to ______.

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Expressing Terms in $a_{1}$ and $r$

Dividing Equations to find $r^{2}$

Solving for $r$

Testing Case 1: $r = 2$

Testing Case 2: $r = - 2$

Substituting Values into $S_{9}$

Calculating the Value of $S_{9}$

Equating to $4 λ$ and Solving

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