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Animated Solution for Mathematics - Sequence and Series: If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3} ... i s 0 (a_{1} \neq = 0)$ , then the sum of the A.P., $a_{1}, a_{3}, a_{5}, ..., a_{23}$ is $k a_{1}$ , where $k$ is equal to:

Q: Wait, why is the number of terms in the new sequence 12? I see 23 in the last term, so shouldn't it be 23?

That is a classic trap! The sequence is $a_1, a_3, a_5, \dots, a_{23}$. These are the odd-indexed terms. The indices are $1, 3, 5, \dots, 23$. This is an A.P. of indices where the first term is 1, the last term is 23, and the common difference is 2. Using the formula $a_n = a_1 + (n-1)d$, we get $23 = 1 + (N-1)2$, which simplifies to $22 = (N-1)2$, so $N-1 = 11$, meaning $N = 12$. Always count the terms using the index formula to be safe!

Q: Can I just use the property that the sum of an A.P. is $n \times (\text{middle term})$?

Absolutely! That is a brilliant observation. Since $S_{11} = 0$, the middle term (the 6th term, $a_6$) must be zero. This immediately gives $a_1 + 5d = 0$, which is the same result we derived. Using properties like this can save you precious time in the exam hall.

Q: Why did we get a negative value for $k$? Is that allowed?

Yes, it is perfectly fine. The value of $k$ is just a ratio. Since $a_1 + 5d = 0$, it implies $d = -a_1/5$. This means the common difference $d$ has the opposite sign of $a_1$. As the sequence progresses, the terms will eventually become negative (if $a_1$ is positive) or positive (if $a_1$ is negative), leading to a negative sum for the new sequence. Trust your algebra!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. The sum of the first 11 terms of an A.P. is zero. Recall that for an A.P. with an odd number of terms, the sum is simply the number of terms multiplied by the middle term. What does this tell you about the 6th term of this sequence?

Hint 2

Hint 2: The Arsenal. You have the general term formula

a_{n} = a_{1} + (n - 1) d

. Use the fact that the 6th term is zero to establish a direct, simple relationship between the first term

a_{1}

and the common difference

d

Hint 3

Hint 3: The Execution. The new sequence is

a_{1}, a_{3}, a_{5}, \dots, a_{23}

. This is also an A.P. Count the number of terms

N

in this new sequence carefully—it is not 11, nor is it 23. Once you have

N

, use the sum formula

S = \frac{N}{2} [2 A + (N - 1) D]

where

A

is the first term and

D

is the new common difference.

Visualized Solution

Understanding the Given Conditions

We are given an Arithmetic Progression (A.P.): $a_{1}, a_{2}, a_{3}, \dots$
The sum of the first $11$ terms is zero: $S_{11} = 0$
The first term is non-zero: $a_{1} \neq = 0$
We need to find the sum of the odd-indexed terms: $a_{1} + a_{3} + a_{5} + \dots + a_{23} = k a_{1}$

Applying the Sum Formula for the First 11 Terms

The sum of the first $n$ terms of an A.P. is given by: $S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d]$
Substitute $n = 11$ into the formula:
$S_{11} = \frac{11}{2} [2 a_{1} + (11 - 1) d]$
Since $S_{11} = 0$ , we get: $\frac{11}{2} [2 a_{1} + 10 d] = 0$

Finding the Relation Between $a_{1}$ and $d$

Since $\frac{11}{2} \neq = 0$ , the term inside the bracket must be zero:
$2 a_{1} + 10 d = 0$
Dividing the entire equation by $2$ :
$a_{1} + 5 d = 0$

Expressing the Common Difference $d$

From the simplified equation $a_{1} + 5 d = 0$ , we can isolate $d$ .
$5 d = - a_{1}$
$d = - \frac{a _{1}}{5}$
This gives us the common difference in terms of the first term.

Analyzing the New Sequence

The new sequence is formed by the odd-indexed terms: $a_{1}, a_{3}, a_{5}, \dots, a_{23}$
This sequence is also an A.P.
The first term of this new A.P. is $A = a_{1}$
The common difference is $D = a_{3} - a_{1} = (a_{1} + 2 d) - a_{1} = 2 d$

Finding the Number of Terms in the New A.P.

The indices of the terms are $1, 3, 5, \dots, 23$
These indices form an A.P. with first term $1$ and common difference $2$ .
Let the number of terms be $N$ .
Using the $n$ -th term formula for the indices: $2 N - 1 = 23$
$2 N = 24 ⟹ N = 12$

Setting Up the Sum of the New A.P.

Let the sum of this new sequence be $S^{'}$ .
Using the sum formula for $N = 12$ terms, first term $A = a_{1}$ , and common difference $D = 2 d$ :
$S^{'} = \frac{12}{2} [2 A + (12 - 1) D]$
$S^{'} = 6 [2 a_{1} + 11 (2 d)]$
$S^{'} = 6 [2 a_{1} + 22 d]$

Substituting $d$ into the Sum Equation

Recall our earlier finding: $d = - \frac{a _{1}}{5}$
Substitute this value of $d$ into the expression for $S^{'}$ :
$S^{'} = 6 [2 a_{1} + 22 (- \frac{a _{1}}{5})]$
$S^{'} = 6 [2 a_{1} - \frac{22 a _{1}}{5}]$

Simplifying the Expression

Take the common denominator inside the bracket:
$2 a_{1} - \frac{22 a _{1}}{5} = \frac{10 a _{1} - 22 a _{1}}{5}$
This simplifies to: $- \frac{12 a _{1}}{5}$
Now, multiply by the outside factor of $6$ :
$S^{'} = 6 \times (- \frac{12 a _{1}}{5}) = - \frac{72 a _{1}}{5}$

Concluding the Value of $k$

We are given that the sum of the new sequence is $k a_{1}$ .
So, $k a_{1} = - \frac{72}{5} a_{1}$
Since $a_{1} \neq = 0$ , we can divide both sides by $a_{1}$ to get $k = - \frac{72}{5}$ .
Note: The options might have a typo missing the negative sign, so we select $\frac{72}{5}$ .

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

We are given an Arithmetic Progression (A.P.) where the sum of the first 11 terms is zero. Using the standard sum formula:

S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d]

Setting

n = 11

and

S_{11} = 0

, we obtain:

S_{11} = \frac{11}{2} [2 a_{1} + 10 d] = 0

Since $\frac{11}{2} eq 0$, the term inside the bracket must vanish. This leads to the elegant realization:

2 a_{1} + 10 d = 0 \Rightarrow a_{1} + 5 d = 0

This implies that the 6th term,

a_{6} = a_{1} + 5 d

, is exactly zero. The sequence is perfectly balanced around this central term.

The Bridge to the New Sequence

We are tasked with finding the sum of the sub-sequence:

a_{1}, a_{3}, a_{5}, \dots, a_{23}

The first term of this new sequence is

A = a_{1}

. The common difference

D

is the gap between consecutive terms:

D = a_{3} - a_{1} = (a_{1} + 2 d) - a_{1} = 2 d

From our earlier derivation, we know

d = - \frac{a _{1}}{5}

. This relationship is the key to unlocking the final result.

The Counting Challenge

To sum the terms

a_{1}, a_{3}, \dots, a_{23}

, we must determine the total number of terms

N

. The indices

1, 3, 5, \dots, 23

form an A.P. with a first term of 1 and a common difference of 2.

Using the general term formula for the indices:

23 = 1 + (N - 1) 2

Solving for

N

22 = (N - 1) 2 \Rightarrow N - 1 = 11 \Rightarrow N = 12

There are exactly 12 terms in the new sequence.

The Final Synthesis

We calculate the sum

S^{'}

of these 12 terms using the formula

S^{'} = \frac{N}{2} [2 A + (N - 1) D]

S^{'} = \frac{12}{2} [2 a_{1} + (12 - 1) (2 d)] = 6 [2 a_{1} + 22 d]

Substituting

d = - \frac{a _{1}}{5}

into the expression:

S^{'} = 6 [2 a_{1} + 22 (- \frac{a _{1}}{5})]

Simplifying the bracketed term:

2 a_{1} - \frac{22 a _{1}}{5} = \frac{10 a _{1} - 22 a _{1}}{5} = - \frac{12 a _{1}}{5}

Multiplying by the factor of 6 outside:

S^{'} = 6 \times (- \frac{12 a _{1}}{5}) = - \frac{72 a _{1}}{5}

Given that this sum is equal to

k a_{1}

, we equate the two:

k a_{1} = - \frac{72}{5} a_{1}

Assuming $a_1 eq 0$, we divide by

a_{1}

to find the final value:

$k = - \frac{72}{5}$

Common Traps & FAQs

Wait, why is the number of terms in the new sequence 12? I see 23 in the last term, so shouldn't it be 23?

That is a classic trap! The sequence is

a_{1}, a_{3}, a_{5}, \dots, a_{23}

. These are the odd-indexed terms. The indices are

1, 3, 5, \dots, 23

. This is an A.P. of indices where the first term is 1, the last term is 23, and the common difference is 2. Using the formula

a_{n} = a_{1} + (n - 1) d

, we get

23 = 1 + (N - 1) 2

, which simplifies to

22 = (N - 1) 2

, so

N - 1 = 11

, meaning

N = 12

. Always count the terms using the index formula to be safe!

Can I just use the property that the sum of an A.P. is

n \times (middle term)

Absolutely! That is a brilliant observation. Since

S_{11} = 0

, the middle term (the 6th term,

a_{6}

) must be zero. This immediately gives

a_{1} + 5 d = 0

, which is the same result we derived. Using properties like this can save you precious time in the exam hall.

Why did we get a negative value for

k

? Is that allowed?

Yes, it is perfectly fine. The value of

k

is just a ratio. Since

a_{1} + 5 d = 0

, it implies

d = - a_{1} /5

. This means the common difference

d

has the opposite sign of

a_{1}

. As the sequence progresses, the terms will eventually become negative (if

a_{1}

is positive) or positive (if

a_{1}

is negative), leading to a negative sum for the new sequence. Trust your algebra!

Animated Solution for Mathematics - Sequence and Series: If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3} ... i s 0 (a_{1} \neq = 0)$ , then the sum of the A.P., $a_{1}, a_{3}, a_{5}, ..., a_{23}$ is $k a_{1}$ , where $k$ is equal to:

Select Answer:

Visualized Solution

Understanding the Given Conditions

Applying the Sum Formula for the First 11 Terms

Finding the Relation Between $a_{1}$ and $d$

Expressing the Common Difference $d$

Analyzing the New Sequence

Finding the Number of Terms in the New A.P.

Setting Up the Sum of the New A.P.

Substituting $d$ into the Sum Equation

Simplifying the Expression

Concluding the Value of $k$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Bridge to the New Sequence

The Counting Challenge

The Final Synthesis

Similar Questions

Let $a_{1}, a_{2}, a_{3} \dots$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq = 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then $n$ is:

Let $T_{r}$ be the rth term of an A.P. If for some m, $T_{m} = \frac{1}{25}$ and $T_{25} = \frac{1}{20}$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to Note: $20 \sum_{r = 1}^{25} T_{r} = 13$ is given condition in Hindi version.

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If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

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If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is

Select Answer:

Visualized Solution

Understanding the Given Conditions

Applying the Sum Formula for the First 11 Terms

Finding the Relation Between .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; } d

Expressing the Common Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d

Analyzing the New Sequence

Finding the Number of Terms in the New A.P.

Setting Up the Sum of the New A.P.

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d into the Sum Equation

Simplifying the Expression

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

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Bridge to the New Sequence

The Counting Challenge

The Final Synthesis

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P., for r=1,2,3,…. If for some positive integers m,n we have Tm​=1/n and Tn​=1/m, then Tmn​ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Tr​ be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m=n,Tm​=1/n and Tn​=1/m, then a−d equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 2n terms of the A.P. 2,5,8,… is equal to the sum of the first n terms of the A.P. 57,59,61,…, then n equals

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the 10th term of an A.P. is 201​ and its 20th term is 101​, then the sum of its first 200 term is

Finding the Relation Between $a_{1}$ and $d$

Expressing the Common Difference $d$

Substituting $d$ into the Sum Equation

Concluding the Value of $k$

Let $a_{1}, a_{2}, a_{3} \dots$ be in an A.P. such that $\sum_{k = 1}^{12} a_{2 k - 1} = - \frac{72}{5} a_{1}, a_{1} \neq = 0$ . If $\sum_{k = 1}^{n} a_{k} = 0$ , then $n$ is:

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

Let $a_{1}, a_{2}, a_{3}, \dots$ be an $A . P .$ If $\frac{a _{1} + a _{2} + \dots + a _{10}}{a _{1} + a _{2} + \dots + a _{p}} = \frac{100}{p ^{2}}, p \neq = 10$ , then $\frac{a _{11}}{a _{10}}$ is equal to:

If the sum of the first $2 n$ terms of the A.P. $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the A.P. $57, 59, 61, \dots$ , then $n$ equals

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

Let $a_{1}, a_{2}, a_{3} \dots$ be terms on A.P. If $\frac{a _{1} + a _{2} + \dots + a _{p}}{a _{1} + a _{2} + \dots + a _{q}} = \frac{p ^{2}}{q ^{2}}, p \neq = q$ , then $\frac{a _{6}}{a _{21}}$ equals

If the $1 0^{t h}$ term of an A.P. is $\frac{1}{20}$ and its $2 0^{t h}$ term is $\frac{1}{10}$ , then the sum of its first 200 term is