JEE Main 2024 (31 Jan Shift 2)

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Animated Solution for Mathematics - Sequence and Series: Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

Q: How do I know which terms to pick for the G.P. condition $b^2 = ac$?

The condition $b^2 = ac$ applies to any three consecutive terms of a G.P. The problem states that $a_2, a_8, a_{44}$ are the $1^{st}, 2^{nd},$ and $3^{rd}$ terms of a G.P. respectively. This means they are consecutive terms in that G.P., so $a_8$ is the middle term.

Q: I keep getting a different sum. Did I mess up the $S_n$ formula?

Double-check your $n$ value! We are looking for the sum of the first $20$ terms, so $n=20$. Ensure you are using $a=1$ and the $d$ you derived. The formula $S_n = \frac{n}{2}[2a + (n-1)d]$ is robust, so if the result differs, re-verify your arithmetic in the final step $10[2 + 95]$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the bridge between two worlds. You are given an Arithmetic Progression (A.P.) and told that specific terms from it form a Geometric Progression (G.P.). The key is to express the terms of the A.P. using the common difference

d

and then apply the fundamental property of three terms in a G.P., which is that the square of the middle term equals the product of the outer two.

Hint 2

Hint 2: Your toolkit for this problem is the definition of the

n^{t h}

term of an A.P.,

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

, and the G.P. condition

b^{2} = a c

. Once you substitute the expressions for

a_{2}

a_{8}

, and

a_{44}

into the G.P. condition, you will obtain a quadratic equation in

d

Hint 3

Hint 3: After expanding the quadratic equation

(1 + 7 d)^{2} = (1 + d) (1 + 43 d)

, you will find terms that cancel out nicely. Remember the constraint that the A.P. is 'non-constant', which is your signal to reject the root

d = 0

and proceed with the non-zero value of

d

to calculate the sum

S_{20} = \frac{n}{2} [2 a + (n - 1) d]

Visualized Solution

The Given Sequence

Given: A non-constant A.P. with first term $a = 1$ .
Terms of interest: $2^{n d}$ , $8^{t h}$ , and $4 4^{t h}$ .
These terms form a G.P. in the given order.

Expressing A.P. Terms

General term of A.P.: $a_{n} = a + (n - 1) d$
Since $a = 1$ :
$a_{2} = 1 + d$
$a_{8} = 1 + 7 d$
$a_{44} = 1 + 43 d$

The G.P. Condition

The problem states these three terms form a Geometric Progression (G.P.).
$a_{2}, a_{8}, a_{44}$ are in G.P.
Condition for G.P.: $(M i dd l e T er m)^{2} = F i r s t \times T hi r d$
Therefore, $(a_{8})^{2} = a_{2} \cdot a_{44}$

Substituting the Expressions

Substitute the A.P. expressions into the G.P. condition:
$(1 + 7 d)^{2} = (1 + d) (1 + 43 d)$

Expanding the Left Side

Expand $(1 + 7 d)^{2}$ using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$(1)^{2} + 2 (1) (7 d) + (7 d)^{2}$
$= 1 + 14 d + 49 d^{2}$

Expanding the Right Side

Expand $(1 + d) (1 + 43 d)$ :
$1 (1 + 43 d) + d (1 + 43 d)$
$= 1 + 43 d + d + 43 d^{2}$
$= 1 + 44 d + 43 d^{2}$

Equating and Simplifying

Equating both sides:
$1 + 14 d + 49 d^{2} = 1 + 44 d + 43 d^{2}$
Subtracting $1$ from both sides:
$49 d^{2} + 14 d = 44 d + 43 d^{2}$
Rearranging terms:
$(49 d^{2} - 43 d^{2}) + (14 d - 44 d) = 0$
$6 d^{2} - 30 d = 0$

Solving for $d$

Factoring the equation:
$6 d (d - 5) = 0$
Possible values for $d$ :
$d = 0$ or $d = 5$
Since the A.P. is non-constant, $d \neq = 0$ .
Therefore, $d = 5$ .

The Goal: Sum of First 20 Terms

We need to find the sum of the first 20 terms ( $S_{20}$ ).
Formula for sum of an A.P.:
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
We know: $n = 20$ , $a = 1$ , $d = 5$ .

Substituting into Sum Formula

Substitute the known values into the formula:
$S_{20} = \frac{20}{2} [2 (1) + (20 - 1) 5]$
$S_{20} = 10 [2 + 19 (5)]$

Final Calculation

Calculate the terms inside the bracket:
$19 \times 5 = 95$
$S_{20} = 10 [2 + 95]$
$S_{20} = 10 [97]$
Final Answer: $S_{20} = 970$

Key Takeaways

Key Takeaway:
Express A.P. terms as $a + (n - 1) d$ .
Use the property $b^{2} = a c$ for terms in G.P.
Always check constraints like non-constant ( $d \neq = 0$ ).

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

Solution Diagram

The Harmony of Sequences

A Mathematical Journey

Imagine you are standing at the edge of a vast mathematical landscape. On one side, we have the steady, rhythmic march of an Arithmetic Progression (A.P.), where each step is a constant addition. On the other, we have the explosive, multiplicative growth of a Geometric Progression (G.P.).

Today, we are going to build a bridge between these two worlds.

Phase 1

Defining the Players

We are given an A.P. where the first term

a = 1

. We don't know the common difference

d

, but we know it's a 'non-constant' sequence, which is our first clue that $d eq 0$.

We are interested in three specific milestones in this sequence: the

2^{n d}

term, the

8^{t h}

term, and the

4 4^{t h}

term. Using the general formula for an A.P.,

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

, we can write these as:

a_{2} = 1 + d

a_{8} = 1 + 7 d

a_{44} = 1 + 43 d

These three values are not just random numbers; they are the first three terms of a G.P. In any G.P., if

x, y, z

are consecutive terms, then the ratio

\frac{y}{x} = \frac{z}{y}

, which leads us to the beautiful property

y^{2} = x z

Since our middle term is

a_{8}

, we must have:

(a_{8})^{2} = a_{2} \cdot a_{44}

Phase 2

The Algebraic Dance

Now, let us substitute our expressions into this condition:

(1 + 7 d)^{2} = (1 + d) (1 + 43 d)

Expanding the left side gives us

1 + 14 d + 49 d^{2}

. Expanding the right side gives us

1 + 43 d + d + 43 d^{2}

, which simplifies to

1 + 44 d + 43 d^{2}

Now, look at the equation:

1 + 14 d + 49 d^{2} = 1 + 44 d + 43 d^{2}

The constant

1

on both sides vanishes. We are left with:

49 d^{2} + 14 d = 43 d^{2} + 44 d

Bringing everything to one side, we get

6 d^{2} - 30 d = 0

. Factoring this, we find

6 d (d - 5) = 0

This gives us two possibilities:

d = 0

d = 5

. Since the sequence is non-constant,

d

cannot be

0

. Thus, we have discovered the heartbeat of our sequence: $d = 5$ .

Phase 3

The Final Ascent

We have conquered the hardest part. Now, we simply need to find the sum of the first

20

terms of this A.P. The sum formula is our trusty companion:

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

Plugging in our values

n = 20

a = 1

, and

d = 5

, we get:

S_{20} = \frac{20}{2} [2 (1) + (20 - 1) 5]

S_{20} = 10 [2 + 19 (5)]

S_{20} = 10 [2 + 95]

S_{20} = 10 [97] = 970

And there it is! Through the interplay of arithmetic and geometric properties, we have arrived at the final answer of $970$ .

Common Traps & FAQs

Wait, why do we reject the

d = 0

root? Is it just a formality?

It is absolutely not just a formality! The problem explicitly states the A.P. is 'non-constant'. If

d = 0

, every term in the sequence would be

1

, making it a constant sequence. In mathematics, constraints are the boundaries of our reality; ignoring them would lead us to a trivial solution that violates the problem's premise.

Why can't we just use the sum formula for a G.P. here?

That is a common trap! The problem tells us that three specific terms (

a_{2}, a_{8}, a_{44}

) form a G.P., not that the entire A.P. is a G.P. We only use the G.P. property to find the common difference

d

of the A.P. Once we have

d

, we return to the A.P. world to calculate the sum.

Is there a faster way to solve

(1 + 7 d)^{2} = (1 + d) (1 + 43 d)

without expanding everything?

You could try substitution or shifting, but for this specific quadratic, expansion is actually quite elegant. Notice how the

1

on both sides cancels out immediately? Sometimes, the most direct path is the safest one to avoid algebraic errors under exam pressure.

How do I know which terms to pick for the G.P. condition

b^{2} = a c

The condition

b^{2} = a c

applies to any three consecutive terms of a G.P. The problem states that

a_{2}, a_{8}, a_{44}

are the

1^{s t}, 2^{n d},

and

3^{r d}

terms of a G.P. respectively. This means they are consecutive terms in that G.P., so

a_{8}

is the middle term.

I keep getting a different sum. Did I mess up the

S_{n}

formula?

Double-check your

n

value! We are looking for the sum of the first

20

terms, so

n = 20

. Ensure you are using

a = 1

and the

d

you derived. The formula

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

is robust, so if the result differs, re-verify your arithmetic in the final step

10 [2 + 95]

Animated Solution for Mathematics - Sequence and Series: Let $2^{n d}, 8^{t h}$ and $4 4^{t h}$ terms of a non-constant A.P. be respectively the $1^{s t}, 2^{n d}$ and $3^{r d}$ terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

Select Answer:

Visualized Solution

The Given Sequence

Expressing A.P. Terms

The G.P. Condition

Substituting the Expressions

Expanding the Left Side

Expanding the Right Side

Equating and Simplifying

Solving for $d$

The Goal: Sum of First 20 Terms

Substituting into Sum Formula

Final Calculation

Key Takeaways

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

The Harmony of Sequences

Phase 1

Phase 2

Phase 3

Similar Questions

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

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

Let $s_{1}, s_{2}, s_{3}, \dots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1, 2, 3, \dots, 10$ and the common differences are $1, 3, 5, \dots, 19$ respectively. Then $\sum_{i = 1}^{10} s_{i}$ is equal to

Let $a, b, c > 1, a^{'}, b^{'}$ and $c^{'}$ be in A.P., and $lo g_{a} b, lo g_{b} a$ and $lo g_{c} b$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a + 4 b + c}{3}$ and the common difference is $\frac{a - 8 b + c}{10}$ is $- 444$ , then $ab c$ is equal to

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. 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.

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

Let $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :

Select Answer:

Visualized Solution

The Given Sequence

Expressing A.P. Terms

The G.P. Condition

Substituting the Expressions

Expanding the Left Side

Expanding the Right Side

Equating and Simplifying

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

The Goal: Sum of First 20 Terms

Substituting into Sum Formula

Final Calculation

Key Takeaways

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

The Harmony of Sequences

Phase 1

Phase 2

Phase 3

Similar Questions

.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

.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 1/20 and its 20th term is 1/10, then the sum of its first 200 terms is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,a3​,… are in A.P. such that a1​+a7​+a16​=40, then the sum of the first 15 terms of this A.P. is :

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​ denote the sum of first n terms an arithmetic progression. If S20​=790 and S10​=145, then S15​−S5​ is :

Solving for $d$

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

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

Consider an A. P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its $1 1^{t h}$ term is :

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. is :

Let $S_{n}$ denote the sum of first $n$ terms an arithmetic progression. If $S_{20} = 790$ and $S_{10} = 145$ , then $S_{15} - S_{5}$ is :