JEE Main 2020 (8 January Shift 2)

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Animated Solution for Mathematics - Sequence and Series: 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:

Q: Is it possible for the first term $a$ and the common difference $d$ to be the same value?

Absolutely! In this specific problem, both $a$ and $d$ turn out to be $\frac{1}{200}$. This is perfectly valid. It just means the sequence starts at $\frac{1}{200}$ and increases by $\frac{1}{200}$ at every step, effectively making the $n^{th}$ term simply $\frac{n}{200}$.

Q: Why do we use $S_n = \frac{n}{2} [2a + (n-1)d]$ instead of $S_n = \frac{n}{2}(a + l)$?

You could use $S_n = \frac{n}{2}(a + l)$ if you first calculated the last term $l$ (which is $T_{200}$). Both formulas are mathematically equivalent. However, the formula $S_n = \frac{n}{2} [2a + (n-1)d]$ is often safer because it relies directly on the parameters $a$ and $d$ we already derived, reducing the risk of a calculation error in finding the $200^{th}$ term separately.

Select Answer:

Visualized Solution

Problem Setup

Given: $T_{10} = \frac{1}{20}$
Given: $T_{20} = \frac{1}{10}$
To find: $S_{200}$

General Term Formula

Formula: $T_{n} = a + (n - 1) d$
Where $a$ is the first term and $d$ is the common difference.

Equations for $T_{10}$ and $T_{20}$

For $n = 10$ : $a + 9 d = \frac{1}{20}$ (Eq 1)
For $n = 20$ : $a + 19 d = \frac{1}{10}$ (Eq 2)

Subtracting the Equations

Subtracting (Eq 1) from (Eq 2):
$(a + 19 d) - (a + 9 d) = \frac{1}{10} - \frac{1}{20}$

Solving for $d$

$10 d = \frac{2 - 1}{20}$
$10 d = \frac{1}{20}$
$d = \frac{1}{200}$

Solving for $a$

Substitute $d = \frac{1}{200}$ into (Eq 1):
$a + 9 (\frac{1}{200}) = \frac{1}{20}$

Isolating $a$

$a = \frac{1}{20} - \frac{9}{200}$
$a = \frac{10 - 9}{200}$
$a = \frac{1}{200}$

Sum of $n$ Terms Formula

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

Substituting Values for $S_{200}$

For $n = 200$ , $a = \frac{1}{200}$ , $d = \frac{1}{200}$ :
$S_{200} = \frac{200}{2} [2 (\frac{1}{200}) + (200 - 1) (\frac{1}{200})]$

Simplifying the Expression

$S_{200} = 100 [\frac{2}{200} + \frac{199}{200}]$
$S_{200} = 100 [\frac{2 + 199}{200}]$

Final Calculation

$S_{200} = 100 [\frac{201}{200}]$
$S_{200} = \frac{201}{2} = 100 \frac{1}{2}$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

An Arithmetic Progression (AP) is defined by its first term

a

and a common difference

d

. The

n^{t h}

term of an AP is given by the formula:

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

We are provided with two specific checkpoints: the

1 0^{t h}

term is

\frac{1}{20}

and the

2 0^{t h}

term is

\frac{1}{10}

The Detective Work

Finding the DNA

To define the sequence, we must solve for

a

and

d

. Using the general term formula, we establish the following system of linear equations:

For

n = 10

a + 9 d = \frac{1}{20} (Equation 1)

For

n = 20

a + 19 d = \frac{1}{10} (Equation 2)

Subtracting Equation 1 from Equation 2 eliminates

a

(a + 19 d) - (a + 9 d) = \frac{1}{10} - \frac{1}{20}

10 d = \frac{1}{20}

d = \frac{1}{200}

Now, we substitute

d = \frac{1}{200}

back into Equation 1 to determine

a

a + 9 (\frac{1}{200}) = \frac{1}{20}

a = \frac{10}{200} - \frac{9}{200} = \frac{1}{200}

Both the starting term and the common difference are identified as

a = \frac{1}{200}

and

d = \frac{1}{200}

The Grand Summation

To find the total distance covered over the first

200

steps, we use the sum formula for an AP:

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

Substituting

n = 200

a = \frac{1}{200}

, and

d = \frac{1}{200}

into the formula:

S_{200} = \frac{200}{2} [2 (\frac{1}{200}) + (200 - 1) (\frac{1}{200})]

Simplifying the expression inside the brackets:

S_{200} = 100 [\frac{2}{200} + \frac{199}{200}]

S_{200} = 100 [\frac{201}{200}]

Performing the final calculation:

S_{200} = \frac{201}{2} = 100.5

The total distance covered over the first

200

steps is $100.5$ (or $\frac{201}{2}$ ).

Common Traps & FAQs

Wait, why do we use the formula

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

instead of just finding the average?

That is a great question! While the average of the terms in an AP is indeed related to the sum, the sequence itself is linear, not constant. The formula

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

is the fundamental definition of an AP. It allows us to translate the 'checkpoints' given in the problem into a precise algebraic system. Without this, we would be guessing rather than calculating.

Is it possible for the first term

a

and the common difference

d

to be the same value?

Absolutely! In this specific problem, both

a

and

d

turn out to be

\frac{1}{200}

. This is perfectly valid. It just means the sequence starts at

\frac{1}{200}

and increases by

\frac{1}{200}

at every step, effectively making the

n^{t h}

term simply

\frac{n}{200}

Why do we use

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

instead of

S_{n} = \frac{n}{2} (a + l)

You could use

S_{n} = \frac{n}{2} (a + l)

if you first calculated the last term

l

(which is

T_{200}

). Both formulas are mathematically equivalent. However, the formula

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

is often safer because it relies directly on the parameters

a

and

d

we already derived, reducing the risk of a calculation error in finding the

20 0^{t h}

term separately.

Animated Solution for Mathematics - Sequence and Series: 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:

Select Answer:

Visualized Solution

Problem Setup

General Term Formula

Equations for $T_{10}$ and $T_{20}$

Subtracting the Equations

Solving for $d$

Solving for $a$

Isolating $a$

Sum of $n$ Terms Formula

Substituting Values for $S_{200}$

Simplifying the Expression

Final Calculation

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

Analyzing the Setup

The Detective Work

The Grand Summation

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

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-

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

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 :

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + .....$ is $(102) m$ , then $m$ is equal to:

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 $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of first $4 n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$ , then the sum of the first $6 n$ terms of the arithmetic progression is equal to:

Let $S_{n}$ denote the sum of the first n terms of an A.P. If $S_{4} = 16$ and $S_{6} = - 48$ , then $S_{10}$ is equal to :

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

Animated Solution for Mathematics - Sequence and Series: .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:

Select Answer:

Visualized Solution

Problem Setup

General Term Formula

Equations for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T10​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T20​

Subtracting the Equations

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

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

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

Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Terms Formula

Substituting Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S200​

Simplifying the Expression

Final Calculation

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

Analyzing the Setup

The Detective Work

The Grand Summation

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; } Let 2nd,8th and 44th terms of a non-constant A.P. be respectively the 1st,2nd and 3rd terms of G.P. If the first term of A.P. is 1 then the sum of first 20 terms is equal to-

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the sum of the first 40 terms of the series, 3+4+8+9+13+14+18+19+.....is (102)m, then m is equal to:

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 the first n terms of an A.P. If S4​=16 and S6​=−48, then S10​ is equal to :

.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

Animated Solution for Mathematics - Sequence and Series: 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:

Equations for $T_{10}$ and $T_{20}$

Solving for $d$

Solving for $a$

Isolating $a$

Sum of $n$ Terms Formula

Substituting Values for $S_{200}$

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

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-

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 :

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + .....$ is $(102) m$ , then $m$ is equal to:

Let $S_{n}$ denote the sum of the first n terms of an A.P. If $S_{4} = 16$ and $S_{6} = - 48$ , then $S_{10}$ is equal to :

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