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Animated Solution for Mathematics - Sequence and Series: In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57,$ then $S_{30} - S_{10}$ is equal to:

Select Answer:

Visualized Solution

Identify Given Information

Given sum of first 40 terms: $S_{40} = 1030$
Given sum of first 12 terms: $S_{12} = 57$
Target: Find the value of $S_{30} - S_{10}$

Recall the Sum Formula for an A.P.

The sum of the first $n$ terms of an A.P. is given by:
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Where $a$ is the first term and $d$ is the common difference.

Set up and Simplify Equation for $S_{40}$

For $n = 40$ : $S_{40} = \frac{40}{2} [2 a + (40 - 1) d] = 1030$
$20 [2 a + 39 d] = 1030$
Dividing by 20: $2 a + 39 d = \frac{103}{2}$ (Equation 1)

Set up and Simplify Equation for $S_{12}$

For $n = 12$ : $S_{12} = \frac{12}{2} [2 a + (12 - 1) d] = 57$
$6 [2 a + 11 d] = 57$
Dividing by 6: $2 a + 11 d = \frac{57}{6} = \frac{19}{2}$ (Equation 2)

Eliminate 'a' to find 'd'

Subtracting Equation 2 from Equation 1:
$(2 a + 39 d) - (2 a + 11 d) = \frac{103}{2} - \frac{19}{2}$
$28 d = \frac{103 - 19}{2}$

Solve for the Common Difference 'd'

$28 d = \frac{84}{2} = 42$
Solving for $d$ : $d = \frac{42}{28}$
$d = \frac{3}{2}$

Substitute 'd' to find 'a'

Substitute $d = \frac{3}{2}$ into Equation 2:
$2 a + 11 (\frac{3}{2}) = \frac{19}{2}$
$2 a + \frac{33}{2} = \frac{19}{2}$

Solve for the First Term 'a'

$2 a = \frac{19}{2} - \frac{33}{2}$
$2 a = \frac{19 - 33}{2} = \frac{- 14}{2} = - 7$
$a = - \frac{7}{2}$

Define the Target Expression $S_{30} - S_{10}$

Target expression: $S_{30} - S_{10}$
Using the formula: $S_{30} = \frac{30}{2} [2 a + 29 d]$
And $S_{10} = \frac{10}{2} [2 a + 9 d]$

Expand the Target Expression

$S_{30} - S_{10} = 15 (2 a + 29 d) - 5 (2 a + 9 d)$
Expanding the brackets:
$= 30 a + 435 d - 10 a - 45 d$

Simplify the Expression in terms of 'a' and 'd'

Combining like terms: $(30 a - 10 a) + (435 d - 45 d)$
Simplified target: $20 a + 390 d$

Substitute the Values of 'a' and 'd' (Standard Method)

Substitute $a = - \frac{7}{2}$ and $d = \frac{3}{2}$ :
Target $= 20 (- \frac{7}{2}) + 390 (\frac{3}{2})$
Target $= - 70 + 195 (3)$

Final Calculation

Calculation: $- 70 + 585$
$- 70 + 585 = 515$
Final Answer: $515$

The JEE Shortcut (Pro-Tip)

Notice the simplified target: $20 a + 390 d = 10 (2 a + 39 d)$
Recall Equation 1: $2 a + 39 d = \frac{103}{2}$
Direct substitution: $10 (\frac{103}{2}) = 5 \times 103 = 515$
Takeaway: $S_{12} = 57$ was redundant information!

00:00 / 00:00

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

Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are uncovering the hidden elegance of Arithmetic Progressions.

When you first look at a problem where

S_{40} = 1030

and

S_{12} = 57

, your instinct might be to immediately dive into the algebra, solving for the first term

a

and the common difference

d

. While that is a perfectly valid, robust path, I want to show you how to see the 'soul' of the sequence.

The Master Key

Every Arithmetic Progression is governed by its sum formula:

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

This formula is your master key. It connects the number of terms

n

, the starting point

a

, and the growth rate

d

When we apply this to

S_{40} = 1030

, we get

20 [2 a + 39 d] = 1030

, which simplifies beautifully to:

2 a + 39 d = 51.5

This is our first anchor.

The Trap of the Second Equation

Then we have

S_{12} = 57

. Applying our key, we get

6 [2 a + 11 d] = 57

, or

2 a + 11 d = 9.5

Most students would now subtract these two equations to find

d

, then find

a

, and finally plug them into the target expression

S_{30} - S_{10}

. But wait! Let us pause.

What is the target? We want

S_{30} - S_{10}

. Let us write it out:

S_{30} - S_{10} = \frac{30}{2} [2 a + 29 d] - \frac{10}{2} [2 a + 9 d]

This becomes

15 (2 a + 29 d) - 5 (2 a + 9 d)

. Expanding this, we get

30 a + 435 d - 10 a - 45 d

, which simplifies to

20 a + 390 d

The JEE Moment

Now, look closely at

20 a + 390 d

. If we factor out a

10

, we get

10 (2 a + 39 d)

Do you see it? That term

2 a + 39 d

is exactly what we found from our very first equation! We do not need

a

d

individually, and we do not need the second equation at all.

The value is simply:

10 \times 51.5 = 515

This is the essence of JEE mastery: recognizing that the path of least resistance is often hidden in plain sight. The second equation was a distraction, a test of your confidence to trust your algebraic intuition.

The final answer is 515. Keep practicing this, and you will start seeing these shortcuts everywhere. You are doing great; keep pushing!

Common Traps & FAQs

Wait, why do we reject the negative root here? Or do we even have a root?

In this specific problem, we are dealing with a linear system of equations, not a quadratic one. There are no roots to reject! We are simply solving for two variables,

a

and

d

, using elimination. If you ever encounter a quadratic in an A.P. problem, always check if the context (like the number of terms

n

) forces a specific sign.

Why can't we just use energy conservation? Wait, this is math, not physics. Why is this so confusing?

I hear your frustration! It is easy to get lost in the formulas. Remember, Arithmetic Progressions are just patterns. Think of

S_{n}

as the total 'accumulation' of the sequence. You don't need energy conservation here, just the 'master key' formula

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

. Keep your focus on the algebraic structure.

Why did the problem give us

S_{12} = 57

if we didn't need it for the shortcut?

That is a classic JEE trap! Examiners often include redundant information to test your ability to identify the most efficient path. If you find a shortcut that bypasses a piece of data, it usually means you have found a deeper structural relationship in the problem. Trust your derivation!

Animated Solution for Mathematics - Sequence and Series: In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57,$ then $S_{30} - S_{10}$ is equal to:

Select Answer:

Visualized Solution

Identify Given Information

Recall the Sum Formula for an A.P.

Set up and Simplify Equation for $S_{40}$

Set up and Simplify Equation for $S_{12}$

Eliminate 'a' to find 'd'

Solve for the Common Difference 'd'

Substitute 'd' to find 'a'

Solve for the First Term 'a'

Define the Target Expression $S_{30} - S_{10}$

Expand the Target Expression

Simplify the Expression in terms of 'a' and 'd'

Substitute the Values of 'a' and 'd' (Standard Method)

Final Calculation

The JEE Shortcut (Pro-Tip)

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

Analyzing the Setup

The Master Key

The Trap of the Second Equation

The JEE Moment

Similar Questions

Let $S_{n}$ denote the sum of first $n$ -terms of an arithmetic progression. If $S_{10} = 530, S_{5} = 140$ , then $S_{20} - S_{6}$ is equal to :

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 :

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15 : 7$ , then $S_{15} - S_{5}$ is equal to:

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 -

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 $a_{1}, a_{2}, a_{3}, \dots$ be in an arithmetic progression of positive terms. Let $A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + \dots + a_{2 k - 1}^{2} - a_{2 k}^{2}$ . If $A_{3} = - 153, A_{5} = - 435$ and $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66$ , then $a_{17} - A_{7}$ is equal to

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n} = 3 S_{2 n}$ , then the value of $\frac{S _{4 n}}{S _{2 n}}$ 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

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:

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this A.P. is \dots.

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; } In an arithmetic progression, if S40​=1030 and S12​=57, then S30​−S10​ is equal to:

Select Answer:

Visualized Solution

Identify Given Information

Recall the Sum Formula for an A.P.

Set up and Simplify Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S40​

Set up and Simplify Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S12​

Eliminate 'a' to find 'd'

Solve for the Common Difference 'd'

Substitute 'd' to find 'a'

Solve for the First Term 'a'

Define the Target Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S30​−S10​

Expand the Target Expression

Simplify the Expression in terms of 'a' and 'd'

Substitute the Values of 'a' and 'd' (Standard Method)

Final Calculation

The JEE Shortcut (Pro-Tip)

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

Analyzing the Setup

The Master Key

The Trap of the Second Equation

The JEE Moment

Similar Questions

.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 of an arithmetic progression. If S10​=530,S5​=140, then S20​−S6​ is equal to :

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sn​ be the sum of the first n terms of an arithmetic progression. If S3n​=3S2n​, then the value of S2n​S4n​​ 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:

Animated Solution for Mathematics - Sequence and Series: In an arithmetic progression, if $S_{40} = 1030$ and $S_{12} = 57,$ then $S_{30} - S_{10}$ is equal to:

Set up and Simplify Equation for $S_{40}$

Set up and Simplify Equation for $S_{12}$

Define the Target Expression $S_{30} - S_{10}$

Let $S_{n}$ denote the sum of first $n$ -terms of an arithmetic progression. If $S_{10} = 530, S_{5} = 140$ , then $S_{20} - S_{6}$ is equal to :

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 :

Let $S_{n}$ be the sum of the first $n$ terms of an arithmetic progression. If $S_{3 n} = 3 S_{2 n}$ , then the value of $\frac{S _{4 n}}{S _{2 n}}$ 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: