JIPMAT 2023 (QA) - Assertion [A]: Sum of the first hundred even natural numbers divisible by 5 is 45050. <br>Reason (R): Sum of the first n-terms of an Arithmetic Progression is given by S = (n/2) *(a + l) where a=first term, l=last term. <br>Choose the correct answer from the options given below. | PYQs + Solutions

JIPMAT 2023

Algebra

Progression & Series

Conceptual

Assertion [A]: Sum of the first hundred even natural numbers divisible by 5 is 45050.
Reason (R): Sum of the first n-terms of an Arithmetic Progression is given by $S = (n/2) *(a + l)$ where a=first term, l=last term.
Choose the correct answer from the options given below.

Both [A] and (R) are true and (R) is the correct explanation of (A).

Both [A] and (R) are true but (R) is NOT the correct explanation of (A).

[A] is true but (R) is false.

[A] is false but (R) is true.

Correct Option: 4

1) First, let's verify if [A] is true:

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- First hundred even numbers divisible by 5: 10, 20, 30,..., 1000

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- This forms an AP with:

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* First term

a = 10

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* Common difference

d = 10

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* Number of terms

n = 100

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* Last term

l = 10 + (100-1)10 = 1000

2) Using formula given in [R]:

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

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\frac{100}{2}(10 + 1000)

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50 × 1010

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50500

3) Therefore, [A] is false as it states sum is 45050.

4) Let's verify if [R] is true:

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- The formula given

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

is indeed correct for arithmetic progression

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- This is a standard formula for sum of AP

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- Therefore [R] is true