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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)S = (n/2) *(a + l) where a=first term, l=last term.
Choose the correct answer from the options given below.

Solution

✅ Correct Option: 4
  1. First, let's verify if [A] is true:
  • First hundred even numbers divisible by 5: 10, 20, 30,..., 1000
    • This forms an AP with:
      • First term a=10a = 10
    • Common difference d=10d = 10
    • Number of terms n=100n = 100
    • Last term l=10+(100−1)10=1000l = 10 + (100-1)10 = 1000
  1. Using formula given in [R]:

S=n2(a+l)S = \frac{n}{2}(a + l)

= 1002(10+1000)\frac{100}{2}(10 + 1000)

= 50×101050 × 1010

= 5050050500

  1. Therefore, [A] is false as it states sum is 45050.
  1. Let's verify if [R] is true:
  • The formula given S=n2(a+l)S = \frac{n}{2}(a + l) is indeed correct for arithmetic progression
    • This is a standard formula for sum of AP
    • Therefore [R] is true

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