The sum of first 45 natural numbers is:(a) 1035(b) 1280(c) 2070(d) 2140

We can use here the concept that whenever the difference between successive consecutive terms of a sequence is constant, it forms an AP and thereafter we will use the formula to find the sum of n terms of an AP to get the sum of first 45 natural numbers. Complete step-by-step answer: The first 45 natural numbers can be written as: 1, 2, 3 , 3, ……. , 45. Here, we can see that the difference between any two successive consecutive terms of this series is 1 which is a constant, so this sequence forms an AP. The constant 1 which is the difference of two terms here is called the common difference of the AP. So, we have an AP with the first term “a”= 1 and also the common difference “d” = 1. Now, to find the sum of this series we will use the formula for the sum of n terms of an AP, which is given as: Sn=n2(2a+(n−1)d) Since, in the given series we have 45 terms, so we have the value of n equal to 45, a = 1 and also d = 1. So, we will substitute these values in this formula to get the sum: S45=452(2×1+(45−1)1)=452(2+44)=452×46=45×23=1035 Therefore, the value of the sum of the first 45 natural numbers is 1035. Hence, option (a) is the correct answer. Note: It should be noted here that we can also use a shortcut formula used to find the sum of n terms of an AP, which is Sn=n2(firstterm+lastterm). Since, here both the first term and the last term are known, using this formula may save our time.