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Arithmetic progression is a series of numbers with a constant difference between consecutive terms in an <\/span>arithmetic sequence<\/span>. For instance, 3, 5, 7, 9, 11, \u2026\u2026 is an arithmetic progression with a common difference of 2 between each succeeding and preceding term. Another example of an AP is <\/span>the sequence of natural numbers whose common difference is 1<\/span>. This article shed light on the <\/span>sum of arithmetic progression.<\/span><\/p>\n Carl Friedrich Gauss invented a method to find the <\/span>sum of n natural numbers<\/b>. However, the truth behind this claim remains unknown as the origin of this method dates back to the 5<\/span>th<\/span> century BC.\u00a0<\/span><\/p>\n <\/span><\/p>\n Suppose you have an arithmetic progression (AP) in front of you with the first term as <\/span>a_1<\/span><\/span> and common difference as d, then you can easily find the <\/span>sum of n terms.<\/b><\/p>\na_{1}, a_{2}, a_{3}, \\ldots \\ldots \\ldots a_{n}<\/span>\n Then,<\/p>\n \na_{1}=a <\/span> \\text { and, } a_{n}=a+(n-1) d<\/span><\/span><\/p>\n Now, sum of AP<\/b> is \\mathrm{S}_{n}=\\mathrm{a}_{1}+\\mathrm{a}_{2}+\\mathrm{a}_{3}+\\ldots \\ldots \\ldots+\\mathrm{a}_{n-1}+\\mathrm{a}_{n}<\/span><\/span><\/p>\n S_{n}=a+(a+d)+(a+2 d)+(a+3 d)+\\ldots+\\{a+(n-2) d\\}+\\{a+(n-1) d\\} \\\\\n\\text { equation } 1<\/span><\/span><\/p>\n After writing equation 1 in reverse order,<\/span><\/span><\/p>\n \nS_{n}=\\{a+(n-1) d\\}+\\{a+(n-2) d\\}+\\{a+(n-3) d\\}+\\ldots+(a+3 d)+(a+2 d)+(a+d)+ \\\\\na-\\text { equation } 2\n<\/span><\/span><\/p>\n Adding equation 1 and equation 2, we get,\u00a0<\/span><\/p>\n 2 S_{n}=\\{2 a+(n-1) d\\}+\\{2 a+(n-1) d\\}+\\{2 a+(n-1) d\\}+\\ldots \\ldots+\\{a+(n-2) d\\}<\/span><\/span><\/p>\n \u21d2 2 S_{n}=n\\{2 a+(n-1) d\\} <\/span><\/span><\/p>\n \u2234 S_{n}=\\frac{n}{2}\\{2 a+(n-1) d\\}<\/span>\u00a0 \u00a0 ……. Equation 3<\/span><\/p>\n Therefore, if you do not know the last term of an AP series, then you can use\u00a0<\/span><\/p>\n S_{n}=\\frac{n}{2}\\{2 a+(n-1) d\\}<\/span> to find the <\/span>sum of AP<\/b>.<\/span><\/p>\n If you are given the last term l,<\/span><\/p>\n then you know that [\\mathrm{l}=n^{\\text {th }} \\text { term }=a+(n-1) d<\/span><\/span><\/p>\n By substituting the value of l in equation 3, we get<\/span><\/p>\n \\mathrm{S}_{\\mathrm{n}}=\\frac{n}{2}\\{\\mathrm{a}+[\\mathrm{a}+(\\mathrm{n}-1) \\mathrm{d}]\\}<\/span><\/span><\/p>\n \\mathrm{S}_{\\mathrm{n}}=\\frac{n}{2}(\\mathrm{a+l})<\/span><\/span><\/p>\n Therefore, the <\/span>sum of terms in AP<\/b> when the last term is \u2018l\u2019 is \\mathrm{S}_{n}=\\frac{n}{2}(\\mathrm{a+l})<\/span><\/span><\/p>\n Consider this series to find the sum of infinite AP,<\/span><\/p>\n 3 + 6 + 9 + 12 + \u2026\u2026\u2026\u2026\u2026\u2026<\/span><\/p>\n Here, first term, a = 3<\/span><\/p>\n The common difference, d = 3<\/span><\/p>\n and, number of terms, n = \u221e<\/span><\/p>\n Substituting these values in the <\/span>sum of AP formula<\/b> from equation 3,<\/span><\/p>\n S_{n}=\\frac{n}{2}\\{2 a+(n-1) d\\} <\/span><\/span><\/p>\n \u21d2 S_{n}=\\frac{\\infty}{2}\\{2(3)+(\\infty-1) 3\\}<\/span><\/span><\/p>\n \u2234 S_{n}=\\infty<\/span><\/span><\/p>\n <\/span><\/p>\n So, when d > 0, then the <\/span>sum of infinite AP<\/b> is \u221e.<\/span><\/p>\n Similarly, when d < 0, then the <\/span>sum of infinite AP<\/b> is -\u221e.<\/span><\/p>\n <\/p>\n <\/strong><\/p>\n <\/b><\/p>\n Consider the first n natural numbers as 1, 2, 3, 4, 5, 6, 7, \u2026\u2026, n<\/span><\/p>\n These natural numbers form an AP sequence with a = 1 and d = 1<\/span><\/p>\n According to equation 3,<\/span><\/p>\n S_{n}=\\frac{n}{2}\\{2 a+(n-1) d\\}<\/span><\/span><\/p>\n Substituting the values of a and d in this equation,\u00a0<\/span><\/p>\n S_{n}=\\frac{n}{2}\\{2(1)+(n-1) 1\\} <\/span><\/span><\/p>\n \u21d2 S_{n}=\\frac{n}{2}(2+n-1)<\/span><\/span><\/p>\n \u21d2 S_{n}=\\frac{n(n+1)}{2}<\/span><\/span><\/p>\n So, the arithmetic progression formula<\/b> for the sum of n natural numbers is S_{n}=\\frac{n(n+1)}{2}<\/span><\/span><\/p>\n <\/span><\/p>\n Here, we must find S for the squares of the first n natural numbers<\/span><\/p>\n S=1^{2}+2^{2}+3^{2}+4^{2}+\\ldots \\ldots \\ldots \\ldots . .+(n-1)^2+n^{2}<\/span><\/span><\/p>\n We will use the identity below\u00a0<\/span><\/p>\n a^{3}-b^{3}=(a-b)\\left(a^{2}+a b+b^{2}\\right)<\/span><\/span><\/p>\n replacing a with n and b with (n \u2013 1) in this equation,<\/span><\/p>\n n^{3}-(n-1)^{3}=(n-n+1)\\left(n^{2}+n(n-1)+(n-1)^{2}\\right) <\/span> \u21d2 n^{3}-(n-1)^{3}=3 n^{2}-3 n+1 \u00a0equation 4<\/span><\/span><\/p>\n <\/p>\n Similarly using (\\text{n}-1)^{\\text{th}}<\/span> and (\\text{n}-2)^{\\text{th}}<\/span> term,<\/span><\/p>\n (n-1)^{3}-(n-2)^{3}=3(n-1)^{2}-3(n-1)+1<\/span><\/span><\/p>\n <\/p>\n Now using (\\text{n}-2)^{\\text{th}}<\/span> and (\\text{n}-3)^{\\text{th}}<\/span> terms,<\/span><\/p>\n (n-2)^{3}-(n-3)^{3}=3(n-2)^{2}-3(n-2)+1 equation 6<\/span><\/span><\/p>\n <\/span><\/p>\n So on,<\/span><\/p>\n 2^{3}-1^{3}=3(2)^{2}-3(2)+1<\/span><\/span><\/p>\n <\/p>\n 1^{3}-0^{3}=3(1)^{2}-3(1)+1 \\ldots \\ldots \\ldots . . \\rightarrow last step<\/span><\/span><\/p>\n <\/span><\/p>\n By adding all the steps above, we get n^{3}-0^{3}=3 \\Sigma n^{2}-3 \\Sigma n+n<\/span><\/span><\/p>\n n^{3}=3 \\Sigma n^{2}-\\frac{3 n(n+1)}{2}+n<\/span> (using the formula for the sum of n natural numbers)<\/span><\/p>\n 3 \\Sigma n^{2}=n^{3}+\\frac{3 n(n+1)}{2}-n<\/span><\/span><\/p>\n <\/span><\/p>\n Taking n as common,<\/span><\/p>\n 3 \\Sigma n^{2}=n\\left[n^{2}+\\frac{3(n+1)}{2}-1\\right]<\/span><\/span><\/p>\n 3 \\Sigma n^{2}=\\frac{n}{2}\\left(2 n^{2}+3 n+1\\right)<\/span><\/span><\/p>\n <\/span><\/p>\n Upon factorizing the quadratic equation in the RHS, we get,<\/span><\/p>\n \\Sigma n^{2}=\\frac{n}{6}(2 n+1)(n+1)<\/span><\/span><\/p>\n \\Sigma n^{2}=\\frac{n}{6}(2 n+1)(n+1)<\/span><\/span><\/p>\n <\/span><\/p>\n Therefore, the sum of squares of first n natural numbers is \\frac{n}{6}(2 n+1)(n+1)<\/span><\/span><\/p>\n <\/p>\n Similar to the above derivations, we can also find the formula for the sum of cubes of first n natural numbers. So, without diving deeper into the complexity of equations, the formula for the sum of cubes of first n natural numbers is \\left\\{\\frac{n(n+1)}{2}\\right\\}^{2}<\/span><\/span><\/p>\n <\/span><\/p>\n <\/span><\/p>\n Example 1<\/b><\/p>\n Find the sum of squares of the first 10 natural numbers.<\/span><\/p>\n Solution<\/b><\/p>\n The sum of squares of first 10 natural numbers<\/span><\/b><\/p>\n \\text {i e., } 1^{2}+2^{2}+3^{2}+4^{2}+\\ldots \\ldots \\ldots \\ldots \\ldots \\ldots . .+10^{2}<\/span><\/span><\/b><\/p>\n Here, n = 10<\/span><\/p>\n And the formula to find the sum is \\frac{n}{6}(2 n+1)(n+1)<\/span><\/span><\/p>\nSum of n terms in an AP<\/strong><\/h2>\n
a_{2}=a+d <\/span>
a_{3}=a+2 d <\/span>
a_{4}=a+3 d<\/span><\/span><\/p>\n<\/b><\/h2>\n
Sum of infinite AP<\/b><\/h2>\n
Examples of AP series<\/strong><\/h2>\n
Sum of n natural numbers<\/b><\/h3>\n
Sum of squares of first n natural numbers<\/strong><\/h3>\n
\u21d2 n^{3}-(n-1)^{3}=1\\left(2 n^{2}-n+n^{2}+1-2 n\\right)<\/span><\/span><\/p>\nSum of cubes of first n natural numbers<\/strong>
<\/span><\/h3>\nSolved examples<\/strong><\/h2>\n