From my book

Number Theory Level pending

If in an A.P. sum of m terms is n and sum of n terms is m then find the sum of m+n terms.

m+n -(m+n) 2(m+n) -2(m+n)

Siddharth Singh by Siddharth Singh , 20, India

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1 solution

Vishnu Bhagyanath
Apr 27, 2015

Let Sm and Sn be the sum of m and n terms respectively. S m = m 2 ( 2 a + ( m − 1 ) d ) = n S_{m} = \frac{m}{2} (2a + (m-1)d) = n S n = n 2 ( 2 a + ( n − 1 ) d ) = m S_{n} = \frac{n}{2} (2a + (n-1)d) = m Subtracting the second equation from the first, n − m = m 2 ( 2 a + ( m − 1 ) d ) − n 2 ( 2 a + ( n − 1 ) d ) n-m = \frac{m}{2} (2a + (m-1)d) - \frac{n}{2} (2a + (n-1)d) n − m = 1 2 ( m ( 2 a + ( m − 1 ) d ) − n ( 2 a + ( n − 1 ) d ) ) n-m = \frac{1}{2} ( m(2a + (m-1)d) - n(2a + (n-1)d)) Simplifying, n − m = 1 2 ( 2 a ( m − n ) + d ( m − n ) ( m + n − 1 ) ) n-m = \frac{1}{2} ( 2a(m-n) +d(m-n)(m+n-1) ) − ( m − n ) = m − n 2 ( 2 a + d ( m + n − 1 ) ) -(m-n) = \frac{m-n}{2} ( 2a +d(m+n-1) ) − 2 = 2 a + d ( m + n − 1 ) -2 = 2a +d(m+n-1) Multiply throughout by (m+n)/2 − ( m + n ) = m + n 2 ( 2 a + d ( m + n − 1 ) ) -(m+n) = \frac{m+n}{2}(2a +d(m+n-1)) − ( m + n ) = S m + n -(m+n) = S_{m+n}

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Thank you.I upvoted your solution.

Siddharth Singh - 6 years, 1 month ago

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