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Show that s³ = 3 ( s2 - s1 ) |
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Answer» If s1,s2 and s3 are the sum of first n,2n,and 3n TERMS of an A.P. RESPECTIVELY then,s1 = n/2[2a + (n-1)d] s2 = 2n/2[2a + (2n-1)d] s3 = 3n/2[2a + (3n-1)d] consider that,s2 - s1 = 2n/2[2a + (n-1)d] - n/2[2a + (n-1)d] = n/2 {[4a + 2(2n - 1)d] - [2a + (n-1)d] s2 - s1 =n/2[2a +(3n - 1)d]3(s2 - s1) = 3 X n/2[2a + (3n - 1)d]3(s2 - s1) = 3n/2[2a + (3n - 1)d]3(s2 - s1) = s3 { ∵ s3 = 3n/2[2a + (3n-1)d]} HENCE, proved. |
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