show that the sum of an A.P whose first term is a , second term b, last term is c is equal to [a+c] [ b-c-2a] by 2 [ b-c]
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first term = a,second term = b,last term = cd= (b-a)
an = a + (n-1) d=> c = a + (n-1) (b-a)=> c = a + nb -an -b +a=> c = 2a -an +nb -b=> n = (b+c-2a) / (b-a)Sn = n/2 (a+ l)=> Sn = (b+c-2a) (a+c) / 2(b-a)
an = a + (n-1) d=> c = a + (n-1) (b-a)=> c = a + nb -an -b +a=> c = 2a -an +nb -b=> n = (b+c-2a) / (b-a)Sn = n/2 (a+ l)=> Sn = (b+c-2a) (a+c) / 2(b-a)
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