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(c) The ratio between the sum of n terms of twoarithmetic progression is (7n +1): (4n +27) Findthe ratio of their 11th terms.V: 70 |
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Answer» Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)Recall the nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes,am : a’m = a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].hence take m=11hence14*11-6/8*11+23=148/111 |
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