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If 3+5+7+........ to n terms/ 5+8+11+.........to 10 terms = 7. Find \'n\' ? |
| Answer» Both the numerator and denominator are the sums of two series up to \'n\' terms:-Since the sum = (n/2)(2a+(n-1)d)For series in the numerator :-a=3, d=2, n=nTherefore numerator = (n/2)*(6+(n-1)2) = (n/2)*(2n+4)And for the denominator series:-a=5, d=3, n=10Therefore denominator = (10/2)*(10+(10-1)3) = (n/2)*(3n+7) = 185Therefore Acc. to question:-(n/2)*(2n+4) / (185) = 7\xa0= n2\xa0+2n- 1295 = 0\xa0= n2\xa0+ 37n - 35n - 1295 = 0=n=35 ( Since ncan not be equal to 37) | |