 | Charles Vyse - 1806 - 342 Seiten
...Share of the eldest ? PROPOSITION III. When the first Term, Ratio, and Numbar of Terms, are given, to find the Sum of all the Terms. RULE. Find the last Term as before, from which take the first : divide the Remainder by the Ratio, less one, and to that Quotient... | |
 | Michael Walsh - 1807 - 290 Seiten
...what was the eldest son's portion ? Ans. .€.25600. The first term, ratio, and number of terms given, to find the sum of all the terms. RULE. Find the last term as before, then subtract the first from it, and divide the remainder by the ratio less one, to the... | |
 | Charles Vyse - 1815 - 340 Seiten
...the share of the eldest ? PROPOSITION IIL When the first term, ratio, and number of terms are given, to find the sum of all the terms. RULE. Find the last term as before, from which take the first ; divide the remainder by the ratio, less one, and to that quotient... | |
 | Charles Hutton - 1818 - 646 Seiten
...terms, when any three of them are given ; as in the following problems : PROBLEM 1. Given the extremes, and the Number of Terms ; to find the Sum of all the Terms. AnD the extremes together, multiply the sum by the number of terms, and divide by 2. EXAMPLES. 1. The... | |
 | Roswell Chamberlain Smith - 1827 - 216 Seiten
...them ? A. $ 171798691,^4 eta. CASE II. — When the first term, ratio, and number of terras are given to find the sum of all the terms. RULE. — Find the last term by the two former cases, then multiply by the ratio, from the product, subtract the first term, then divide... | |
 | Michael Walsh - 1828 - 318 Seiten
...what was the eldest son'* portion I Ans. £25600. The first term, ratio, and numl.T of terms given, to find the sum of all the terms. RULE. Find the last term as before, then subtract the first from it. and divide the remainder by the ratio less one, to the... | |
 | Francis Walkingame - 1832 - 224 Seiten
...what was the eldest son's portion ? Ant. ££5600. The first term, ratio, and number of terms given, to find the sum of all the terms. RULE. Find the last term as before, then subtract the firxt from it, and divide the remainder by the ratio, less 1 ; to the... | |
 | Benjamin Greenleaf - 1839 - 356 Seiten
...cent. for 30 years ? Ans. $574.34.91172913250116264106332310802645846357252196069357387776. PROBLEM II. Given the first term, the ratio, and the number of terms, to find the sum of the series. RULE. Find the last term, as before, multiply it by the ratio, and from the product subtract... | |
 | Nathan Daboll - 1839 - 220 Seiten
...ratio ; 5th, the sum of all the terms, any 3 of which being given, the other two may be found. CASE I. Given, the first term, the ratio, and the number of 'terms, to fmd the last term. RULE. 1. Write down a few of the leading powers of the ratio, placing their indices,*... | |
 | Benjamin Greenleaf - 1841 - 334 Seiten
...cent. for 30 years ? Ans. $574.34.91172913250116264106332310802645846357252196069357387776. PROBLEM II. Given the first term, the ratio, and the number of terms, to find the sum of the series. RULE. Find the last term, as before, multiply it by Ike ratio, and from the product subtract... | |
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Hence, when we have given the first term, the ratio, and the number of terms, to find the last term, we have this RULE. 
