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" ... the terms, RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the terms. "
The Youth's Assistant in Theoretick and Practical Arithmetic - Σελίδα 127
των Zadock Thompson - 1826 - 164 σελίδες
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The American Arithmetic: In which the Principles of Numbers are Explained ...

James Robinson - 1850 - 342 σελίδες
...series of numbers in arithmetical progression, when we have the extremes and number of terms given. • RULE. Multiply the sum of the extremes by the number of terms, and divide the product by 2, the quotient will be the sum of all the terms. 1. The extremes of* an arithmetical...

Introduction to the National Arithmetic ...

Benjamin Greenleaf - 1851 - 332 σελίδες
...multiplied by the number of terms, the product will be double the sum of either series. Hence, RDLE I. — Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series. Or, RULE II. — Multiply the sum of the extremes by half the number of terms, and...

North American Arithmetic: part third

Frederick Emerson - 1851 - 342 σελίδες
...two may be readily found. Theirs/ term. The last term. The number of terms. The common difference. The sum of all the terms. RULE. Multiply the sum of the extremes by the number of the terms, and half the product will be the sum of all the terms. See Theorem 4th. PROBLEM. I. The...

Engineers' and Mechanics' Pocket-book ...

Charles Haynes Haswell - 1851 - 346 σελίδες
...what is the common difference 1 15— 3-H7— 1) = 2 Ans. When the Extremes and Number of Terms are given, to find the Sum of all the Terms. RULE. — Multiply the number of terms by half the sum of the extremes. EXAMPLE. — How many times does the hammer of a clock...

Elementary Algebra: For the Use of Schools

William Smyth - 1851 - 272 σελίδες
...To find, therefore, the sum of all the terms, when the extremes and the number of terms are given, Multiply the sum of the extremes by the number of terms, and take one-half of the product. Ex. 1. In a progression by difference, the first term is 5, the last...

Engineers' and Mechanics' Pocket-book ...

Charles Haynes Haswell - 1853 - 318 σελίδες
...7; what is the common difference ? 15— 3-H(7~-l)=2 Ans. When the Extremes and Number of Terms are given, to find the Sum of all the Terms. RULE. — Multiply the number of terms by half the sum of the extremes. EXAMPLE. — How many times does the hammer of a clock...

Indroduction to the National Arithmetic ...

Benjamin Greenleaf - 1854 - 342 σελίδες
...multiplied by the number of terms, the product will be double the sum of either series. Hence, > RULE I. — Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series. Or, RULE II. — Multiply the sum of the extremes by half the number of terms, and...

Introduction to The National Arithmetic: On the Inductive System : Combining ...

Benjamin Greenleaf - 1857 - 336 σελίδες
...multiplied by the number of terms, the product will be double the sum of either series. Hence, RULE 1. — Multiply the sum of the extremes by the number of terms, and half the product will be the sum of the series. Or, RULE 2. — Multiply the sum of the extremes by half the number of terms, and...

Arithmetic, in which the Principles of Operating by Numbers are Analytically ...

Daniel Adams - 1858 - 354 σελίδες
...extremes of the series ; hence, when the extremes and the number of terms are given to find the sum of tin terms, RULE. Multiply the sum of the extremes by the...number of terms, and half the product will be the sum of the terms. EXAMPLES FOR PRACTICE. 2. If the extremes be 5 and 605, and the number of terms 151,...

The North American Practical School Arithmetic: Particularly Adapted to the ...

David Price - 1858 - 264 σελίδες
...third $10: what was the last payment? Ans. $37. II. The two extremes, and the number of terms being given, to find the sum of all the terms. RULE. — Multiply the sum of the extremes by half the number of terms. PROOF. — Divide the sum of all the terms by half the number of terms, for...




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