 | Francis Henney Smith - 1845 - 300 σελίδες
...there are terms, that is, eight times ; and hence, in an arithmetical progression, the sum of its terms is equal to half the sum of the extremes, multiplied by the number of terms, EXAMPLES. 1 . Find the sum of the terms of an arithmetical series whose first term is 3, last term... | |
 | George Roberts Perkins - 1846 - 266 σελίδες
...is the common difference of their ages ? Ans. 6. CASE VI. By Case II., we know that the sum of all the terms of an arithmetical progression is equal to half the sum of the extremes multiplied into the number of terms ; therefore, the number of terms is equal to the sum of all the terms divided... | |
 | Elias Loomis - 1846 - 380 σελίδες
...we obtain Eepresent the number of terms in the series by n ; then II o Hence S= Therefore, The sum of an Arithmetical Progression is equal to half the sum of the two extremes, multiplied by the number of terms; It also appears from the above, that the sum of the... | |
 | George Roberts Perkins - 1848 - 234 σελίδες
...the terms will average half the sum of the extremes ; consequently, the sum of all the terms equals half the sum of the extremes multiplied by the number of terms. Representing the sum of n terms by s, we have s=a-±lxn. (18) From (18), we easily obtain •=-/. (19)... | |
 | George Roberts Perkins - 1849 - 344 σελίδες
...the common difference of their ages ? Ans. 6 years. Case VI. By Case II. we know that the sum of all the terms of an arithmetical progression is equal to half the sum of the extremes multiplied into the number of terms ; therefore, the number of terms is equal to the sum of all the terms divided... | |
 | George Roberts Perkins - 1850 - 356 σελίδες
...the common difference of their ages ? Ans. 6 years. Case VI. By Case II. we know that the sum of all the terms of an arithmetical progression is equal to half the sum of the extremes multiplied into the number of terms; therefore, the number of terms is equal to the sum of all the terms divided... | |
 | 1851 - 260 σελίδες
...which is equal to 16 — 1. III. The sum of all the terms is equal to the product of the mean, or of half the sum of the extremes, multiplied by the number of terms. As above, the mean is 9 ; which, multiplied by 7, the number of terms, gives 63=15+13+11+9+7+5+3. And... | |
 | James B. Dodd - 1853 - 398 σελίδες
...progression, is equal to twice the mean or middle term. The Sum of all the Terms. § 347. The sum of all the terms of an arithmetical progression, is equal to half the sum of the two extremes multiplied into the number of terms. To prove this we add the several terms of an arithmetical... | |
 | Elias Loomis - 1855 - 356 σελίδες
...by term, we obtain Represent the number of terms in the series by n ; then Hence Therefore, The sum of an Arithmetical Progression is equal to half the sum of the two extremes, multiplied by the number of terms. It also appears from the above, that the sum of the... | |
 | Elias Loomis - 1856 - 280 σελίδες
...we represent the number of terms of the serie* l>yw, we shall have 2S=n(l+a) ; , _ whence <S= RULE. The sum of the terms of an arithmetical progression is equal to half the sum of the two extremes, multiplied by the number of terms. We also see from the preceding demonstration, that... | |
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That is, the sum of the terms of an arithmetical progression is equal to half the sum of the two extremes multiplied by the number of terms. 
