Page images

RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the answer.

EXAMPLES. 1. The first terin of an arithmetical series is $, the last terms, and the number of terms 11; required the sum of the series.

23+3 86 sum of the extremos.

Then 26x11-228 143 the Answer. 2. How many strokes does the laminer of a clock trike, in twelve hours ?

Ans. 78. 3. A merchant sold 100 yards of cloth, via, the first yarul for 1 ct. the second for 2 cts. the third for 3 cts. &c. I demand what the cloth came to at that rate ?

Ans. $501. 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave 1s. and for the last yd. H. 178. what did the whole come to ? Ans. £18 is.

5. A draper sold 100 yards of broadcloth, at 5 cts. for the first yard, 10 cts. for the second, 15 for the third, &c. increasing 5 cents for every yard ; what did the whole aniount tri, and what did it average per yard ?

Ans. Annount $252), and the arerage price is 82, 52cts. 5 mills per yard.

. Suppose 144 oranges were laid 2 yards distant from each other, in a right line, and a basket placed two yards from the first orange, what length of greund will that boy travel over, who gathers them up singly, returning with them one by one to the basket ?

Ans. 23 miles, 5 furlongs, 180 yds.

PROBLEM II. The first term, the last term, and the number of terms given, in find the coinnon difference.

' RULE. Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common dif. ference.

EXAMPLES 1. The extremes are 3 and 29, and the number ca terms 14, what is the common difference?

29] Extremes.

-35 Number of terms less 1=13)26(2 Ans.

2. A man had 9 sons, whose several ages differed alike: che youngest was 3 years old, and the oldest 35 ; whai was the common difference of their ages ?

Ans. 4 years 3. A man is to travel from New-London to a certain place in 9 days, and to go but 3 miles the first day, incrcasing every day by an equal excess, so that the last day's journey may be 43 miles : Required the daily increase, and the length of the whole journey ?

Ans. The daily increase is 5, and the whole journey 207 miles.

4. A debt is to be discharged at 16 different payment (in arithmetical progression, the first payment is to be 141. the last 100l.: What is the common dillerence, and the sum of the whole debt ?

Ans. 51. 14s. 8d. common difference, and 9121. the whole debt.

PROBLEM III. Given the first term, last term, and common difference to

find the number of terms.

RULE. . Divide the difference of the extremes by the common difference, and the quotient increased by 1 is the number of terms.

EXAMPLES. 1. If the extremes be 3 and 45, and the common dif ference 2; what is the number of terms ? Ans. 22.

2. A man going a journey, travelled the first day five miles, the last day 45 miles, and each dav increased his journry by 4 miles; how many days did he travel and how far: 15. 11 days, and the whole distance travelled 275 miles

GEOMETRICAL PROGRESSION, Is when any rank or series of numbers increased by one common multiplier, or decreased by one common divisor; as 1, 2, 4, 8, 16, &c. increase by the multiplier 2; and 27, 9, 3, 1, decrease by the divisor 3.

PROBLEM I. The first term, the last term (or the extremes) and the ratio given, to find the sum of the series.

* RULE. Multiply the last term by the ratio, and from the product subtract the first term; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms.

EXAMPLES. 1. If the scries be 2, 6, 18, 54, 162, 486, 1458, and the ratio 3, what is its sum total ?


=2186 the Answer.

3-1 2. The extremes of a geometrical series are 1 and 65536, and the ratio 4; what is the sum of the series?

Ans. 87381. PROBLEM II. Griven the first term, and the ratio, to find any other term


CASE I. When the first term of the series and the ratio are equal./

As the last term in a long series of numbers is very tedious to be found by continual multiplications, it will be necessary for the readier finding it out, to have a series of numbers in arithmetical proportion, called indices, whose common difference is 1.

When the first term of the series and the ratio are equal, the indices must begin with the unit, and in this case, the 1. Write down a few of the seading terms of the series, and place their indices over them, beginning the indices with an unit or 1.

2. Add together such indices, whose sum shall make up the entire index to the sum required.

3. Multiply the terms of the geometrical series belong, ing to those indices together, and the product will be the term sought.

EXAMPLES · 1. If the first be 2, and the ratio 2; what is the 13th term. 1, 2, 3, 4, 5, indices. Then 5+5+3=13 2, 4, 8, 16, 52, leading terms. 32X32X8=8192 Arse

2. A draper sold 20 yards of superfine cloth, the first yard for Sd. the second for 9d. the third for 27d. &c. in triple proportion geometrical ; what did the cloth come to at that rate ?

The 20th, or last term is 3486784401d. Then 3+3486784401-S

--- =5280176600d. the sum of all

S- 1 the terms (by Prob. I.) equal to £21792402 10s. Ans.

3. A rich miser thought 20 guineas a price too much for 12 Sine horses, but agreed to give 4 cents for the first, 16 cents for the second, and 64 cents for the third horse, and so on in quadruple or fourfold proportion to the last. what did they come to at that rate, and how much did they cost per head, one with another ?

Ans. The 12 horses came to 8223696, 20cts. and the average price was $18641, 35cts. per head.

product of any two terms is equal to that term, signified by the sum of their indices. Thure S 1 2 3 4 5 &c. Indices or arithmetical series.

" {2 4.8 16 32 &c. geometrical series. Vom 3+2 = 5 = the index of the fifth term, and

4x8 = 32 = the fifth term

CASE II. When the first term of the series and the ratio are diffe

rent, that is, when the first term is either greater or less than the ratio. *

1. Write down a few of the leading terms of the series, and begin the indices with a cypher: Thus, 0, 1, 2, 3, &c.

2. Add together the most convenient indices to make an index less by 1 than the number expressing the place , of the term sought.

3. Multiply the terms of the geometrical series together belonging to those indices, and make the product & dividend.

4. Raise the first term to a power whose index is one ess than the number of the terms multiplied, and make the resulta divisor. 5. Divide, and the quotient is the term sought.

EXAMPLES. 4. If the first of a geometrical series be 4, and the ratie 3, what is the 7th term ?

0, 1, 2, 3, Indices.
4, 12, 36, 108, leading terms.

3+2+1=6, the index of the 7th term.

---=2916 the 7th term required,

16 Here the number of terms multiplied are three; therefore the first term raised to a power less than three, is the 21 power or square of 4=16 the divisor.

*When the first term of the series and the ratio are different, the indices must begin with a cypher, and the sum of the indices made choice of must be one less than the number of terms given in the question: because I in the indices stands over tire second term, and 2 in the indices over the third term, fc. and in this case, the product of any two 'teris, divided by the first, is equal to that term beyond the first, signified by the sum of their indices. there. So, 1, 2, 3, 4, 8c. Indices. Thus, 31. 3. 9. 27. 81, &c. Geometrical series. Here 4+3=the index of the 8th term.,

81X27=2187 the 8th term, or the 7th beyond the 1st.

« PreviousContinue »