CHAPTER XIII. ARITHMETICAL PROGRESSION. GEOMETRICAL PROGRESSION. GEOMETRIC MEAN AND ARITHMETIC MEAN. SIMPLE INTEREST. COMPOUND INTEREST. DISCOUNT. - d, Arithmetical Progression.-Quantities are said to be in arithmetical progression when they increase or decrease by the addition or subtraction of the same quantity. Thus the numbers 1, 2, 3, 4, which increase by the addition of 1 to each successive term; 21, 18, 15, 12, which decrease by the subtraction of 3 from each successive term; a, a+d, a +2d, etc.; and a, aa-2d, increasing or diminishing by the addition or subtraction of a quantity d—are in arithmetical progression. It will be seen that if a be the first term, a+d the second, a +2d the third, etc., any term, such as the eighth, is equal to a added to d repeated (8-1) times, or a+7d; therefore, if I be the nth or last term, l=a+(n − 1)d. ..(i) If s denote the required sum, then using the same letters as before, s=a+(a+d)+(a+2d) +...+(l − 2d)+(l− d) +l; also writing the series in the reverse order, s=l+(l-d)+(-2d)+...+(a+2d)+(a+d)+a. Adding these two series together, we obtain or in words, the sum of a number of terms in arithmetical progression is found by multiplying the sum of the first and last terms by half the number of terms. In this form, when the first, last, and number of terms are known, the sum can be found. The sum can also be obtained when the first term and the number of terms are given, as follows: Ex. 1. Find the 11th term of the series 1, 3, 5, etc. Here a=1, d=2, n=11; .. the 11th term =1+(11−1)2=21. Ex. 2. Find the sum of the series 1, 3, 5, etc., to 120 terms. Ex. 3. s=120{2×1+(120 − 1)2} =240 × 60=14400. Find the sum of the series 15, 11, 7, 3 – 1, − 5, to 20 terms. s=20{2 × 15, +(20 − 1) − 4 } 8= =-46 × 10-460. 1 2 Ex. 4. Find the sum of 3, 3 ... to 150 terms. Here a=1, d= }}; .. s=15°{2×3+ (150 − 1) } } = 3775. Arithmetic Mean.-The middle term of any three quantities in an arithmetical progression is the arithmetic mean of the other two. Thus if a and b are the two quantities, and A the arithmetic mean, then a, A, and b form three terms of an arithmetical progression, and or . A-a-b- A, Hence the arithmetical mean of two quantities is half their sum. It is always possible between any two given quantities to insert a number of terms such that the whole series are in arithmetical progression. Thus from (i) d= gives the common difference. n-1 Ex. 1. Insert 6 arithmetic means between 1 and 43. Including the two given terms, the number of terms will be 8. Hence, we have to find an arithmetical progression of which 1 is the first term and 43 the last. 1. The 1st term is 5 and the 7th is 23, find the 20th. 2. The sum of 20 terms is 500, and the last term is 45, find the 1st term. 3. Sum the following series to 12 terms: 1 3 B 1- etc., to 20 terms. 5. Find the sum of 1 - 1 - 3 to 21 terms. ... 6. The 1st term is 3 and the 3rd term 9, find the sum of the first 20 terms. 7. The sum of the first 7 terms is 49, and the sum of the next 8 is 176. What is the series? 8. The 3rd term is 7 and the 7th is 3. 9. Find the 30th term, the 1st term term - 16. What is the series? being 17 and the 100th 10. If the 1st term is 50 and the 5th term 42, find the sum of 21 terms. 11. The sum is 1625, the 2nd term 21, and the 7th term 41; find the number of terms. 12. If the 20th term is 15 and the 30th term is 20, find the sum of the first 25 terms. 13. How many strokes are struck in a day by a clock that tolls the hours? Sum the series : 14. 14.2, 12.3, 104, etc., to 15 terms. 15. 13.1, 11.4, 9.7, etc., to 15 terms. Sum the series : 21. −1, −1, − to 17 terms. 22. 1.2-2.1-5.4 to 10 terms. 23. The sum of n terms of the series 2, 5, 8, etc., is 950; find n. 24. Sum 7+32 +57 + etc. to 20 terms. Geometrical Progression.—Quantities which increase or decrease by a constant factor are said to be in geometrical progression. Thus, 1, 3, 9, 27, etc.; 4, 2, 1, 1⁄2, etc.; a, ar, ar2, etc.; are quantities respectively in geometrical progression. The constant factors 3, §, and r respectively are called the common ratios, and may be found by dividing any term by the term preceding it. If we examine the series a, ar, ar2, etc., where a is the first term and r the common ratio, it is obvious that any term, such as the 6th, is equal to a multiplied by raised to the power . (6-1), or is equal to ará. or Hence if I denote the last or nth term, l=arn-1 If s denote the sum of n terms, then s=a+ar+ar2 + ... + arn-2+ apn-1. Multiplying every term by r, we obtain rs=ar+ar2+ar3+ +ap-2+arn-1+arn. ... Hence, subtracting (ii) from (iii), rs-s=arn - α, .(i) ..(ii) .(iii) By changing the signs in both numerator and denominator, (iv) may be written a (1 - p) 1-r Note.-312 is most easily obtained by using logarithms; obtain 312 in this way, subtract unity from it, and divide the difference by 2. Geometric Mean.-The middle term of any three quantities in a geometrical progression is said to be the geometric mean of the other two, and is written G.M. To insert a given number of geometric means between two given quantities. From l-arm-1, we obtain Ex. 1. Insert three geometric means between 2 and 32. Including the two given terms the number of terms will be 5. Hence we have 5 numbers in a geometrical progression, the first 2 and the last 32. Hence the means are 4, 8, 16. We have found that the sum of n terms is given by a(1-) 1-r Now if r be a proper fraction, it is evident that " decreases as n increases. Thus let r, then r2=100, 3=1000, etc.; hence by making n sufficiently great, the sum of n terms differs from small a quantity as desired. This is perhaps better expressed by writing α by as 1 and in the limit when n is indefinitely great, the value of |