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Find the number of terms and the common difference when 31. The first term is 3, the lilst term 90, and the sum 1395. 32. The first term is 79, the last term 7, and the sum 1075. 33. The sum is 24, the first term 9, the last term - 6. 34. The sum is 714, the first term 1, the last term 581. 35. The last terın is – 16, the sum – 133, the first term - 3. 36. The first term is – 75, the sum – 740, the last term 1. 37. The first term is u, the last 13u, and the sum 49a. 38. The sum is – 320x, the first term 3x, the last term - 35x.

346. If any two terms of an Arithmetical Progression be given, the series can be completely determined; for the data furnish two simultaneous equations, the solution of which will give the first term and the common difference.

Example. Find the series whose 7th and 51st terms are - 3 and - 355 respectively. If a be the first term, and d the common difference,

- 3=the 7th term

=a +60; and

- 355=the 51st term

=a +500; whence, by subtraction, - 352 =44d; .. d=-8; and consequently a=45.

Hence the series is 45, 37, 29 ......

347. DEFINITION. When three quantities are in Arithmetical Progression the middle one is said be the arithmetic mean of the other two.

Thus a is the arithmetic mean between a-d and a+d..

348. To find the arithmetic mean between two given quantities.

Let a and b be the two quantities; A the arithmetic mean. Then since a, 4, b are in A.P. we must have

1-A= A -a each being equal to the common difference;

a+b whence

2

A=

349. Between two given quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A.P.; and by an extension of the definition in Art. 347, the terms thus inserted are called the arithmetic means.

Example. Insert 20 arithmetic means between 4 and 67.

Including the extremes the number of terms will be 22; so that we have to find a series of 22 terms in A.P., of which 4 is the first and 67 the last.

Let d be the common difference; then

67=the 22nd term

=4+21d; whence d=3, and the series is 4, 7, 10, ......61, 64, 67; and the required means are 7, 10, 13, ......58, 61, 64.

350. To insert a given number of arithmetic means between two given quantities.

Let a and b be the given quantities, n the number of means.

Including the extremes the number of terms will be n+2; so that we have to find a series of n+2 terms in A.P., of which a is the first, and b is the last.

Let d be the common difference; then

b=the (12+2)th term
=a+(n+1)d;

b whence

d=

a

n+1

and the required means are

2(b a) at

at n+1

n+1

b-a

...at

n (b a)

2+1

Example 1. Find the 30th term of an A.P. of which the first term is 17, and the 100th term – 16.

Let d be the common difference; then

- 16=the 100th term
=17+99d;

1
... d=

3'

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Example 2. The sum of three numbers in A.P. is 53, and their product is 792; find them.

Let a be the middle number, d the common difference; then the three numbers are a d, a, a + d. Hence

a-d+a+a+d=33; whence a=11; and the three numbers are 11 - d, 11, 11+ d.

.. 11 (11+d) (11 – d)=792,

121-də=72,

d=+7; and the numbers are 4, 11, 18.

Example 3. How many terms of the series 24, 20, 16, ...... must bo taken that the sum may be 72?

Let the number of terms be n; then, since the common difference is 20 – 24, or – 4, we have from (3), Art. 314,

72 (2 x 24+ (n − 1) ( - 4);

=24n – 2n (n-1); whence

m2 - 13n+36=0,
(n – 4)(n-3)=0;

.. n=4 or 9. Both these values satisfy the conditions of the question; for if wo write down the first 9 terms, we get 24, 20, 10, 12, 8, 4, 0, - 4, -8; and, as the last five terms destroy each other, the sum of 9 terms is the same as that of 4 terms.

or

Example 4. An A.P. consists of 21 terms; the sum of the three terms in the middle is 129, and of the last three is 237; find the series, Lct a be the first term, and d the common difference. Then

237=the sum of the last three terms

=a+20d +a+19d+a+18d

= 3a +57d; whence a+19d=79

.(1). Again, the three middle terms are the 10th, 11th, 12th; henco

129=the sum of the three middle terms

=a+9d+a+10d +a+11d

=3a + 300; whence a+100=43

...(2). From (1) and (2), we obtain d=4, a=

=3. Hence the series is 3, 7, 11, ......83.

EXAMPLES XXXVI. b.

Find the series in which 1. The 27th term is 186, and the 45th term 312. 2. The 5th term is 1, and the 31st term – 77. 3. The 15th term is – 23, and the 23rd term –41. 4. The 9th term is – 11, and the 102nd term – 150). 5. The 15th term is 25, and the 29th term 46. 6. The 16th term is 214, and the 51st term 739. 7. The 3rd and 7th terms of an A.P. are 7 and 19; find the 15th

term. 8. The 54th and 4th terms are – 125 and 0; find the 42nd term. 9. The 31st and 2nd terms are 1 and 74; find the 59th term. 10. Insert 15 arithmetic means between 71 and 23. 11. Insert 17 arithmetic means between 93 and 69. 12. Insert 14 arithmetic means between -7} and – 27. 13. Insert 16 arithmetic means between 7.2 and - 6.4. 14. Insert 36 arithmetic means between 8} and 23. How many

terms must be taken of 15. The series 42, 39, 36, ... to make 315 ? 16. The series – 16, – 15, - 14,... to make – 100 ? 17. The series 153, 153, 15,... to make 129 ? 18. The series 20, 184, 17,... to make 162!? 19. The series – 103, -1, -73, ... to make – 42? 20. The series – 63, -63, -6, ... to make – 524? 21. The sum of three numbers in A.P. is 39, and their product

is 2184; find them. 22. The sum of three numbers in A.P. is 12, and the sum of

their

squares is 66; find them. 23. The sum of five numbers in A.P. is 75, and the product of

the greatest and least is 161; find them. 24. The sum of five numbers in A.P. is 40, and the sum of their

squares is 410; find them. 25. The 12th, 85th and last terms of an A.P. are 38, 257, 395

respectively; find the number of terms.

GEOMETRICAL PROGRESSION.

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351. DEFINITION. Quantities are said to be in Geometrical Progression when they increase or decrease by a constant factor.

Thus each of the following series forms a Geometrical Progression :

3, 6, 12, 24,

1 1 1 1,

3' 9 27

a, ar, ara, ar}, The constant factor is also called the common ratio, and it is found by dividing any term by that which immediately precedes it. In the first of the above examples the common ratio is 2;

1 in the second it is ; in the third it is r.

3 352. If we examine the series

a, ar, ara, ar), art, we notice that in any term the index of r is always less by one than the number of the term in the serics. Thus

the 3rd term is ar;
the 6th term is aro;

the 20th term is arl); and, generally, the pth term is arp-1.

If n be the number of terms, and if l denote the last, or nth term, we have

l=arn-1,

1 1 Example. Find the 8th term of the series

'2' 1

3 The common ratio is

3

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or

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... the 8th term

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