Please answer the following problems:

Illustrative Problems.

Problem No. 1.

Given the following numbers in Arithmetic Progression such as 4, 10, 16, .......

Solve and find the following.

a) the 4th term of the AP c) the sum of all the first four terms

b) the 10th term of the AP

SOLUTION:

Given: a = 4 (first term),

d = common difference (fixed no). = 10-4 = 16 -10 = 6

d = succeeding minus preceding = 10 - 4 = 6

n = 4 (number of terms) = 4, 10, 16? (4th term)

a) 4th term = L4 = L = a + (n -1) d = 4 + (4 - 1) (6) = 4 + (3)(6) = 4 + 18 = 22

b) 10th term = L10 = a + (n - 1) d = 4 + (10 - 1) (6) = 4 + (9)(6) = 4 + 54 = 58

c) the sum of all the first four terms which are 4, 10, 16, 22

S4= S = n/2 [ 2a + (n-1) (d)] = (4/2) [2(4) + (4 -1) (6)] = (2) [ 8 + (3)(6)]

S4 = S = (2) [8 +(3)(6)] = (2) (8 + 18) = (2) (26) = 52

Check S = 4 + 10 + 16 + 22 = 14 + 38 = 52

Another Solution: S = Sn = (n/2) [ 2a + (n-1) (d)] = (4/2) [ 2(4) + (4 -1) (6)

S = Sn = (2) [ 8 + (3) (6)] = (2) [ 8 + 18] = (2)(26) = 52 (same)

Problem No. 2.

Given the following numbers in Arithmetic Progression (AP) -36, - 16, 4, 24 ..........

Solve and find the following.

a) the fifth (5th term)

b) the sum of the first five terms

SOLUTION:

Given: a = - 36 n = 5 d = succeeding - preceding = latter - former

d = - 16 - (-36) = - 16 + 36 = 20

a) Fifth Term = L5

L5 = a + (n -1) d = -36 + (5 - 1) (20) = -36 + (4)(20) = - 36 + 80 = 44

b) The sum of the first five terms.

S5 = (n/2) [a + L5] = (5/2) [-36 + (44)] = (5/2) (8) = (5)(4) = 20

Check: S = - 36 + (-16) + 4 + 24 + 44= - 52 + 72 = 20 (Same)

Problem No. 3

The 3rd term of an Arithmetic Progression is 25 and the 6th term is 40.

Solve the following:

a) the first term "a".

b) the common difference "d".

_____, _____, 25, _____, _____, 40