PROBLEMS ON ARITHMETIC PROGRESSION

Problem 1 :

Find the middle term(s) of the following arithmetic progression.

9, 15, 21, 27,…,183.

Solution :

In order to find the middle term of the sequence, first we have to know how many terms are in the given sequence.

n = [(l - a)d] + 1

a = 9, d = 15 - 9  ==>  6 and l = 183

n = [(183 - 9)/6] + 1

 n = (174 / 6) + 1

n = 29 + 1  =  30

n = 30(even)

So, the middle terms are

(n/2)^th term and [(n/2) + 1]^th term

Problem 2 :

If nine times ninth term is equal to the fifteen times fifteenth term of an arithmetic progression, show that six times twenty fourth term of the same arithmetic progression is zero.

Solution :

9t₉  =  15t₁₅

9(a + 8d)  =  15(a + 14d)

9a + 72d  =  15a + 210 d 

15a - 9a + 210d - 72d  =  0

6a + 138d  =  0

6(a + 23d)  =  0

a + 23d  =  0

t₂₄  =  0

Hence proved.

Problem 3 :

If 3 + k, 18 - k, 5k + 1 are in arithmetic progression, then find the value of k. 

Solution :

If a, b, c are in A.P then 2b = a + c

a = 3 + k, b = 18 - k and c = 5k + 1

2(18 - k)  =  3 + k + 5k + 1

2(18 - k)  =  6k + 4

18 - k = 3k + 2

18 - 2  =  3k + k

4k  =  16

k = 4

So, the value of k is 4.

Problem 4 :

Find x, y and z, given that the numbers x, 10, y, 24, z are in arithmetic progression. 

Solution :

Since the given sequence is arithmetic progression,

t₂ - t₁  =  10 - x  ---(1)

t₃ - t₂  =  y - 10 ---(2)

t₄ - t₃  =  24 - y ---(3)

t₅ - t₄  =  z - 24 ---(4)

By applying the value of y in x + y  =  20

x + 17  =  20

x  =  3

(3)  =  (4)

24 - y  =  z - 24

y + z  =  24 + 24

y + z  =  48

17 + z  =  48

z  =  31

So, the values of x, y and z are 3, 17, 31 respectively.

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