# HOW TO FIND THE MIDDLE TERM OF AN ARITHMETIC SEQUENCE

How to Find the Middle term of an Arithmetic Sequence :

Here we are going to see some practice questions on arithmetic sequence.

Question 1 :

Find the middle term(s) of an A.P. 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)

middle terms are (n/2)th term and [(n/2) + 1]th term

 15th term  =  a + 14d=  9 + 14(6)=  93 16th term  =  a + 15d=  9 + 15(6)=  99

Question 2 :

If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term 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.

Question 3 :

If 3 + k, 18 - k, 5k + 1 are in A.P. then find 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

Hence the value of k is 4.

Question 4 :

Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.

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)

 (1)  =  (2)10 - x  =  y - 10x + y  =  20 (2) = (3)y - 10  =  24 - yy + y  =  24 + 102y  =  34y  =  17

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

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

After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Middle term of an Arithmetic Sequence".

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