In the arithmetic progression 2,5,8....upto 50 terms,and 3,5,7,9....upto 60 terms ,find how many pairs are identical and find the identical pairs.
Given, first A. P. = 2, 5, 8, ... up to 50 terms
and second A. P. = 3, 5, 7, 9, ... up to 60 terms
Now, the last term of Ist A. P. is:
a50 = a + (50 – 1) d = 2 + (49) × 3
= 149
and last term of IInd A. P. is:
a60 = a + (60 – 1) d = a + 59 d = 3 + 59 × 2 = 121
So, first A. P. = 2, 5, 8, ..., 121
and second A. P. = 3, 5, 7, 9, ..., 149
Let us suppose that the pth term of first A.P be equal to the qth term of second A.P. Then.
From the above results, we get
k ≤ 20, i.e.: k = 1, 2, 3, ... 20
So, there are 20 identical pairs.
Now, in the given A.P´s,
first identical term = 5,
second identical term = 11
third identical term = 17
So, the last identical term is find out by using the formula
Tn = a + (n – 1) d
Tn = 5 + (20 – 1) × 6 [Here, first term = a = 5 common difference, d = 11 – 5 = 6]
= 5 + 19 × 6
= 119
Hence there are 20 identical term viz. 5, 11, 17, ... 119.