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.

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