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 I^{st} A. P. is:

*a*_{50} = *a* + (50 – 1) *d* = 2 + (49) × 3

= 149

and last term of II^{nd} A. P. is:

*a*_{60} = *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 *p*^{th} term of first A.P be equal to the *q*^{th} 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

T* _{n}* =

*a*+ (

*n*– 1)

*d*

T* _{n}* = 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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