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Finding common terms in two arithmetic progressions

Posted on October 24, 2022 by ahmed ismail

Finding common terms in two arithmetic progressions

We welcome you to answer me today and answer an important account question that everyone is looking for, and now here is the solution in detail.

Finding common terms in two arithmetic progressions

The number of common terms in two A.P.’s 2,7,12,17 500 terms and 1,8,15,22 300 terms is The number of common terms in two A.P's
2,7,12,17........... 500 terms and 1,8,15,22,.
......... 300 terms is
1) 58 2) 60 3) 61 4) 63

Find the number of terms common to the two arithmetic progression 5,9,13,17,…,216 and 3,9,15,21,…..,321

The first common term in two Aps is 9.
LCm of their common difference = LCM (4,6) =12
New AP of common elements is 9,21,33,45,….,
Let is contain n terms. Then,
Tn>216⇒9+(n−1)×12≤216⇒(n−1)×12≥207Tn>216⇒9+(n-1)×12≤216⇒(n-1)×12≥207
⇒(n−1)×12=204⇒n=18⇒(n-1)×12=204⇒n=18
So, there are 18 common terms in two APs.

The number of terms common to the two arithmetic progressions
3,7,11,…,407 and 2,9,16,…,709 is

Solution

Given A.P.s are

3,7,11,15,19,23…,407 (common difference is 4)

2,9,16,23…,709 (common difference is 7)

We can observe that the first common term is 23.

( 6th term in 1st A.P. and 4th term in 2nd A.P. are same.)

Let the number of common terms be ′n′.

They will be in A.P. with common difference =28 ( i.e. (LCM (4,7) )

Now, the last common term can’t be greater than 407. [Last term of the first A.P]

⇒23+(n−1)28≤407

⇒n≤38428+1

⇒n≤14.7⇒n=14

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