Find the last two digits of 7*19*31*.....*1999 . (Here 7, 9,31,........ 1999 form an arithmetic sequence of common difference 12.)
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Answer: 89
Step-by-step explanation:
- We can see clearly that the given sequence is is in A.P.
- Given that , a=7,d=12 and tn=1999
- tn= a+(n-1)d
- 1999= 7+(n-1) 12
⇒1999-7=(n-1)12
⇒(n-1)=
⇒n=166+1
∴ n=167
- Now we need to find the product of 167 terms of the above A.P.
- Since all the terms here are odd numbers ending with 7,9,1,3 and 5.
- So,the product of these numbers are 189.
- Hence,the last two digits of the given sequence will be 89.
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