Question

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 1

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)=$\frac{1992}{12}$

    ⇒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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