Find the next number in given series 79, 64, 26, 15.
Answer
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Hint: To find the next number in given series 79, 64, 26, 15, we have to find a common method in which each of the terms in this series is obtained. When we multiply the digits of the first number and add one to it, we will get the second term. Similarly, when we multiply the digits of the second number and add two to it, we will get the second term. Likewise, we can find the required number.
Complete step by step answer:
We have to find the next number in the given series 79, 64, 26, 15. We can see that the first term is 79. When we multiply the digits of this number and add one to it, we will get the second term.
Second term \[=7\times 9+1=63+1=64\]
Now, let us check whether all the numbers in the series are obtained in the above manner.
Third term $=6\times 4+2=24+2=26$
Fourth term $=2\times 6+3=12+3=15$
So, in a similar way, we can find the required number.
Fifth term $=1\times 5+4=5+4=9$
Therefore, the next number in the 79, 64, 26, 15 is 9.
Note: Students must note that each term is obtained by multiplying the digits of the previous term and adding $\left( n-1 \right)$ to the result, where n is the current position of the term. To be clear, let us examine how we obtained the second ter. We added one to the result, that is, $\left( 2-1 \right)$ , where 2 is the position of this term (second).
Complete step by step answer:
We have to find the next number in the given series 79, 64, 26, 15. We can see that the first term is 79. When we multiply the digits of this number and add one to it, we will get the second term.
Second term \[=7\times 9+1=63+1=64\]
Now, let us check whether all the numbers in the series are obtained in the above manner.
Third term $=6\times 4+2=24+2=26$
Fourth term $=2\times 6+3=12+3=15$
So, in a similar way, we can find the required number.
Fifth term $=1\times 5+4=5+4=9$
Therefore, the next number in the 79, 64, 26, 15 is 9.
Note: Students must note that each term is obtained by multiplying the digits of the previous term and adding $\left( n-1 \right)$ to the result, where n is the current position of the term. To be clear, let us examine how we obtained the second ter. We added one to the result, that is, $\left( 2-1 \right)$ , where 2 is the position of this term (second).
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