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3 years ago

How many numbers are equal to the sum of two odd, one-digit numbers?

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

4 0

Answer:

Seven numbers.

Step-by-step explanation:

Finding the numbers, which are equal to the sum of two odd number and it has to be single digit number.

Lets look into numbers which are odd and single digit.

1 = $1+3, 1+5, 1+7, 1+9$

∴ Sum of the number is $4,6,8\ and\ 10$

3 = $3+5, 3+7, 3+9$

∴ Sum of above number is $8,10\ and\ 12$

5 = $5+7, 5+9$

∴ Sum of above number is $12\ and\ 14$

7= $7+9$

∴ Sum of above number is $16$

Now, accumlating numbers which are fullfiling the criteria, however, making sure no number should get repeated.

∴ Numbers are: $4,6,8,10,12,14\ and\ 16$

Hence, there are total 7 numbers, which are equal to the sum of two odd, one-digit numbers.

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Given:

$\bold{t_2=5}\\\\\bold{t_8=15}\\\\$

To find:

$\bold{T_n=?}\\\\\bold{S_{10}=?}$

Solution:

$\to \bold{t_2=5}\\\\ \bold{a+d=5} \\\\ \bold{a=5-d} ................(i)\\\\ \to \bold{t_8=15}\\\\ \bold{a+ 7d=15}.................(ii)\\\\$

Putting the equation (i) value into equation (ii):

$\to \bold{5-d+7d=15}\\\\\to \bold{5+6d=15}\\\\\to \bold{6d=15-5}\\\\\to \bold{6d=10}\\\\\to \bold{d=\frac{10}{6}}\\\\\to \bold{d=\frac{5}{3}}\\\\$

Putting the value of (d) into equation (i):

$\to \bold{a=5-\frac{5}{3}}\\\\\to \bold{a=\frac{15-5}{3}}\\\\\to \bold{a=\frac{10}{3}}\\\\$

Calculating the $\bold{t_n :}$

$\to \bold{t_n=a+(n-1)d}\\\\\to \bold{t_n=\frac{10}{3}+(n-1)\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}+\frac{5}{3}n-\frac{5}{3}}\\\\\to \bold{t_n=\frac{10}{3}-\frac{5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{10-5}{3} +\frac{5}{3}n}\\\\\to \bold{t_n=\frac{5}{3} +\frac{5}{3}n}\\\\ \to \bold{t_n=\frac{5}{3}(1+n)}\\\\$

Formula:

$\bold{S_n = \frac{n}{2}[2a + (n - 1)d ] }\\$

Calculating the $\bold{S_{10}} :$

$\to \bold{S_{10} = \frac{10}{2}[2\frac{10}{3} + (10- 1)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + (9)\frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 9 \times \frac{5}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 3\times 5 ] }\\\\\to \bold{S_{10} = 5[\frac{20}{3} + 15 ] }\\\\\to \bold{S_{10} = 5[\frac{20+ 45}{3} ] }\\\\\to \bold{S_{10} = 5[\frac{65}{3} ] }\\\\\to \bold{S_{10} = 5 \times \frac{65}{3} }\\\\\to \bold{S_{10} = \frac{325}{3} }\\\\\to \bold{S_{10} =108.33 }$

Therefore, the final answer is " $\bold{\frac{5}{3}(1+n)\ and\ 108.33}$ "

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