The sum of two natural numbers is 20 find the chance that their product will be less than 50

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Find the sum of :i all odd natural numbers less than 50 .ii first 12 natural numbers each of which is a multiple of 7 .

Solution

(i) For odd numbers $a=1d=2$

last term (less than 50) = 49

$a_n=a+(n-1)d1+(n-1)2=49(n-1)2=48n-1=24n=25$

Sum of n terms, $S_n=\frac{n}{2}[a+l]S_{25}=\frac{25}{2}[1+49]=\frac{25}{2}\times 50=625$

(ii) For multiples of 7 $a=7d=7n=12$

Sum of n terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]S_{12}=\frac{12}{2}\left[2\right(7)+11\times 7]=6\times [14+77]=6\times 91=546$

Mathematics

Concise Mathematics

Standard X