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Sia ? 6 years, 6 months ago
a, 2a, 3a, 4a, ....
a2, - a1 = 2a - a = a
a3 - a2 = 3a - 2a = a
a4 - a3 = 4a - 3a = a
i.e. ak+1 - ak is the same every time.
So, the given list of numbers from an AP with the common difference d = a.
The next three terms are:
4a + a = 5a, 5a + a = 6a and 6a + a = 7a
Posted by Aman Singh 7 years, 6 months ago
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In the given AP
a=2, l =29 and Sn=155
We know that
Sum of n terms = Sn={tex}\frac{n}{2}{/tex}(a+l)=155
{tex}\therefore{/tex} {tex}\frac{n}{2}{/tex}(2+29)=155
{tex}\mathrm n=\frac{155\times2}{31}=10{/tex}
{tex}\text{Now l=a+(n-1)d}{/tex}
or 29=2+(10-1)d
29=2+9d
29-2=9d
27 = 9d
d = 3
Posted by Shristi Siddharath 7 years, 6 months ago
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Posted by Shweta Shah 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
let us assume that {tex}\sqrt 3{/tex} be a rational number.
{tex}\sqrt { 3 } = \frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \neq{/tex}0
Squaring both sides, we have
{tex}\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}
or, {tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)
a2 is divisible by 3.
Hence a is divisible by 3..........(ii)
Let a = 3c ( where c is any integer)
squaring on both sides we get
(3c)2 = 3b2
9c2 = 3b2
b2 = 3c2
so b2 is divisible by 3
hence, b is divisible by 3..........(iii)
From equation(ii) and (iii), we have
3 is a factor of a and b which is contradicting the fact that a and b are co-primes.
Thus, our assumption that {tex}\sqrt 3{/tex} is rational number is wrong.
Hence, {tex}\sqrt 3{/tex} is an irrational number.
Let us assume that 7 - 2{tex}\sqrt 3{/tex} is a rational number
7 - 2{tex}\sqrt 3{/tex} = {tex}\frac { p } { q }{/tex} (q {tex}\neq{/tex}0 and p and q are co-primes)
{tex}\style{font-family:Arial}{\begin{array}{l}7-2\sqrt3=\frac pq\\-2\sqrt3=\frac pq-7\\2\sqrt3=7-\frac pq\\2\sqrt3=\frac{7q-p}q\\\sqrt3=\frac{7q-p}{2q}\end{array}}{/tex}
Here 7q-p and 2q both are integers, hence {tex}\sqrt 3{/tex} is a rational number.
But this contradicts the fact that {tex}\sqrt 3{/tex} is an irrational number.
This contradict is due to our assumption that {tex}\style{font-family:Arial}{7-2\sqrt3}{/tex} is rational.
Hence, 7 - 2{tex}\sqrt3{/tex} is irrational.
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Total number from 1 to 100 = 100
Two digit numbers that are multiple of 5 = {10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95} = 18
P(Two digit number multiple of 5 from 1 to 100) = {tex}\frac{18}{100} = \frac{9}{50}{/tex}
Posted by Aman Bhatt 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Least prime factor of {tex}"a"{/tex} is {tex}3{/tex} and least prime factor of {tex}"b"{/tex} is {tex}7{/tex}
Therefore, sum of least prime factors of {tex}a\ and\ b = 3+7 = 10{/tex}
and least factor of {tex}10\ is\ 2{/tex}
Therefore, least factor of {tex}a + b{/tex} is also {tex}2{/tex}
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