Question

6. Find the number among natural number 1
to 288 for which the sum of the numbers
smaller than it is same as the sum of the
numbers greater than it.​

Answer 1

Answer:Answer

5.0/5

10

Deleted account

Let the number be n.

So:

1+2+3................n-1 = n+1+n+2......................288

In the first AP,

Sn=n/2*[2a+(n-1)d]

no. of terms=n-1

a=1

d=1

Sn=(n-1)/2*[2+n-1-1]

=(n-1)/2*[n]................(1)

Second AP

a=n+1

d=1

no. of terms=288-n

Sn=n/2*[2a+(n-1)d]

Sn=(288-n)/2*[2+288-n-1]

=(288-n)/2*[n+289]..............(2)

(1)=(2)

So:

(n-1)/2*[n]=(288-n)/2*[n+289]

Multiplying both sides by 2.

==) n(n-1)=(288-n)(289+n)

==)n²-n=83232+288n-289n-n²

==)2n²-n=83232-n

==)2n²=83232

==)n²=83232/2

==)n²=41616

==)n=204

n cannot be negative so n is not -204

The term hence is 204

Hope it helps you.

plz mark as brainliast answer

Step-by-step explanation:

Answer 2

Step-by-step explanation:

$\bf{AnsWeR:} \\ \sf \: trigonometry \: table \\ \\ \purple{\fcolorbox{red}{pink}{{{{\: Table {{{{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}}}}}}}}}}$