Math, asked by asilaryo10, 2 months ago

find the first quantity as a percentage of second quantity​

Answers

Answered by lavaterupali4
0

Given

\sf \to \: S_p = S_q→S

p

=S

q

To Proved

\sf \to \: S_{p + q} = 0→S

p+q

=0

Formula

\sf \to \: S_n = \dfrac{n}{2} \{2a + (n - 1)d \}→S

n

=

2

n

{2a+(n−1)d}

So

\sf \to \: S_p = \dfrac{p}{2} \{2a + (p - 1)d \}→S

p

=

2

p

{2a+(p−1)d}

and

\sf \to \: S_q = \dfrac{q}{2} \{2a + (q - 1)d \}→S

q

=

2

q

{2a+(q−1)d}

According to question, we can write as

\sf \to \: \dfrac{p}{2} \{2a + (p - 1)d \} = \dfrac{q}{2} \{2a + (q - 1)d \}→

2

p

{2a+(p−1)d}=

2

q

{2a+(q−1)d}

\sf \to \: {p} \{2a + (p - 1)d \} = {q}\{2a + (q - 1)d \}→p{2a+(p−1)d}=q{2a+(q−1)d}

\sf \to \: p \{2a + pd - d \} = q \{2a + dq - d \}→p{2a+pd−d}=q{2a+dq−d}

\sf \to \: 2ap + {p}^{2} d - dp = 2aq + d {q}^{2} - dq→2ap+p

2

d−dp=2aq+dq

2

−dq

\sf \to 2ap - 2aq + {p}^{2} d - {q}^{2} d - dp + dq = 0→2ap−2aq+p

2

d−q

2

d−dp+dq=0

\sf\to2a(p - q) + d( {p}^{2} - {q}^{2}) - d(p - q) = 0→2a(p−q)+d(p

2

−q

2

)−d(p−q)=0

Now Simplify

\sf \to \: ( {a}^{2} - {b}^{2} ) = (a + b)(a - b)→(a

2

−b

2

)=(a+b)(a−b)

we get

\sf \to \: 2a(p - q) + d(p - q)(p + q) - d(p - q) = 0→2a(p−q)+d(p−q)(p+q)−d(p−q)=0

Now Take a common

\sf \to \: (p - q) \{2a + d(p + q) - d \} = 0→(p−q){2a+d(p+q)−d}=0

\sf \to \: 2a + \{ ( p + q) - 1 \}d = 0 \: \: \: \: \: \: \: \: \: \: \: (i)→2a+{(p+q)−1}d=0(i)

Now Take

\sf \to \: S_{p + q} = \dfrac{(p + q)}{2} \{2a + (p + q - 1)d \}→S

p+q

=

2

(p+q)

{2a+(p+q−1)d}

Now Put the value

\sf \to \: 2a = - \{(p + q) - 1 \}d→2a=−{(p+q)−1}d

we get

\sf \to \: S_{p + q} = \dfrac{(p + q)}{2} \{ - (p + q - 1)d + (p + q - 1)d \}→S

p+q

=

2

(p+q)

{−(p+q−1)d+(p+q−1)d}

\sf \to \: S_{p + q} = \dfrac{(p + q)}{2} \{ 0 \}→S

p+q

=

2

(p+q)

{0}

\sf \to \: S_{p + q} = 0→S

p+q

=0

Hence Proved

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