find the first quantity as a percentage of second quantity
Answers
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