Question

Question 1:
(a) If the sum of p terms of an A.P. is equal to sum of its q terms. Prove that the sum of
(p + q) terms of it is equal to zero.​

Answer 1

Given

$\sf \to \: S_p = S_q$

To Proved

$\sf \to \: S_{p + q} = 0$

Formula

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

So

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

and

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

According to question, we can write as

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

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

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

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

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

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

Now Simplify

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

we get

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

Now Take a common

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

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

Now Take

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

Now Put the value

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

we get

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

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

$\sf \to \: S_{p + q} = 0$

Hence Proved

Answer 2

Answer:

Given :-

• The sum of p terms of an A.P is equal to sum of its q term.

Prove That :-

• The sum of (p + q) terms of it is equal to zero.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]}}}$

Solution :-

Given :

$\mapsto \sf S_p =\: \dfrac{p}{2}\bigg[2a + (p - 1)d\bigg]$

And,

$\mapsto \sf S_q =\: \dfrac{q}{2}\bigg[2a + (q - 1)d\bigg]$

According to the question by using the formula we get,

$\implies \sf S_p =\: S_q$

$\implies \sf \dfrac{p}{2}\bigg[2a + (p - 1)d\bigg] =\: \dfrac{q}{2}\bigg[2a + (q - 1)d\bigg]$

$\implies \sf p[2a + (p - 1)d] =\: q[2a + (q - 1)d]$

$\implies \sf 2ap + (p - 1)pd =\: 2aq + (q - 1)qd$

$\implies \sf 2ap + {p}^{2}d - pd =\: 2aq + {q}^{2} - qd$

$\implies \sf 2ap - 2aq =\: {q}^{2}d - qd - {p}^{2}d + pd$

$\implies \sf 2a(p - q) =\: d(q^2 - p^2) + d(p - q)$

$\implies \sf 2a(p - q) =\: d(q + p)(q - p) + d(p - q)$

$\implies \sf 2a(p - q) =\: d\bigg[(q + p)(q - p) + (p - q)\bigg]$

$\implies \sf 2a(p - q) =\: d\bigg[- (q + p)(p - q) + (p - q)\bigg]$

$\implies \sf 2a(p - q) =\: d(p - q)\bigg[1 - q - p\bigg]$

$\implies \sf 2a =\: d(1 - q - p)$

$\implies \sf\bold{\purple{2a =\: d(1 - p - q)\: ----\: (Equation\: No\: 1)}}$

Now,

$\leadsto \sf S_{p + q} =\: \bigg(\dfrac{p + q}{2}\bigg) \bigg[2a + (p + q - 1)d\bigg]$

$\implies \sf S_{p + q} =\: \bigg(\dfrac{p + q}{2}\bigg) \bigg[(1 - q - p)d + (p + q - 1)d\bigg][\because\: From\: equation\: no\: 1]$

$\implies \sf S_{p + q} =\: \bigg(\dfrac{p + q}{2}\bigg)d(\cancel{1} \cancel{- q} \cancel{- p} \cancel{+ p} \cancel{+ q} \cancel{- 1})$

$\implies \sf\bold{\red{S_{p + q} =\: 0}}$

${\large{\bold{\purple{\underline{\diamondsuit\: HENCE, PROVED}}}}}$