Question

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show

that the sum of its first (p + q) terms is zero. (p ¹ q)​

Answer 1

Answer:-

Given:-

sum of first p terms = Sum of first q terms of an AP.

We know that,

Sum of first n terms of an AP (Sₙ) = n/2 [ 2a + (n - 1)d ]

So,

According to the question,

⟹ p/2 [ 2a + (p - 1)d ] = q/2 [ 2a + (q - 1)d ]

⟹ p (2a + pd - d) = q(2a + qd - d)

⟹ 2ap + p²d - pd = 2aq + q²d - qd

⟹ 2ap - 2aq + p²d - q²d = pd - qd

⟹ 2a(p - q) + d(p² - q²) = d (p - q)

using a² - b² = (a + b)(a - b) in LHS we get,

⟹ 2a(p - q) + d(p + q)(p - q) = d(p - q)

⟹ (p - q) [ 2a + (p + q)d ] = d (p - q)

⟹ 2a + (p + q)d = d

Now, subtract d from both sides.

⟹ 2a + (p + q)d - d = d - d

⟹ 2a + (p + q - 1)d = 0

Now,

Multiply (p + q)/2 both sides.

⟹ (p + q)/2 [ 2a + (p + q - 1)d ] = 0 * (p + q)/2

⟹ S_(p + q) = 0.

[ ∵ n = p + q ]

Hence , Proved.

Answer 2

Given : The sum of first p terms of an A.P. is equal to the sum of first q terms .

Exigency To Prove : The sum of its first (p + q) terms is zero .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given that ,

⠀⠀⠀⠀⠀▪︎⠀⠀The sum of first p terms of an A.P. is equal to the sum of first q terms .

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\bf Sum\:of\:an\:A.P.\:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ S _{(n)} \:: \dfrac{n}{2} \:[ 2a + ( n - 1 ) d] }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , S is the sum of $\sf n^{th}$ term , a is the first term of an A.P , n is the$\sf n^{th}$ term of an A.P d is the common Difference.

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \:, \: According \:to\: the \: Question \::}}$

⠀⠀⠀⠀⠀━━━━The sum of first p terms of an A.P. is equal to the sum of first q terms .

$\qquad:\implies \sf Sum \: of \:first \: p \: terms \: = Sum \: of \:first \: q \: terms$

$\qquad:\implies \sf S_{p}\: =S_{q}\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf S_{p}\: =S_{q}\:$

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

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

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

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

$\qquad:\implies \sf p \:[ 2a + pd - d] \: =\:q \:[ 2a + qd - d]\:$

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

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

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

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese \: \bf Algebraic\: \: Indentity\:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ (a^2 -b ^2 ) = ( a + b ) ( a - b ) }\bigg\rgroup$

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

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

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

$\qquad:\implies \sf \cancel{(p -q)} [ 2a + (p+q)d ] \: =\: d\cancel{(p - q)}\:$

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

$\qquad \bigstar \:\sf Subtracting \: \:\:\:\bf d \:\:\sf both\: \: sides \::$

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

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

$\qquad:\implies \sf [ 2a + (p+q-1)d ] \: =\: 0\:$

$\qquad \bigstar \:\sf Multiplying \: \:\:by\:\:\bf \dfrac{p +q}{2} \:\:\sf both\: \: sides \::$

$\qquad:\implies \sf [ 2a + (p+q-1)d ] \: =\: 0\:$

$\qquad:\implies \sf \dfrac{p +q}{2} [ 2a + (p+q-1)d ] \: =\: 0\: \times \dfrac{p +q}{2}$

$\qquad:\implies \sf \dfrac{p +q}{2} [ 2a + (p+q-1)d ] \: =\: 0\:$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ S _{(n)} \:: \dfrac{n}{2} \:[ 2a + ( n - 1 ) d] }\bigg\rgroup$

Therefore,

$\qquad:\implies \sf \dfrac{p +q}{2} [ 2a + (p+q-1)d ] \: =\: 0\:$

$\qquad:\implies \bf S_{(p+ q)} \: =\: 0\:$

⠀⠀⠀⠀⠀⠀Here n or $\sf n^{th}$ term is p + q .

⠀⠀━━━━ As , We can see that here , the sum of its first (p + q) terms is zero .

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Proved \:}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀