Question

If the sum of the first p terms of an ap is the same as the sum of its first q terms
then show that the sum of first p+q terms is 0​

Answer 1

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

$let \: \: a \: \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: ap \: \\ then \\ s _{p} = s _{q} \implies \frac{p}{2} (2a + (p - 1)d) = \frac{q}{2} (2a + (q - 1)d \\ \implies(p - q)(2a) = (q - p)(q + p - 1) \\ \implies2a = (1 - p - q)d \: \: \: \: \: \: \: .....(1) \\ sum \: of \: the \: first \: (p + q) \: terms \: of \: the \: given \: ap \\ = \frac{(p + q)}{2} (2a + (p + q - 1)d) \\ = \frac{(p + q)}{2} .(1 - p - q)d + (p + q - 1)d \: \: \: \: \: \: \: \: (using \: 1) \\ = 0$

${\huge{\underline{\underline{\underline{\orange{\mathtt{❢◥ ▬▬▬▬▬▬ ◆ ▬▬▬▬▬▬ ◤❢}}}}}}}$

Answer 2

Answer:

Formula :

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

Consider sum of first p terms .

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

Consider sum of first q terms .

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

Given ,

Sp =Sq

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

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

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

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

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

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

(p- q ) is not equal 0

=> [2a +( p+q -1 ) d ] = 0

Now,

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

p+q/2 ×0

= 0

Therefore the sum of ( p+q ) terms is 0