Question

please solve Qno.12 ,13 ,given in the photo...

Answer 1

12. Given : Sp =Sq

Prove : Sp+q =0
Proof :
Sn = n/2 [2a + (n-1)d]
Sn = p/2 [2a + (p+1) ____ equation 1
Similarly,
Sq = q/2 [2a + (q+1) ____ equation 2

.°. Sp+q = p+q/2 [2a + (p+q-1)d]___ equation 3


° . ° Sp=Sq (given)

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


Multiplying both the sides by 2,

p[2a+(p-1)d] = q[2a + (q-1)d]
p(2a + pd - d) = q (2a +qd-d)
2ap + p(square) - pd = 2aq + q(square) - qd

2ap+ p(square) - pd - 2ap - q(sqare) + qd =0

2ap - 2aq + p(square) - q(square) - pd +qd =0

2a(p-q) + d (p square - q square) - d ( p-q) =0

2a(p-q) + d(p+q) (p-q) - d(p-q) =0

Dividing throughout by p-q we get,

2a+d(p-q)-d =0

2a + d (p+q-1) =0______ equation 4

Substituing 4 in 3,
Sp+q = p+q/2 [0]


Therefore,

Sp+q = 0

Hence proved.



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