यदि `S_(1),S_(2),S_(3)` तीन A.P. के n पदों के योग हैं जिनके ( प्रत्येक का ) पद प्रथम पद 1 हैं और उनके c.d. क्रमशः 1,2,3 हैं , तो सिद्ध करें कि `S_(1) + S_(3) = 2 S_(2)`.

For an A.P., S_(1)=6 and S_(7)= 105 . Prove that S_(n): S_(n-3) = (n+ 3): (n-3)

If S_(n) denotes the sum of n terms of an A.P., S_(n +3) - 3S_(n + 2) + 3S_(n + 1) - S_(n) =

Let sum of n , 2n , 3n , terms of an A.P are S_(1), S_(2), S_(3) respectively. Prove that S_(3) = 3 (S_(2) - S_(1)) .

Let each of the circles, S_(1)=x^(2)+y^(2)+4y-1=0 , S_(2)=x^(2)+y^(2)+6x+y+8=0 , S_(3)=x^(2)+y^(2)-4x-4y-37=0 touches the other two. Let P_(1), P_(2), P_(3) be the points of contact of S_(1) and S_(2), S_(2) and S_(3), S_(3) and S_(1) respectively and C_(1), C_(2), C_(3) be the centres of S_(1), S_(2), S_(3) respectively. Q. The co-ordinates of P_(1) are :

Let each of the circles, S_(1)=x^(2)+y^(2)+4y-1=0 , S_(2)=x^(2)+y^(2)+6x+y+8=0 , S_(3)=x^(2)+y^(2)-4x-4y-37=0 touches the other two. Let P_(1), P_(2), P_(3) be the points of contact of S_(1) and S_(2), S_(2) and S_(3), S_(3) and S_(1) respectively and C_(1), C_(2), C_(3) be the centres of S_(1), S_(2), S_(3) respectively. Q. The co-ordinates of P_(1) are :

If the sums of n, 2n and 3n terms of an A.P. be S_(1), S_(2), S_(3) respectively, then show that, S_(3) = 3(S_(2) - S_(1)) .

If the sum of n, 2n, 3n terms of an A.P are S_(1), S_(2), S_(3) , respectively, prove that S_(3) = 3 (S_(2) -S_(1)).