Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \bf \: S_{x + y} \: = \: - \: (x + y) \: }}\\ \\ [/tex]

Step-by-step explanation:

Let assume that first term and common difference of an AP be a and d respectively.

Now, it is given that

[tex]\sf \: S_x = y \\ \\ [/tex]

[tex]\sf \: \dfrac{x}{2} \bigg(2a + (x - 1)d \bigg) = y \\ \\ [/tex]

[tex]\sf\implies \sf \: 2a + (x - 1)d =\dfrac{2y}{x} - - - (1) \\ \\ [/tex]

Further given that

[tex]\sf \: S_y = x \\ \\ [/tex]

[tex]\sf \: \dfrac{y}{2} \bigg(2a + (y - 1)d \bigg) = x \\ \\ [/tex]

[tex]\sf\implies \sf \: 2a + (y - 1)d =\dfrac{2x}{y} - - - (2) \\ \\ [/tex]

On Subtracting equation (2) from equation (1), we get

[tex]\sf \: (x - y)d =\dfrac{2y}{x} - \dfrac{2x}{y} \\ \\ [/tex]

[tex]\sf \: (x - y)d =\dfrac{2 {y}^{2} - 2 {x}^{2} }{xy} \\ \\ [/tex]

[tex]\sf \: (x - y)d =\dfrac{2({y}^{2} -{x}^{2})}{xy} \\ \\ [/tex]

[tex]\sf \: (x - y)d =\dfrac{2(y + x)(y - x)}{xy} \\ \\ [/tex]

[tex]\sf \: (x - y)d = - \dfrac{2(y + x)(x - y)}{xy} \\ \\ [/tex]

[tex]\sf\implies \sf \:d = - \dfrac{2(y + x)}{xy} \\ \\ [/tex]

Now, Consider

[tex]\sf \: S_{x + y} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(2a + (x + y - 1)d \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(2a + (x - 1)d + yd \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(\dfrac{2y}{x} - \dfrac{2y(x + y)}{xy} \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(\dfrac{2y}{x} - \dfrac{2(x + y)}{x} \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(\dfrac{2y - 2x - 2y}{x} \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg(\dfrac{ - 2x}{x} \bigg)\\ \\ [/tex]

[tex]\sf \: = \: \dfrac{x + y}{2} \bigg( - 2 \bigg)\\ \\ [/tex]

[tex]\sf \: = \: - \: (x + y)\\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: S_{x + y} \: = \: - \: (x + y) \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formula used :-

↝ Sum of n  terms of an arithmetic progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}[/tex]

Wʜᴇʀᴇ,

a is the first term of the progression.

n is the no. of terms.

d is the common difference.

Verified answer

Answer:

-(x+y)

Step-by-step explanation:

Given:

• Sum of x terms (Sx) = y
• Sum of y terms (Sy) = x

To Find:

• Sum of (x+y) terms

Formula used:

• Sn = n(2a+(n-1)d)/2

where,

Sn is sum of n terms

n is no. of terms

a is the first terms

d is the common difference

Using the above formula, we have

Sx = x(2a+(x-1)d)/2 = y

So, (2a+(x-1)d) = 2y/x...(1)

Sy = y(2a+(y-1)d)/2 = x

So, (2a+(y-1)d) = 2x/y...(2)

Now, S(x+y) = (x+y)(2a+(x+y-1)d)/2 .... (3)

Eq(2) - Eq(1), we get

d(y-x) = 2x/y - 2y/x

d(y-x) = 2(x^2-y^2)/xy

d(y-x) = 2(x+y)(x-y)/xy

Cancelling common term, we get

d = -2(x+y)/xy

Putting value of d in eq(3), we get

S(x+y) = (x+y)(2a+(x-1)d+yd)/2

S(x+y) = (x+y)(2y/x - 2(x+y)/x)/2

S(x+y) = (x+y)(-x/x)

S(x+y) = -(x+y)

So, the answer is -(x+y).