Question

The second term of an AP is (x - y) and
fifth term is (x + y), its first term is .......​

Answer 1

$\huge \red{\mathfrak{solution}}$

◆ Given:

● 2nd term of an AP is (x-y)

and

●5th term is (x+y)

◆To find :

first term (a).

✧a+d= x-y------(i)

✧a+4d= x+y----(ii)

subtract both the equation we get ,

=> -3d= -2y

=> d=2y/3

Now, put the value of d in equation (i)

$\leadsto \: a + d = x - y \\ \\ \leadsto \: a + \frac{2y}{3} = x - y \\ \\ \leadsto \: a = x - y - \frac{2y}{3} \\ \\ \leadsto \: a = \frac{3x - 3y - 2y}{3} \\ \\ \leadsto \: a = \frac{3x - 5y}{3}$

Hence the first term is

$\huge\boxed{ \boxed{ \purple{ \bold{ \frac{3x - 5y}{3} }}}}$

Answer 2

Given :-----

• second term of AP = (x-y)
• 5th term = (x+y)

Question :----- we have to find first term of AP ?

we know that ,

an = a+(n-1)d

so,

a2 = a+d = x-y ---------------(1)

a5 = a+4d = x+y ------------(2)

Adding both equation we get,

2a+5d = 2x

2a = 2x - 5d

a = (2x-5d)/2 ----------------(3)

Subtracting Equation (1) from (2) we get,

3d = 2y

d = 2y/3 -----------------------(4)

Putting value of d from Equation (4) in Equation (3) we get,

$a = \frac{2x - (5 \times \frac{2y}{3} )}{2} \\ \\ a = \frac{(6x - 10y)}{6} \\ \\ a = \frac{(3x - 5y)}{3}$

So, First term of AP is

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:</strong><strong>a = \frac{(3x - 5y)}{3}</strong><strong>}}}}}}}}}}$

(Hope it helps you)