Question

in an ap sum of 4th and 8th term is 22 and product 2nd and 6th term is 33 find the AP​

Answer 1

Answer:

AP will be 1,3,5,7,--- and so on

solution is given in image so check image also:-)

Answer 2

in an ap sum of 4th and 8th term is 22 and product 2nd and 6th term is 33 then the AP​ is 1, 3, 5, 7, 9, 11, 13, 15, 17,..............

Given,

a4 + a8 = 22

a2 × a6 = 33

Formula: an = a + (n-1) d

a4 + a8 = 22     ................condition 1

a4 = a + 3d

a8 = a + 7d

(a + 3d) + (a + 7d) = 2a + 10d

= 2 (a + 5d)

= 2 × 6th term

= 22

a6 = 22/2 = 11

a6 × a2 = 33 ................condition 2

11 × a2 = 33

a2 = 33/11 = 3

a6 = a + 5d

a2 = a + d

11 - 3 = (a + 5d) - (a + d)

8= 4d

d = 8/4

d = 2

we have,

a4 + a8 = (a+3d) + (a+7d) = 22

2a + 10d = 22

2a + 10(2) = 22

2a + 20 = 22

2a = 2

a = 1

Therefore, we have the series,

1, 3, 5, 7, 9, 11, 13, 15, 17,..............