Question

the AP 12 th term is -13 ,sum of 1st 4 terms is 24 find AP​

Answer 1

It is given that a12= -13

=> a + 11d =-13----(i)

and it is also given that sum of first four terms is 24

== >

a - 3d + a - d + a + d + a+3d=24

4a = 24

a = 6

Therefore from (i),

a +11d= -13

6 + 11d = -13

11d = -19

d = -19/11

So, a1= 6

a2 = 6 -19/11

= 66-19/11

= 47/11

Therefore the required AP is 6,47/11.....

Answer 2

We have,

a12 = -13 and

S4 = 24

we want to find the AP.

so,

a + 11d = -13.............(1)

and

n/2 {2a + (n-1)d} = 24

n = 4

4/2 {2a + (4-1)d} =24

2 {2a + 3d} = 24

2a + 3d = 12............(2)

Multiplying eq. (1) by 2

2 × (a + 11d) = -13 × 2

2a + 22d = -26..........(3)

Equation (3) - equation (2)

2a + 22d = -26

2a + 3d = 12

(-) (-) (-)

19d = -38

d = -38/19

d = -2................(4)

substituting value of d in equation (3)

2a + 3(-2) = 12

2a - 6 = 12

2a = 12 + 6

2a = 18

a = 9

Hence The required AP will 9, 7, 5,........an.

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