Question

jldddddiii denaaaa anssss

Answer 1

Heya☺

an = a+ (n-1)d

sn = n/2 ( 2a + (n-1) d)

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

24 = 2a + 10d
12 = a+ 5d
a= 12 - 5d. ----- eq 1


a6 + a10 = a+5d + a+ 9d
34 = 2a + 14d
17 = a + 7d --. eq2


Substituting the value of a in eq2

17 = 12- 5d + 7d
17 = 12 +2d.
5 = 2d.

d = 2/5

Therefore
a = 12 - 5 × 2/5
a = 12 - 2
a = 10

First term (a). = 10

Second term (a2) = a+d = 10+ 2/5 = 52/5

Third term (a3) = a+2d = 10 + 2× 2/5 = 54/5

Answer 2

Let the first term of an A.P = a

and the common difference of the given  A.P = d
As we know that 
a n = a+(n-1) d
a 4 = a +( 4-1) d
a 4 = a+3d
Similarly , 
a 8 = a + 7 d 
a 6  = a + 5 d
a 10 = a+ 9d
Sum of 4 th and 8th terms of an A.P = 24 ( given )
a 4 +a 8 = 24
a + 3d + a + 7d = 24
2a + 10 d = 24
a +5d = 12  .....................(i)
Sum of 6 th and 10 th term  of an A.P = 34 ( given )
a 6 +a 10 = 34
a + 5d +a+ 9d = 34
2a + 14 =34
a + 7d = 17  .....................(ii)
Solving (i) & (ii)
a +7 d  = 17
a + 5d = 12
-   -       -

2d  = 5
d = 2.5
From equation (i) , 
a + 5d = 12
a + 5 (2.5) = 12
a+12.5= 12
a = -0.5
a 2 = a+d = -0.5+5 = 4.5
a 3 = a 2 + d = 4.5+5 = 9.5
-0.5,4.5,9.5