Question

plz solve fast and help me to solve the problem ​

Answer 1

t4 = 36 , n=4

(t10 + t9 + t8) = t3 = 105 , n = 3

now,

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

36 = 4/2 [ 2a + (4 -1)d]

36 = 2 [2a + 3d]

18 = 2a + 3d ........1

And,

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

105 = 3/2 [2a + (3 -1)d]

105 = 3/2[2a + 2d]

105 = 3/2 ×2 (a + d)

105 = 3(a + d)

105 = 3a + 3d ........2

subtracting eq 1 from 2

3a + 3d = 105

-(2a + 3d = 18 )

a = 87✅

substituting a = 87 in 1

•°• 2 × 87 + 3d = 18

174 + 3d = 18

3d = -156

d = -156/3

d = -52✅

The common difference of A.P is -52