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parth your question
The A.P. in which 4th term is -15 and 9th term is -30. Find the sum of the firm numbers.
Answer

It is given that,

t_{4} = - 15

t_{9} = - 30

Now,

t_{n} = a + (n - 1) * d

t_{4} = a + (4 - 1) * d

Rightarrow - 15 = a + 3d

Rightarrow a=-15 - 3d ...(1)

t_{9} = a + (9 - 1) * d

Rightarrow - 30 = a + 8d

Rightarrow a + 8d = - 30

Rightarrow - 15 - 3d + 8d =-30(from(1)

Rightarrow - 15 + 5d = - 30

Rightarrow 5d = - 30 + 15

Rightarrow 5d = - 15 Rightarrow d = - 3

Rightarrow a = - 15 - 3(- 3) (from (1))

Rightarrow a = - 15 + 9

Rightarrow a = - 6

Hence, the given A.P. is -6, -9, -12,....

Now,

S_{n} = n/2 * (2a + (n - 1) * d) S_{10} = 10/2 * (2a + (10 - 1) * d)

= 5(2(- 6) + 9(- 3)) 5(-12-27)

= 5(-39)

= -195

Hence, the sum of the first 10 numbers

is - 195

Hope this helps you second one is coming.​