The A.P. in which 4th term is – 15 and 9th term is – 30. Find the sum of the first 10 numbers.

t₄ = – 15 and t₉ = – 30

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

t₄ = – 15

⇒ a + (4 – 1)d = – 15

⇒ a + 3d = – 15 …..(1)

t₉ = – 30

⇒ a + (9 – 1)d = – 30

⇒ a + 8d = – 30 …..(2)

Subtracting eq. (1) from eq. (2)

⇒ [a + 8d] – [a + 3d] = – 30 – ( – 15)

⇒ 5d = – 30 + 15 = – 15

Substituting, “d” in eq. (1)

⇒ a + 3 × ( – 3) = – 15

⇒ a + – 9 = – 15

⇒ a = – 15 + 9 = – 6

Thus, By using sum of n^th term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_n = sum of n terms

We need to find S₁₀

⇒S₁₀ = 5 [ – 12 + 9 × ( – 3)]

⇒S₁₀ = 5 [ – 12 – 27]

⇒S₁₀ = 5 × ( – 39) = – 195