Question

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

Answer 1

It is given that,

t₄ = –15

t₉ = –30

Now,

t_n = a + (n - 1)d

t₄ = a + (4 - 1)d

⇒ -15 = a + 3d

⇒ a = -15 - 3d    ...(1)

⇒ t₉ = a + (9 - 1)d

⇒ -30 = a + 8d

⇒ a + 8d = -30

⇒ -15 - 3d + 8d = -30         ...(From 1)

⇒ -15 + 5d = -30

⇒ 5d = -30 + 15

⇒ 5d = -15

⇒ d = -3

⇒ a = -15 - 3(-3)       ...(From 1)

⇒ a = -15 + 9

⇒ a = -6

t₁ = a = -6

t₂ = t₁ + d = -6 - 3 = -9

t₃ = t₂ + d = -9 - 3 = -12

t₄ = t₃ + d = -12 - 3 = -15 

Hence, the given A.P. is –6, –9, –12, -15, ....

Now,

\[S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right)\]

\[ S_{10} = \frac{10}{2}\left( 2a + \left( 10 - 1 \right)d \right)\]

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

= 5 (-12 - 27)

= 5 (-39)

= -195

Hence, the sum of the first 10 numbers is –195.