Question

The sum of the first five terms of an AP is 25 and the sum of its next five terms is -75. find the 10th term of the AP.

Answer 1

SolutioN :-

Sum of first five terms of an AP is 25 , Therefore

→ Sₙ = n/2 [ 2a + ( n - 1 ) d ]

→ S₅ = 5/2 [ 2a + ( 5 - 1 ) d ]

→ 25 = 5/2 [ 2a + 4d ]

→ 25 × 2/5 = 2a + 4d

→ 2a + 4d = 10 .... ( 1 )

Sum of its next five terms is -75 , Therefore

Sum of first 10 terms = 25 + (- 75 ) = - 50

→ Sₙ = n/2 [ 2a + ( n - 1 ) d ]

→ S₁₀ = 10/2 [ 2a + ( 10 - 1 ) d ]

→ - 50 = 5 [ 2a + 9d ]

→ - 50/5 = 2a + 9d

→ 2a + 9d = - 10 .... ( 2 )

Subtract ( 2 ) from ( 1 )

→ 2a + 4d = 10

→ - 2a - 9d = 10

→ - 5d = 20

→ d = - 20/5

→ d = - 4

Put value of d in ( 1 )

→ 2a + 4d = 10

→ 2a + 4 × ( - 4 ) = 10

→ 2a - 16 = 10

→ 2a = 26

→ a = 13

Now 10th term of the AP is

→ A₁₀ = a + 9d

→ A₁₀ = 13 + 9 × ( - 4 )

→ A₁₀ = 13 - 36

→ A₁₀ - 23

∴ 10th term of AP is - 23