Question

The 104 th term and 4 th term of an ap are 125 and 0. Find the sum of first 35 terms

Answer 1

Solution :

⇒ a + 103d = 125

⇒ a = 125 - 103d .......eq1

⇒ a + 3d = 0

put value of a from eq1

⇒ 125 - 103d + 3d = 0

⇒ 125 - 100d = 0

⇒ d = - 125 / - 100

⇒ d = 1.25

now

⇒ a + 3d = 0

⇒ a + 3 × 1.25 = 0

⇒ a = 3.75

Now sum of 35 terms

★ Sn = n/2 [ 2a + ( n - 1 ) d ] ★

⇒ S₃₅ = 35/2 [ 2 × 3.75 + ( 35 - 1 ) 1.25 ]

⇒ S₃₅ = 35/2 [ 7.5 + 34 × 1.25 ]

⇒ S₃₅ = 35/2 [ 7.5 + 42.5 ]

⇒ S₃₅ = 35/2 × 50

⇒ S₃₅ = 875

Answer 2

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