Question

The fourth term of Ap is 10. If eleventh term is one more three times of the fourth term find the sum of its 25 terms
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Answer 1

Solution!!

The concept of arithmetic progression has to be used here. The 4th term of an AP is given. It is given that the 11th term is one more than 3 times the 4th term. We have to find the sum of first 25 terms.

T₄ = 10

T₁₁ = 3T₄ + 1 = 30 + 1 = 31

a + 3d = 10...(1)

a + 10d = 31...(2)

Here: a is the first term and d is the common difference between the terms.

Subtracting (1) and (2), we get,

a - a + 3d - 10d = 10 - 31

-7d = -21

d = 3

Putting the value of d in (1),

a + 3d = 10

a + 3(3) = 10

a + 9 = 10

a = 1

Sₙ = n/2 [2a + (n - 1)d]

Here: We had to find the sum of 25 terms. Hence, n is 25.

S₂₅ = 25/2 [2(1) + (25 - 1)(3)]

S₂₅ = 25/2 [2 + (24)(3)]

S₂₅ = 25/2 [2 + 72]

S₂₅ = 25/2 [74]

S₂₅ = (25/2) × 74

S₂₅ = 25 × 37

S₂₅ = 925