The fourth term of Ap is 10. If eleventh term is one more three times

1.	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
Answer» 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

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

Answer»

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