Question

please answer this question ​

Answer 1

Answer:

15/2×(10+56)

15/2×66

495

the required no is 495

Answer 2

Question:-

The 16th term of an AP is 5 times it's 3rd term. If it's 10th term is 41 , find the sum of first 15 terms.

Answer:-

Given:

10th term of an AP (a₁₀) = 41

a₁₆ = 5 * a₃

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

So,

★ a + (10 - 1)d = 41

⟶ a + 9d = 41

⟶ a = 41 - 9d -- equation (1)

Similarly,

★ a + 15d = 5(a + 2d)

⟶ a + 15d = 5a + 10d

Substitute the value of a from equation (1)

⟶ 41 - 9d + 15d = 5(41 - 9d) + 10d

⟶ 41 + 6d = 205 - 45d + 10d

⟶ 41 - 205 = - 45d + 10d - 6d

⟶ - 164 = - 41d

⟶ - 164/ - 41 = d

⟶ 4 = d

substitute the value of d in equation (1),

⟶ a = 41 - 9(4)

⟶ a = 41 - 36

⟶ a = 5

Now,

Sum of first n terms of an AP – Sₙ = n/2 [ 2a + (n - 1)d

⟶ S₁₅ = 15/2 * [ 2(5) + (15 - 1)(4) ]

⟶ S₁₅ = 15/2 * [ 10 + 14 * 4 ]

⟶ S₁₅ = 15/2 * (66)

⟶ S₁₅ = 495

∴ The sum of first 15 terms of the given AP is 495.