Question

solve above question​

Answer 1

Answer:

2,10,18,26..

Step-by-step explanation:

∴ Sum of first n terms = (n/2)[2a + (n - 1) * d]

(i) Sum of 1st seven terms is 182.

∴ S₇ = 182

⇒ (7/2)[2a + (7 - 1) * d = 182

⇒ (7/2)[2a + 6d] = 182

⇒ 7[a + 3d] = 182

⇒ a + 3d = 26

(ii) 4th and 17th terms in ratio of 1:5

∴ a₄ : a₁₇ = 1:5

⇒ (a + 3d)/(a + 16d) = 1/5

⇒ 5(a + 3d) = a + 16d

⇒ 5a + 15d = a + 16d

⇒ d = 4a

Substitute d = 4a in (i), we get

⇒ a + 3d = 26

⇒ a + 3(4a) = 26

⇒ a + 12a = 26

⇒ 13a = 26

⇒ a = 2

Substitute a = 2 in (i), we get

⇒ a + 3d = 26

⇒ 2 + 3d = 26

⇒ 3d = 24

⇒ d = 8

Therefore, the AP is 2, 10, 18, 26....

Hope it helps!

Answer 2

Let the 1st term be a and the common difference be d

ii) Sum to 7 terms = (7/2)(2a + 6d) = 7(a + 3d) = 182; ==> a + 3d = 26 ---------- (1)

iii) 4th term = a + 3d and 17th term = a + 16d

==> from the given ratio, a + 16d = 5a + 15d; ==> d = 4a ---- (2)

Solving (1) & (2): a = 2 and d = 8

Thus the AP is: 2, 10, 18, 26, ----