Math, asked by nisha1456, 1 year ago

please answer this guys...
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Answered by siddhartharao77
4

Answer:

3

Step-by-step explanation:

nth term of an AP a(n) = a + (n - 1) * d.

(i)

8th term of an AP is half its second term.

a₈ = (1/2)[a + (2 - 1) * d]

⇒ a + 7d = (1/2)[a + d]

⇒ 2(a + 7d) = a + d

⇒ 2a + 14d = a + d

⇒ a = -13d


(ii)

Given that 11th term exceeds one-third of its fourth term by 1.

a₁₁ = (1/3)a₄ + 1

⇒ a + (11 - 1) * d = (1/3)[a + (4 - 1) * d] + 1

⇒ a + 10d = (1/3)[a + 3d] + 1

⇒ 3(a + 10d) = a + 3d + 3

⇒ 3a + 30d = a + 3d + 3

⇒ 2a + 27d = 3


Substitute (i) in (ii), we get

2a + 27d = 3

2(-13d) + 27d = 3

-26d + 27d = 3

d = 3


Substitute d = 3 in (i), we get

⇒ a = -13d

      = -13(3)

      = -39.


15th term of the AP:

= a + 14d

= -39 + 14(3)

= -39 + 42

= 3.


Hope it helps!


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Answered by Arey
2
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