Question

The 17th term of an ap is 5 more than twice it's 8th term if the 11th term of an ap is 43 find the 15th term​

Answer 1

Step-by-step explanation:

17th term = a + 16 d

8th term = a + 7d

11th term = a + 10 d

acc to question

a + 16 d = 2 (a + 7d) + 5

a + 16 d = 2a + 14 d + 5

a - 2d = -5 ------(1)

a + 10 d = 43------(2)

Solving both equation simultaneously, you get

a = 3

d = 4

15th term = a + 14 d = 59

hope it helps

Answer 2

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✤ Required Answer:

✒ GiveN:

• 17th term of an AP is 5 more than twice of 8th term.
• 11th term is 43

✒ To FinD:

• Find its 15th term....?

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✤ How to solve?

For the above question, We need to know the formula of nth term of an AP, That is:

$\large{ \boxed{ \sf{a_n = a + (n - 1)d}}}$

Here, an is the last term, a is the first term, n is the no. of terms and d is the common difference. ☃️ So, let's solve this question...

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✤ Solution:

By using the above formula,

• a17 = a + 16d
• a8 = a + 7d
• a11 = a + 10d

According to condition-1)

➝ a17 = 2a8 + 5

➝ a + 16d = 2(a + 7d) + 5

➝ a + 16d = 2a + 14d + 5

➝ a - 2a + 16d - 14d = 5

➝ - a + 2d = 5

➝ - a = 5 - 2d

➝ a = 2d - 5.........(1)

According to condition-2)

➝ a11 = 43

➝ a + 10d = 43

Substituting value of a in this eq.

➝ 2d - 5 + 10d = 43

➝ 12d - 5 = 43

➝ 12d = 48

➝ d = 48/12

➝ d = 4

Putting the value of d in eq.(1),

➝ a = 2(4) - 5

➝ a = 8 - 5

➝ a = 3

We have to find, a15

➝ a15 = a + 14d

➝ a15 = 3 + 14(4)

➝ a15 = 3 + 56

➝ a15 = 59

☀️ 15th term of the AP = 59

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