Question

The 17th term of an AP is 5 more than twice its 8th term if the 11th term of the AP os 43 find the nth term

Answer 1

a17 = 2(a8) + 5
a + 16d = 2( a + 7d) +5
a + 16d = 2a +14d + 5
2a - a + 14d - 16d = -5
a - 2d = -5 ----------eqn1
a11 = 43 (given)
a11 = a + 10d
a + 10d = 43 ------------ eqn2
On subtracting eqn1 from eqn2
We get
a + 10d = 43
- a - 2d = -5
___________
12d = 48
___________

12d = 48
d = 48/12
d = 4
By putting the value of d in eqn1
a = -5 + ( 2x4 )
a = -5 + 8
a = 3
an = a + ( n - 1 ) d
an = 3 + ( n - 1 ) 4
an = 3 + 4n - 4
an = 4n -1

Answer 2

✤ 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}}}

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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