Question

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. find the nth term.

Answer 1

Answer:

The nth term is 4n - 1.

Step-by-step explanation:

Given :  

a₁₇ = 5 + 2(a₈) and a11 = 43

 

a₁₇ = 5 + 2(a₈)

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

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

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

2d = a + 5

a = 2d - 5………….(1)

 

a11 = 43

a + (n - 1) d = 43

a + 10d = 43………….(2)

On Putting the value of a = 2d - 5 from eq 1  in eq 2,

(2d - 5) + 10d = 43

12d - 5 = 43

12d = 43 + 5

12d = 48

d = 48/12

d = 4

Common Difference , d = 4  

 

On Putting the value of d = 4  in eq 1

a = 2d - 5

a = 2d - 5  

a = 2(4) - 5  

a = 8 - 5  

a = 3

∴ first term,  a = 3 and common Difference , d = 4

 

nth term of AP, an  = a + (n-1)d

an = 3 + (n - 1)(4)

an = 3 + 4n - 4

an = 4n - 1

Hence, the nth term is 4n - 1.

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Answer 2

Given: T17 = 2× T18 +5

T11= 43 (11th term of the A.P. is 43.)

To find: Tn( the nth term)

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

as Tn =a+( n-1 )d

so, -a + 2d= 5

Likewise, a+10d= 43

On solving the two equations we have,

12d= 48 i.e. d=3

and a=3 .

SO, nth term= 3+ (n-1) 4

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