Question

Q.11 The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

90

87

85

55

​

Answer 1

Answer:

MARK AS BRAINLIEST

Step-by-step explanation:

HOPE THIS WILL HELP YOU

Answer 2

Answer:

40th term of the AP is 87.

Step-by-step explanation:

Given :-

• The 6th and 17th terms of an AP are 19 and 41 respectively.

To find :-

• The 40th term.

Solution :-

Formula used :

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

• a = 1st term
• d = common difference

Now find the 6th term and 17th term.

$\sf{T_6=a+(6-1)d}$

$\implies\sf{19=a+5d}$

$\implies\sf{a+5d=19}$

$\implies\sf{a=19-5d.............(i)}$

&

$\sf{T_{17}=a+(17-1)d}$

$\implies\sf{41=a+16d}$

• Put a = 19-5d from eq (i)

$\implies\sf{41=19-5d+16d}$

$\implies\sf{41=19+11d}$

$\implies\sf{19+11d=41}$

$\implies\sf{11d=41-19}$

$\implies\sf{11d=22}$

$\implies\sf{d=2}$

Now put d = 2 in eq (i) for getting the value of a.

a = 19-5d

→ a = 19-5×2

→ a = 9

Now find the 40th term of the AP.

$\to\sf{T_{40}=9+(40-1)\times2}$

$\to\sf{T_{40}=87}$

Therefore, the 40th term of the AP is 87.