Question

The seventh term of an A.P is 41 and the thirteenth term is 77. find the twentieth term?​

Answer 1

Answer:-

$a_{20} = 119$

Given :-

$a_7 = 41$

$a_{13} = 77$

To find :-

The 20 th term .

Solution:-

Let the first term be a and common Difference be d .

A/Q

$a_7= 41$

→$a + 6d = 41 -------1$

$a_{13}= 77$

$a + 12 d = 77 -------2$

Subtract eq.1 and eq.2

We get,

→$a + 6d - ( a + 12d) = 41 - 77$

→$a - a + 6d -12 d = 41 - 77$

→$-6d = - 36$

→$d = \dfrac{-36}{-6}$

→$d = 6$

Put the value of d in eq.1,

a + 6d = 41

a + 6 × 6 = 41

a + 36 = 41

a = 41 - 36

a = 5

Now,

→$a_{20} = a + 19 d$

→$a_{20}= 5 + 19 \times 6$

→$a_{20} = 5 + 114$

→$a_{20} = 119$

hence,

The 20 th term of AP will be 119 .

Answer 2

GIVEN :

Seventh term of an AP = 41

=> a + 6d = 41 ----(1)

Thirteenth term of the AP = 77

=> a + 12d = 77 -----(2)

Solve eq - (1) and (2) to find (d) common difference.

a + 6d = 41

a + 12d = 77

(-)

------------------

-6d = -36

------------------

6d = 36

d = 36/6

Common Difference = 6

Substitute (d) in eq - (1) to find first term(a)

a + 6d = 41

a + 6(6) = 41

a + 36 = 41

a = 41 - 36

First term = 5

We got both a and d. Now,

Twentieth term :

a20 = a + 19d

= (5) + 19(6)

= 5 + 144

a20 = 119

Therefore, twentieth term is 119.