Question

the sum of 5th and 12th term of an ap is 81 and the sum of 8th and 15th term is 111 find the 13th term​

Answer 1

Answer

63

$\rule{200}2$

Solution

Given : Sum of 5th and 12th term of an A.P. is 81 and sum of 8th and 15th term is 111.

To find : The 13th term

Let the first term of the A.P. be a and common difference be d

We know that nth term of A.P is given as

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

Hence, 5th term would be a + 4d

12th term = a + 11d

Their sum = a + 4d + a + 11d = 81

⇒2a + 15d = 81 .........(i)

And, 8th term = a + 7d

15th term = a + 14d

Their sum = a + 7d + a + 14d = 111

⇒ 2a + 21d = 111.........(ii)

Subtracting (i) from (ii)

⇒ 2a + 21d - 2a - 15d = 111 - 81

⇒ 6d = 30

⇒ d = 5

Put value of d in (i)

2a + 15d = 81

⇒ 2a + 15(5) = 81

⇒ 2a + 75 = 81

⇒ 2a = 81 - 75

⇒ 2a = 6

⇒ a = 3

Hence, the 13th term of the given A.P. would be a + 12d

= 3 + 12(5)

= 3 + 60

= 63

Answer 2

$\bold{\underline{\underline{Answer:}}}$

13th term of the AP = 63.

$\bold{\underline{\underline{Step\:by\:step\:explanation:}}}$

Given :

• Sum of 5th and 12th term of an AP = 81
• Sum of 8th and 15th term of an AP = 111

To find :

• 13th term of the AP.

Solution :

Let the first term of the AP be a.

Let the common difference of the AP be d.

For fifth term :

We know the formula for nth term of an AP.

Forumla :

$\bold{\boxed{\large{\pink{\rm{T_n=\:a+\:(n-1)\:d}}}}}$

Block in the values,

$\bold{t_5\:=\:a+\:(5-1)\:d}$

$\bold{t_5\:=\:a+\:(4)\:d}$

$\bold{t_5\:=\:a+\:4d}$ --> (1)

For 12th term :-

Block in the values in the formula for nth term.

$\bold{t_{12}\:=\:a+\:(12-1)\:d}$

$\bold{t_{12}\:=\:a+\:(11)\:d}$

$\bold{t_{12}\:=\:a+\:11d}$ --> (2)

Sum of 5th and 12th term = 81.

Add equation 1 to equation 2,

a + 4d + a + 11d = 81

2a + 15d = 81 ---> (3)

For 8th term :-

$\bold{t_8\:=\:a+\:(8-1)\:d}$

$\bold{t_8\:=\:a+\:(7)\:d}$

$\bold{t_8\:=\:a+\:7d}$ --> (4)

For 15th term :-

$\bold{t_{15}\:=\:a+\:(15-1)\:d}$

$\bold{t_{15}\:=\:a+\:(14)\:d}$

$\bold{t_{15}\:=\:a+\:14d}$ --> (5)

Sum of 8th and 15th term = 111

Add equation 4 to equation 5,

a + 7d + a + 14d = 111

2a + 21d = 111 ---> (6)

Solve equation 3 and equation 6 simultaneously by elimination method.

Subtract equation 3 from equation 6,

2a + 15d = 81 ---> (3)

2a + 21 d = 111 --> (6)

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

- 6d = - 30

$\bold{d={\dfrac{-30}{-6}}}$

$\bold{d={\dfrac{30}{6}}}$

$\bold{d=5}$

•°• Common difference, d = 5.

Substitute d = 5 in equation 6,

2a + 21d = 111

2a + 21(5) = 111

2a + 105 = 111

2a = 111 - 105

2a = 6

$\bold{a={\dfrac{6}{2}}}$

$\bold{a=3}$

•°• First term, a = 3.

To find 13th term :

$\bold{t_{13}\:=\:a+\:(13-1)\:d}$

$\bold{t_{13}\:=\:3+\:(12)\:5}$

$\bold{t_{13}\:=\:3+\:60}$

$\bold{t_{13}\:=63}$

•°• 13th term of AP = 63.