Question

13th term of an AS is 48. 17th term of an AS is 64.
a) find the common difference?
b) find the 1st term ?
c) Find the 20th term ?
please write answer in a paper
if anyone tell the right answer I choose it the brainliest answer​

Answer 1

Answer:

a=0 hence the first term is 0

Step-by-step explanation:

a+12d=48

a+16d=64

4d=16

d=4

a+12d=48

a+12(4)=48

a+48=48

a=0

Answer 2

Required Answer :

• Common difference = 4

• First term of A.P. = 0

• The 20th term of A.P. = 76

Given :

• 13th term of A.P. = 48
• 17th term of A.P. = 64

To find :

• a) find the common difference?
• b) find the 1st term ?
• c) Find the 20th term ?

Solution :

$\\ \longrightarrow \sf t_{13} = a + (n - 1)d = 48$

$\\ \longrightarrow \sf t_{13} = a + (13 - 1)d = 48$

$\\ \longrightarrow \bf t_{13} = a + 12d = 48 \dots \dots \dots (1)$

$\\ \longrightarrow \sf t_{17} = a + (n - 1)d = 64$

$\\ \longrightarrow \sf t_{17} = a + (17 - 1)d = 64$

$\\ \longrightarrow \bf t_{17} = a + 16d = 64 \dots \dots \dots (2)$

Solving (1) and (2) :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a + 12d = 48

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀a + 16d = 64

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀-⠀-⠀⠀⠀⠀-

⠀⠀⠀⠀⠀⠀⠀____________________

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀- 4d = - 16

⠀⠀⠀⠀⠀⠀⠀____________________

⇒ - 4d = - 16

⇒ 4d = 16

⇒ d = 4

Answer →

• Common difference = 4

Substituting the value of d in equation (1) :

⇒ a + 12d = 48

⇒ a + 12(4) = 48

⇒ a + 48 = 48

⇒ a = 48 - 48

⇒ a = 0

Answer →

• First term of A.P. = 0

$\\ \longrightarrow \sf t_{20} = a + (n - 1)d$

$\\ \longrightarrow \sf t_{20} = 0 + (20 - 1)4$

$\\ \longrightarrow \sf t_{20} = 0 + 19 \times 4$

$\\ \longrightarrow \sf t_{20} = 19 \times 4$

$\\ \longrightarrow \bf t_{20} = 76$

Answer →

• The 20th term of A.P. = 76