Question

The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41th term of that A.P.

Answer 1

ANSWER :–

41 th term = 55

EXPLANATION :–

GIVEN :–

• 11 th term of A.P. = 16

• 21 th term of A.P. = 29

TO FIND :–

41 th term

SOLUTION :–

● 11 th term of A.P. = a + 10d = 16 ------(1)

● 21 th term of A.P. = a + 20d = 29 -------(2)

➪ Here , a = first term

d = common difference

⮕ by equation (1) and (2) –

=> 10d = 13

=> d = 13/10

• Now put the value of 'd' in equation (1) –

=> a = 3

⮕ Now , 41 th term = a + 40d

=> 41th term = 3 + 40 (13/10)

=> 41th term = 55

Answer 2

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$\huge\tt{GIVEN:}$

• The 11th term and the 21st term of an A.P. are 16 and 29 respectively.

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$\huge\tt{TO~FIND:}$

• the 41th term of that A.P.

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$\huge\tt{SOLUTION:}$

Let a be the First Term

& Let d be the Common Differences

↪11th Term of A.P = a + 10d = 16 ___(EQ.1)

↪21th Term of A.P = a + 20d = 29 ___(EQ.2)

BY COMPARING THE EQUATIONS,

↪10D= 13

↪D = 13/10

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Putting the value of D in (EQ.1)

↪a=3

Then,

41th Term would be a + 40D

↪41th Term = 3 + 40 ( 13/10)

↪41th Term = 44(13/10)

↪41th Term = 55

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