Question

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Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose 10th term is 77.

Answer 1

Answer:

Answer is 47

Step-by-step explanation:

given :- 3rd term is 35

10th term is 77

by the tn formula

t3=a+(n-1) d

35=a+2d --------1

t10=a+(n-1) d

77=a+9d --------2

subtracting 2 and 1

77=a+9d

35=a+2d

__________

42=7d

d=42/7

d=6

and

a=23

therefore the 5th term is

t5= 23+(5-1)6

=23+24

=47

hence the 5th term of arithmetic sequence is 47

Answer 2

Given: The 3rd term of an arithmetic sequence is 35 and 10th term of the arithmetic sequence is 77.

To find: 5th term of the arithmetic sequence

Explanation: Let the first term of this arithmetic progression be a and its common difference be d.

The nth term of an A.P can be given by:

a +(n-1) d

Therefore, 3rd term can be written as:

a + d(3-1) = 35

=> a+2d = 35 equation(i)

10th term can be written as:

a+ d(10-1)

=> a +9d= 77 equation(ii)

Subtracting equation(i) from equation(ii),

a+9d - (a+2d) = 77-35

=> 7d = 42

=> d = 42/7

=>d = 6

Putting d=6 in equation (i):

a + 2*6 = 35

=> a = 35-12

=>a = 23

Therefore, a = 23 and d =6

5th term= 23 + 6×(5-1)

= 23 + 6×4

= 23+24

= 47

Therefore, the 5th term of the arithmetic sequence is 47.