What is the 21st term in the arithmetic sequence in which a3 is 59 and a7 is 103 (enter only the term)

257

Step-by-step explanation:

Given that:

a3 = 59

a7 = 103

The nth term of an Arithmetic Progression (A. P) is given by:

an = a1 + (n-1)d

Where a1 = first term of the sequence ;

d = common difference

Therefore a3 = 59 can be represented thus:

a3 = a1 + (3-1)d = 59

a3 = a1 + 2d = 59 - - - - (1)

a7 = 103

a7 = a1 + (7-1)d = 103

a7 = a1 + 6d = 103 - - - - - (2)

Subtracting (2) from (1)

(2d - 6d) = (59 - 103)

-4d = - 44

d = 11

Substitute d= 11 into (1)

a1 + 2d =59

a1 + 2(11) = 59

a1 + 22 = 59

a1 = 59 - 22

a1 = 37

The 21st term:

a21 = a1 + (21 - 1)d

a21 = 37 + 20(11)

37 + 220 = 257

10.416 meters per second

step-by-step explanation:

distance=1 00 meters

time= 9.6 seconds in decimal form or 9 3/5 as a mixed number

average speed= 100/9.6

=10.416 meters per second or 10 52/125 as a mixed number.

if y=4x then you will substitute 4x in for y in the second equation.

8x + 4x = 18 combine like terms

12x = 18 divide by 12 on both sides

x = 18/12 or 3/2

now replace x in the first equation with 3/2 to get

y = 4 3/2

y = 6

the answer is a

(3/2,6)

hope that .

brainliest is always apprectiated