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

figure c

step-by-step explanation:

hope this .

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4