Question

The 3 rd and 6 th term of an arithmetic progression are 19 and 37 respectively.What is the 13 th term?

Answer 1

Answer:

HERE IS UR ANSWER:

Explanation:

a3= 19

a6= 37

an= a+(n-1)d

so,

19= a+2d    ---1

37=a+5d     ---2

So, subtract 1 from 2

(a gets cancelled out by elimination)

So,

3d=18

d=6

Putting d in eq.1

a+2(6)=19

a=7

Now finding a13:

a13= a+12d

a13=7+12(6)

therefore,

a13=79

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