Question

Find the 20th term of an AP whose 3rd term is 7 and the seventh term exceeds three times the 3rd term by 2. Also find its nth term.
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Answer 1

Answer:

an=4n-5

Step-by-step explanation:

3rd term =a3

a3=7

7=(a+(3-1)d)

7=a+2d---------(1)

a7 exceeds three times the third term by 2

(7)3+2

=23

a7=23

23=a+(7-1)d

23=a+6d--------(2)

(2)-(1)

16=4d

d=4

Substituting in (1),

7=a+2(4)

a= -1

an= (-1+(n-1)4)

an= -1+4n-4

an= 4n-5

Answer 2

since nth term of AP

= a+ (n-1)d

third term= a+2d= 7.........(1). (given)

7nth term= a+ 6d= 3/2× third term(. also given)

= 3/2×7

a+ 6d = 21/2 .......(2)

from both eqns

d= 4/7

from (1)

a+ 2×4/7= 7

a= 41/7

therefore 20th term

= a+19d= 41/7+19×4/7

= 117 / 7