How many terms are in A.P. 7, 10, 13,…43?

Answer:

Given,

AP of 7, 10, 13,…

a₁ = a = 7

a₂ = 10

Common difference,

d = a₂ – a₁

= 10 – 7

= 3

a_n = a + (n – 1) d

a_n = 7 + (n – 1)3

= 7 + 3n – 3

= 3n + 4

Put, a_n = 43

3n + 4 = 43

3n = 43 – 4

3n = 39

n = 39/3

n = 13

Total 13 terms exists in the A.P.