Question

Show that a1, a2, . . ., an, . . . form a_n AP where an is defined as below :
(i) a_n = 3 + 4n (ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

Answer 1

In the  cases where 

(i) a_n=3+4n

a₁=3+4×1=3+4=7

a₂=3+4×2=3+8=11

a₃=3+4×3=3+12=15

d=a₂-a₁=11-7=4

  =a₃-a₂=15-11=4

Hence a_n=3+4n forms an A.P

a₁₅=3+4×15=3+60=63

Sum of 15 terms is

S₁₅= n(a₁+a₁₅)/2=15 (7+63)/2=15×70/2=15×35=525

(ii) a_n=9-5n

a₁=9-5×1=4

a₂=9-5×2=-1

a₃=9-5×3=-6

d=a₂-a₁=-1-4=-5

  =a₃-a₂=-6+1=-5

Hence a_n=9-5n forms an A.P

a₁₅=9-5×15=-66

Sum of 15 terms is 

S₁₅=n (a₁₅-a₁)/2=15 (4-66)/2=15×(-62)/2=15×(-31)=-465