Fewpal
+2 votes
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in Mathematics by (127k points)

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

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1 Answer

+1 vote
by (519 points)

In the  cases where 

(i) an=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 an=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) an=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 an=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

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