find the tenth term and last term of the sequence of positive integer in A.P. whose terms are less than 400 and divisible by 8
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a1=8,a2=8+8=16...and so on..till An= 392(as it's given the last term is less than 400)
d=a2-a1=16-8=8
an= a+(n-1)d=8+n-1×8=392
n-1=392-8÷ 8= 48
n=49(there are totally 49 terms in the A.P)
a10=a+9d= 8+9×8=8+72=80.
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