The sum of 3 terms of an a.p. is 21 and the product of the 1st and 3rd term exceeds the second term by 6
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let the three terms in a.p be (a-d),a,(a+d)
then,
(a-d)+a+(a+d)=21
3a=21
a=7
also acc. to question
(a-d)(a+d)=a+6
a^2-d^2=a+6.....................(1)
substituting a=7 in (1) we get
49-d^2=7+6
d^2=36
d=6
hence the three no.s are 1 ,7, 13.
then,
(a-d)+a+(a+d)=21
3a=21
a=7
also acc. to question
(a-d)(a+d)=a+6
a^2-d^2=a+6.....................(1)
substituting a=7 in (1) we get
49-d^2=7+6
d^2=36
d=6
hence the three no.s are 1 ,7, 13.
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