determine the AP who's third term is 16 and the seventh term is exceedsthe fifth term by 12
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a₃ =16
⇒a +(3-1)d =16
⇒a+2d =16............[1]
a₇=a₅+12
⇒a +(7-1) d =a +(5-1) d +12
⇒a +6d =a +4d+12 [in both side a will cancel]
6d-4d =12
2d=12
d=6
put d value in eq[1]
a+2x6 =16
a +12 =16
a=16-12
a=4
so,AP =4 ,10,16,22......
⇒a +(3-1)d =16
⇒a+2d =16............[1]
a₇=a₅+12
⇒a +(7-1) d =a +(5-1) d +12
⇒a +6d =a +4d+12 [in both side a will cancel]
6d-4d =12
2d=12
d=6
put d value in eq[1]
a+2x6 =16
a +12 =16
a=16-12
a=4
so,AP =4 ,10,16,22......
CUTEBARBIE:
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