if the 3rd and 7th term of an ap is 18 and 30 respectively find the series
Answers
Answered by
0
Answer:
12, 15, 18, 21,..........
Step-by-step explanation:
3rd term = a + 2d = 18
7th term = a + 6d = 30
subtract 3rd term from 7th term
a + 6d = 30
- a - 2d = -18
we get
4d = 12
d = 12/4
d = 3
substitute d = 3 in 3rd term
a + 2d = 18
a + 2(3) = 18
a + 6 = 18
a = 18 - 6
a = 12
the series will be
a, a+ d, a + 2d, a + 3d, .........
12, 12 + 3, 12 + 2(3), 12 + 3(3) ...........
12, 15, 18, 21 .............
Hope you get your answer
Answered by
0
Answer:
12,15,,18,and soon
Step-by-step explanation:
in this
The Starting value is 12 and the difference is 3
we will take in 2 equations first is
IN it the first equation is
nth is 3
a+(3-1)d=18 -(1)
a+(7-1)d=30-(2)
add the eq 1+2
it gives 4d=12
d=3
sub in eq-1
a+2*3=18
a=18-6
a=12
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