The 10Th Term Of An A.P Is 41 And The 18Th Term Is 73 Find The A.P.
Answers
Step-by-step explanation:
10th term=a+9d=41...........(1)
18th term=a+17d=73.............(2)
equation (2)-(1),
8d=32
d=4
substitute the value of d in (1)
a+9×4=41
a+36=41
a=5
the required AP is 5, 5+4, 5+2×4, 5+3×4
ie, 5, 9, 13, 17
hope it helps....
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5 ,9 ,13 .....................
Step-by-step explanation:
Given:
10th term of A.P = 41
18th term of A.P = 73
Computation:
A10 = a + 9d
41 = a + 9d.........................Eq1
A18 = a +17d
73 = a + 17d.........................Eq2
From Eq2 - Eq1
32 = 8d
32/8 = d
4 = d
on putting the value of d in Eq1
41 = a + 9(4)
41 = a + 36
41 - 36 = a
5 = a
A.P = a , a + d , a + 2d , a + 3d..........................
= 5 , 5+4 , 5+2(4) ...................
= 5 ,9 ,13 .....................
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