Question

the sum of the first 4 terms of an ap is 360. the fourth term of ap is double the first term. find all 4 terms​

Answer 1

Answer:

answer of these question is as given below

Step-by-step explanation:

sn= n/2 (2a + (n-1)d)

360 = 2(2a+ 3d)

180 = 2a + 3d .........(1)

2a = forth term of ap

2a= a+ 3d

a = 3d

putting a = 3d in ....(1)

180 = 9d

d = 20

fourth term of ap = a+ 3d

= 6d

= 120

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Answer 2

Answer:

let a & d be the first term & the common difference respectively

ATQ

$t_{4}=2t_{1}$

$a+(n-1)d=2[a+(n-1)d]\\\\a+(4-1)d = 2a+2d(1-1)\\\\a+3d=2a+2d(0)\\\\a+3d=2a\\\\-a=-3d\\\\a=3d$

$S_{4} = 360$

$\frac{n}{2}(a+(n-1)d)=360\\\\\frac{4}{2}(3d+(4-1)d)=360\\\\2(3d+3d)=360\\\\6d=180\\\\d=\frac{180}{6}\\\\d=30$

putting the value of d in the value of a so obtained in terms of d

a = 3d = 3(30) = 90

thus the AP is in the form :-

a , a + d , a +2d , a+3d ............

90 , 90 + 30 , 90 + 2(30) , 90 + 3(30) .............

90 , 120 , 150 , 180 ...........