Question

check whether 140 is a term of AP - 7 - 4 - 1 , 2​

Answer 1

Answer:

let the last term of an AP be 140

therefore an=140

an=a+(n-1)(d)

140=(-7)+(n-1)(3)

140=(-7)+3n-3

-7-3+3n=140

3n-10=140

3n=150

n=150/3

n=50

therefore we can say that 140 is from of the given arthematic progression

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Answer 2

Step-by-step explanation:

let a= -7

d= 3

An= 140

then,

An= a+ (n-1) d

140 = -7 +(n-1) 3

140 +7 =(n-1) 3

147 = (n-1) 3

147/3=n-1

49 = n-1

n= 49+1

n= 50

verification= 140= -7+(n-1)3

140 = -7+(50-1)3

140= -7 + 49×3

140 = -7 + 147

140 = 140

hence , it is verified that 140 is a term of given AP