Question

plz solve it fast........ any one.. according to ur Choice..

Answer 1

heya

here is ur answer

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(a) to check whether 68 is a term of AP - 7 , 10 , 13.

let's check :-

♦ an = a+ (n-1) d

here

a = 7 ,

d = a2-a1

d = 10-7

d = 3

♦ substituting Values

➡ 68 = 7+(n-1) 3

➡ 68 = 7+3n-3

➡ 68 = 4+3n

➡ 68-4 = 3n

➡ 64 = 3n

➡ n = 64/3

➡ n = 21.33...

since ' n' cannot be in decimal expansion , so 68 is not a term of given ap.

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(b) to Check whether 302 is a term of given AP- 3 ,8 , 13

♦ here "a " = 3

♦ d = A2 - A1

♦ d = 8-3 = 5

an = 302.

➡ an = a + ( n- 1) d

➡ 302 = 3 +(n-1) 5

➡ 302 = 3 + 5n -5

➡ 302 = -2 + 5n

➡ - 2 + 5 n = 302

➡ 5n = 302 + 2

➡ n = 304/5

➡ n = 60.8.

⭐ so the given term is not in the ap

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(c) to check whether the (-150) is in the given AP- 11 , 5 , 8 , 2

here

♦ a = 11

♦ d = A2 - A1

♦ d = 5-11

♦ d = -6

➡ an = a +( n-1) d

➡ (-150) = 11 + ( n-1) -6

➡ (-150) = 11 -6n +6

➡ (-150) = 17 - 6n

➡ (-150 -17 ) = -6n

➡ ( -167) = -6n

➡ n = -167/-6

➡ n = 27.833..

⭐ value of"n" cannot be in decimal expansion.
so the given number is not an term in the given AP.

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hope it helps u..!!

thnkq