Question

Find the ap whose third term is 16 and 7th term exceeds 5th term by 12

Answer 1

this is the correct answer.

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Third term of an Ap = 16

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• 7th term exceeds the 5th term by 12

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$\sf{\bold{\green{\underline{\underline{To\: Find}}}}}$

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• Ap = ???

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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$\sf{\red{\boxed{\bold{a_n = a + ( n - 1 ) d}}}}$

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Acc. to the 1st statement :-

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a3 = a + ( 3 - 1 ) d

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a3 = a + 2d ---- ( i )

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Acc. to the 2nd statement :-

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a7 = a5 + 12

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a + ( 7 - 1 ) d = a ( 5 - 1 ) d + 12

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a + 6d = a + 4d + 12

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a - a + 6d - 4d = 12

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2d = 12

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d = 12/2

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d = 6

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• Putting the value of d in Eq (i)

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a + 2d = 16

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a + 2× 6 = 16

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a + 12 = 16

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a = 16 - 12

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a = 4

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Therefore ;

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A.P. = a , a + d , a + 2d , a + 3d ,.......

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A.P. = 4 , 4 + 6 , 4 + 2×6 , 4 + 3×6 ......

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A.P. = 4 , 10 , 16 , 22.......

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Required AP = 4 , 10 , 16 , 22 ............n