For what n , are the nth terms of two APs: 63,65,67,.. and 3,10,17,.. equal?

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Solution

The correct option is D
13

In the first AP; a = 63 & d = 2

In the second AP; a = 3 & d = 7

As per question,

63 + 2(n – 1 )= 3 + 7( n – 1)

60 = 5 ( n-1)

n-1 = $\frac{60}{5}$

n = 13

As the result ,there will be $13^{t h}$ term with equal value for both APs.

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For what value of n, the nth terms of the arithmetic progression 63, 65, 67,... and 3, 10, 17,.. are equal?

For what value of $n,$ the $n^{t h}$ term of A.P. $63, 65, 67, . . . . . . . .$ and $n^{t h}$ term of A.P. $3, 10, 17, . . . . . . . . . .$ are equal to each other?

Q. For what value of n, are the nth terms of 2 APs 63,65,67,... and 3,10,17,... are equal?

For what n , are the nth terms of two APs: 63,65,67,.. and 3,10,17,.. equal?

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For what n , are the nth terms of two APs: 63,65,67,.. and 3,10,17,.. equal?

The correct option is D 13 In the first AP; a = 63 & d = 2 In the second AP; a = 3 & d = 7 As per question, 63 + 2(n – 1 )= 3 + 7( n – 1) 60 = 5 ( n-1) n-1 = 605 n = 13 As the result ,there will be 13th term with equal value for both APs.

The correct option is D
13

In the first AP; a = 63 & d = 2

In the second AP; a = 3 & d = 7

As per question,

63 + 2(n – 1 )= 3 + 7( n – 1)

60 = 5 ( n-1)

n-1 = $\frac{60}{5}$

n = 13

As the result ,there will be $13^{t h}$ term with equal value for both APs.