Question

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

Answer 1

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◆We know that

◆An=a+(n-1)d

◆From 1st series

==> a=63

◆d=65-63

◆d=2

◆An=63+(n-1)2------(1)

√√ From 2nd series

◆a=3

◆d=10-3=7

◆An=3+(n-1)7-------(2)

====> According to statement equation (1) is equal to (2)

◆63+(n-1)2=3+(n-1)7

◆63+2n-2=3+7n-7

◆61+2n=-4+7n

◆61+4=7n-2n

◆65=5n

◆n=65/5

◆n=13

====>so the value of n is 13,so the AP will be equal of both

Answer 2

(1)

Given AP is 63,65,67..

Let a be the first term and d be the common difference.

a = 63, d = 65 - 63 = 2

We know that an = a + (n - 1) * d

= > 63 + (n - 1) * (2)

= > 63 + 2n - 2

= > 61 + 2n    ----- (1)

(2)

Given AP is 3,10,17...

here, a = 3, d = 10 - 3 = 7.

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

= > 3 + (n - 1) * 7

= > 3 + 7n - 7

= > 7n - 4     ------ (2)

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Given that (1) = (2)

= > 61 + 2n = 7n - 4

= > 61 + 4 = 7n - 2n

= > 65 = 5n

= > n = 13.

Therefore, the value of n = 13.

Hope this helps!