Question

If nth term of the AP 3,10,17.... is same as the nth term of the AP 63,65,67.... find n.​

Answer 1

n = 13

Explanation:

`a_(n)` =
`a_(1)` + (n - 1)d

`a_(n)` = 3 + 7(n - 1) (for the first AP)

`a_(n)` = 63 + 2(n - 1) ( for the second one)

3 + 7(n - 1) = 63 + 2(n - 1)

3 + 7n - 7 = 63 + 2n - 2

5n = 65

n = 13

Answer 2

n = 13

Explanation:

AP = 2, 10, 17 ...

AP1 = 63,65, 67

Find the general term of AP

L = a + (n - 1)*d

a = 2

d = 7

Find the general term of AP1

L = 63 + (n - 1)*d

a = 63

d = 2

Equation to equate both of them

L = 3 + (n-1)7

L1 = 63 + (n - 1)*2

Equate L and L1

2 + (n - 1)*7= 63 + (n-1)*2 Subtract 3 from both sides

(n - 1)*7 = 63 - 3 + (n - 1)*2 Remove the brackets

7n - 7 = 60 + 2n - 2 Combine the like terms on the right

7n - 7 = 58 + 2n Subtract 2n from both sides

7n - 2n -7 = 58 Add 7 to both sides

5n = 58 +7 Divide by 5

n = 65/5

n = 13

So the 13th term on each series is equal.

L_13 = 3 + (13 - 1)*7

L_13 = 3 + (12)*7

L_13 = 87

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L'_13 = 63 + (13 -1 ) * 2

L'_13 = 63 + 12*2

L'_13 = 87