# Which is the value of n for the nth term of the series " 3+10+17+..." and " 63+65+67+... " are equal?

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The difference between terms for the first series is 7 (d1 = 7):

10 -3 = 7

17 - 10 = 7

The difference between terms for the first series is 2 (d2 = 2)

65 - 63 = 2

67 - 65 = 2

The term equation for each series is:

1: an = 3 + 7(n-1)

2: an = 63 + 2(n-1)

When they are equation an(1) = an(2):

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

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

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

5n = 65

n = 13

Therefore the 13th term in sequence 1 is equal to the 13th term in sequence 2

3,10,17....

We notice that the difference between terms is 7

==> r = 7

==> an = a1 + (n-1)*r

==> an = 3 + (n-1)*r........(1)

63, 65, 67, ....

We notice that the difference is 2

==> r= 2

==> an = a1 + (n-1)*r

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

But we know that (1) and (2) are equals:

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

==> 5(n-1) = 60

==> n-1 = 60/5 = 12

==> n = 13

We'll have to apply the formula of the general terms for both given series:

an = a1 + (n-1)*d1

bn = b1 + (n-1)*d2

We'll calculate a1 and d1 for the first sequence:

a1 = 3

10 - 3 = 17 - 10 = ...= 7

d1 = 7

an = 3 + (n-1)*7

We'll calcuate the first term b1 and the common difference d2, for the second sequence:

b1 = 63

65-63 = 67-65 = ... = 2

d2 = 2

bn = 63 + (n-1)*2

We know, from enunciation, that:

an = bn

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

We'll remove the brackets:

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

-4 + 7n = 61 + 2n

We'll move all terms to one side:

7n - 2n - 4 - 61 = 0

5n - 65 = 0

5n = 65

n = 65/5

**n = 13**

The 1st AP :

a1 =3, common difference d = obvioudly 10=3 =17-10 =7

The nth term Tn = a1+(n-1)d = 3+(n-1)5........(1)

The 2nd AP:

a1 = 63, d = 65-63 = 67-65 =2.

Therefore Tn = a1+(n-1)d = 63 +(n-1)3...................(2)

Since the nth terms of both series are equal the values at (1) and (2) are equal.

3+(n-1)5 = 63 +(n-1)2

(n-1)5 - (n-1)2 = 63-3

(n-1)(5-2) = 60

n-1 = 60/3 = 20

n = 20+1 =21