How to find the common difference of AP when the first and last term are given

Question:

The first and last terms of an AP are 6.7 and 17.1 respectively. If there are 14 terms in the sequence, find its common difference.

Solution

First term of the sequence a = 6.7

Last term of the sequence Tn = 17.1

Since the sequence has 14 terms, it means that the 14th term is the last term

Therefore:

Tn = T14 = 17.1

Recall the nth term formula

Tn = a + (n-1)d

Substitute Tn for T14, a for 6.7 and n for 14 to get the formula below

T14 = 6.7 + (14-1)d

T14 = 6.7 + 13d

Substitute T14 for 17.1

T14 = 6.7 + 13d

17.1 = 6.7 + 13d

Subtract 6.7 from both sides to get

17.1 - 6.7 = 6.7-6.7 +13d

10.4 = 13d

Divide through by 13

10.4/13 = 13d/13

d = 104/130

The common difference is 104/130

CHECK

Substitute 104/130 for d in T14 = 6.7 + 13d

T14 = 6.7 + 13*104/130

= (6.7*130 + 13*104)/130

= (871 + 1352)/130

= 2223/130

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