If the nth term of a progression is (4n-10) show that it is an AP. Find its (i) first term, (ii) common difference, and (iii) 16th term.

Answer

To show that the sequence defined by the nth term \( a_n = 4n - 10 \) is an arithmetic progression (AP), we need to demonstrate that the difference between consecutive terms is constant. Let's go through the steps to solve the problem.

### Step 1: Find the first term
The first term of the progression can be found by substituting \( n = 1 \) into the nth term formula.

\[ a_1 = 4(1) - 10 = 4 - 10 = -6 \]

### Step 2: Find the second term
Next, we find the second term by substituting \( n = 2 \).

\[ a_2 = 4(2) - 10 = 8 - 10 = -2 \]

### Step 3: Find the third term
Now, we find the third term by substituting \( n = 3 \).

\[ a_3 = 4(3) - 10 = 12 - 10 = 2 \]

### Step 4: Check the common difference
To confirm that this is an AP, we need to calculate the common difference \( d \) between the consecutive terms.

\[ d = a_2 - a_1 = -2 - (-6) = -2 + 6 = 4 \]

\[ d = a_3 - a_2 = 2 - (-2) = 2 + 2 = 4 \]

Since the common difference \( d \) is the same between all consecutive terms, we can conclude that the sequence is indeed an AP.

### Step 5: Find the 16th term
To find the 16th term, substitute \( n = 16 \) into the nth term formula.

\[ a_{16} = 4(16) - 10 = 64 - 10 = 54 \]

### Summary of Results
(i) First term \( a_1 = -6 \)
(ii) Common difference \( d = 4 \)
(iii) 16th term \( a_{16} = 54 \)

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