Question

In an A.P., the first term is 4 and the last term is 31. If sum of all the terms is 175, find the number of terms and the common difference.

Hint:

Using the sum formula \( S_n = \frac{n}{2}(a+l) \) is much faster than using \( S_n = \frac{n}{2}[2a + (n-1)d] \) when the last term is explicitly given.
It allows you to find \( n \) directly without dealing with quadratic equations or simultaneous equations.

Answer 1

Step 1: Note the given information.
First term $a = 4$, last term $l = 31$, sum of all terms $S_n = 175$. We need the number of terms $n$ and the common difference $d$.

Step 2: Find n using the sum as the number of terms times the average term.
The sum of an AP always equals the number of terms multiplied by the average of the first and last term:
\[ S_n = n \times \left(\frac{a + l}{2}\right) \]
The average of the first and last term here is:
\[ \frac{a + l}{2} = \frac{4 + 31}{2} = \frac{35}{2} \]
So:
\[ 175 = n \times \frac{35}{2} \]
\[ n = \frac{175 \times 2}{35} = \frac{350}{35} = 10 \]

Step 3: Find d using the full expanded sum formula instead of the last term formula.
The sum of the first $n$ terms can also be written by expanding $S_n = \frac{n}{2}[2a + (n-1)d]$ into:
\[ S_n = na + \frac{n(n-1)}{2}d \]
Substitute $n = 10$, $a = 4$, $S_n = 175$:
\[ 175 = 10(4) + \frac{10(9)}{2}d \]
\[ 175 = 40 + 45d \]

Step 4: Solve this equation for d.
\[ 175 - 40 = 45d \]
\[ 135 = 45d \]
\[ d = \frac{135}{45} = 3 \]

Final Answer:
The number of terms is $n = 10$ and the common difference is
\[ \boxed{d = 3} \]