Question

The sum of the first \(10\) terms of an A.P. is \(150\). If the first term is \(10\), what is the common difference?

• A. \(1\)
• B. \(10/9\)
• C. \(2\)
• D. \(5/9\) \bigskip

Hint:

Recall that the sum of an AP equals the number of terms multiplied by the average of the first and last terms. You can either expand the standard sum formula directly, or first work out the average term as sum divided by number of terms, and then connect that average to a and d.

Answer 1

The Correct Option is B

Concept: The sum of the first \(n\) terms of an arithmetic progression is given by: \[ S_n = \frac{n}{2}\big(2a + (n-1)d\big) \] where \begin{itemize} \item \(a\) = first term \item \(d\) = common difference \item \(n\) = number of terms \end{itemize} Step 1: {\color{red}Substitute the given values.} \[ S_{10} = 150, \quad a = 10, \quad n = 10 \] \[ 150 = \frac{10}{2}\big(2(10) + (10-1)d\big) \] Step 2: {\color{red}Simplify the equation.} \[ 150 = 5(20 + 9d) \] \[ 150 = 100 + 45d \] Step 3: {\color{red}Solve for \(d\).} \[ 50 = 45d \] \[ d = \frac{50}{45} \] \[ d = \frac{10}{9} \] Thus, the common difference is: \[ d = \frac{10}{9} \] \bigskip

Answer 2

Concept:

• The sum of an AP can also be written as Sum equals (number of terms) times (average value of the terms), since the terms are evenly spaced and their average equals the average of the first and last terms.
• For an AP with first term $a$ and common difference $d$, the average value of the first $n$ terms equals $a + \dfrac{(n-1)d}{2}$.

Step 1: Find the average value of the 10 terms directly from the given sum.
Average $= \dfrac{S_{10}}{10} = \dfrac{150}{10} = 15$

Step 2: Express this average in terms of $a$ and $d$.
Average of the first $n$ terms of an AP $= a + \dfrac{(n-1)d}{2}$
For $n=10$: Average $= 10 + \dfrac{9d}{2}$

Step 3: Equate the two expressions for the average and solve for $d$.
$10 + \dfrac{9d}{2} = 15$
$\dfrac{9d}{2} = 5$
$d = \dfrac{10}{9}$

Final Answer: $d = \dfrac{10}{9}$