Question

If \(S_n\) denotes the sum of the first \(n\) terms of an Arithmetic Progression, and \(S_1 : S_4 = 1 : 10\), then the ratio of the first term to the fourth term is:

• A. 1 : 3
• B. 2 : 3
• C. 1 : 4
• D. 1 : 5

Hint:

Write S1 = a and S4 = 4a + 6d using the AP sum formula, then use the given ratio S1:S4 = 1:10 to relate a and d.

Answer 1

The Correct Option is C

Step 1: Write \(S_1\) and \(S_4\) in terms of the first term and common difference.
Let the first term be \(a\) and the common difference be \(d\).
\(S_1\) is simply the first term itself:
\[ S_1 = a \]
For \(S_4\), use the sum formula \(S_n = \frac{n}{2}\big(2a+(n-1)d\big)\) with \(n=4\):
\[ S_4 = \frac{4}{2}\big(2a+3d\big) = 2(2a+3d) = 4a+6d \]

Step 2: Use the given ratio to form an equation.
\[ \frac{S_1}{S_4} = \frac{1}{10} \]
\[ \frac{a}{4a+6d} = \frac{1}{10} \]
Cross multiplying:
\[ 10a = 4a + 6d \]

Step 3: Solve for the relation between \(a\) and \(d\).
\[ 10a - 4a = 6d \]
\[ 6a = 6d \]
\[ a = d \]
So the common difference equals the first term.

Step 4: Find the fourth term.
The fourth term of an AP is \(a + 3d\). Since \(d = a\):
\[ T_4 = a + 3a = 4a \]

Step 5: Find the required ratio.
\[ \frac{T_1}{T_4} = \frac{a}{4a} = \frac{1}{4} \]
Options (A), (B) and (D) would result from an arithmetic slip while cross multiplying, or from using the wrong sum formula for \(S_4\).

Final Answer:
The ratio of the first term to the fourth term is 1 : 4. \[ \boxed{1:4} \]