NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.4
Question 8

Question. 8

Solve the equation \(-4+(-1)+2+\cdots+x=437\).

Answer:

\(x=50\)

Detailed Answer with Explanation:

Step 1: Identify the sequence.

The numbers are: \(-4, -1, 2, ... , x\).

This is an arithmetic progression (AP), because the difference between terms is constant.

Step 2: Find the first term (a) and common difference (d).

First term: \(a = -4\).

Second term: \(-1\).

Common difference: \(d = -1 - (-4) = 3\).

Step 3: General formula for the nth term.

In an AP, the nth term is:

\(T_n = a + (n-1) \cdot d\).

Here, the last term is \(x\). So:

\(x = -4 + (n-1) \cdot 3\).

Step 4: Rearrange to find n.

\(x + 4 = (n-1) \cdot 3\).

\(n - 1 = \dfrac{x+4}{3}\).

\(n = \dfrac{x+7}{3}\).

Step 5: Use the sum formula for an AP.

The sum of n terms is:

\(S_n = \dfrac{n}{2}(a + \text{last term})\).

Here, \(S_n = 437, a = -4, \text{last term} = x\).

So: \(437 = \dfrac{n}{2}(-4 + x)\).

Step 6: Substitute n.

\(437 = \dfrac{1}{2} \cdot \dfrac{x+7}{3} (x-4)\).

Simplify:

\(437 = \dfrac{(x+7)(x-4)}{6}\).

Step 7: Eliminate the fraction.

Multiply both sides by 6:

\(437 \times 6 = (x+7)(x-4)\).

\(2622 = x^2 + 3x - 28\).

\(x^2 + 3x - 2650 = 0\).

Step 8: Solve the quadratic equation.

Equation: \(x^2 + 3x - 2650 = 0\).

Using factorization (or quadratic formula), we get:

\(x = 50\) or \(x = -53\).

Step 9: Choose the valid answer.

The sequence starts from \(-4\) and increases by 3 each time. So a negative value like \(-53\) cannot appear in this sequence.

Therefore, the valid solution is:

\(x = 50\)

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.4 | Detailed Answers