Class 10 Sequence and Series Exercise 6.2 Solutions | Maths Links Readmore Publishers & Distributors

Given:

• The sum of the first 10 terms of an arithmetic series is \( \rm S_{10} = 27.5 \)
• The 10th term of the series is \( \rm a_{10} = 5 \)

We need to determine the value of its first term \( \rm a \) and common difference \( \rm d \).

We will use the formulas for the sum of the first \( \rm n \) terms of an arithmetic series:

\( \rm S_n = \frac{n}{2}(2a + (n - 1)d) \)

Given \( \rm S_{10} = 27.5 \) and \( \rm a_{10} = 5 \), we'll first find the value of \( \rm a \) and \( \rm d \).

From the given information:

\( \rm S_{10} = 27.5 \)

\( \rm a_{10} = 5 \)

We know:

\( \rm S_{10} = \frac{10}{2}(2a + (10 - 1)d) \)

\( \rm 27.5 = 5(2a + 9d) \)

\( \rm 5.5 = 2a + 9d \)

Also, we know:

\( \rm a_{10} = a + (10 - 1)d = a + 9d = 5 \)

Now, we have a system of two equations with two variables:

\( \rm 5.5 = 2a + 9d \) ... \( \rm (i) \)

\( \rm a + 9d = 5 \) ... \( \rm (ii) \)

Solve these equations simultaneously to find the values of \( \rm a \) and \( \rm d \).

Substitute \( \rm a = 5 - 9d \) from equation \( \rm (ii) \) into equation \( \rm (i) \):

\( \rm 5.5 = 2(5 - 9d) + 9d \)

\( \rm 5.5 = 10 - 18d + 9d \)

\( \rm 5.5 = 10 - 9d \)

\( \rm 9d = 10 - 5.5 \)

\( \rm 9d = 4.5 \)

\( \rm d = \frac{4.5}{9} \)

\( \rm d = 0.5 \)

Substitute \( \rm d = 0.5 \) back into equation \( \rm (ii) \) to find \( \rm a \):

\( \rm a + 9(0.5) = 5 \)

\( \rm a + 4.5 = 5 \)

\( \rm a = 5 - 4.5 \)

\( \rm a = 0.5 \)

So, the first term \( \rm a \) of the arithmetic series is 0.5, and the common difference \( \rm d \) is 0.5 as well.