# Solution to #1-7

Paper 2 Difficulty: Medium

The sum of the first 6 terms of a sequence is 39.75. Also, the 7th term is $\frac{6}{5}$ times the 2nd term. Find the first term and the common difference of the sequence.

[Maximum mark: 4]

Since the sum of the first six terms is 39.75, $6 \cdot u_{1} + \left (1 + 2 + 3 + 4 + 5 \right ) \cdot d = 39.75$ $u_{1} + \frac{15}{6} \cdot d = \frac{39.75}{6}$ (a)

We also know that the 7th term is $\frac{6}{5}$ times the 2nd term.

Therefore: $\left ( u_{1} + 6d \right ) = \frac{6}{5} \left ( u_{1} + d \right ) \rightarrow u_{1}-24d=0$ (b)

Solving (a) and (b) simultaneously gives $u_{1} = 6$ and $d = \frac{1}{4}$.