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Sum of n Terms

u1, u2, u3, ....

Question

$u_{1}, u_{2}, u_{3}, . . . .$ are the terms of an A.P. and $S_{n}$ is the sum of first $n$ terms. If $u_{m} = 4, u_{4 m} = 24$ and $S_{4 m} = 44 S_{m}$ , then find the value of $u_{1}$ and $m$ .

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Solution

$S_{m} = u_{1} + u_{2} + \dots + u_{m} = \frac{m}{2} (u_{1} + u_{m})$
Similarly, $S_{4 m} = u_{1} + u_{2} + \dots + u_{4 m} = \frac{4 m}{2} (u_{1} + u_{4 m})$
Given, $S_{4 m} = 44 S_{m}$

$\Rightarrow 2 m (u_{1} + 24) = 22 m (u_{1} + 4) \Rightarrow u_{1} + 24 = 11 u_{1} + 44 \Rightarrow u_{1} = - 2$

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