First slide

Arithmetic progression

Question

$Let S_{1}, S_{2}, S_{3} \dots and t_{1}, t_{2}, t_{3} \dots are two Arithmatic Sequence such that$ $S_{1} = t_{1} \neq 0, S_{2} = 2 t_{2} and \sum_{i = 1}^{10} S_{i} = \sum_{i = 1}^{15} t_{i} then \frac{S_{2} - S_{1}}{t_{2} - t_{1}} = \frac{d_{1}}{d_{2}} then d_{1} - d_{2} =$

Easy

A

10
B

11
C

9
D

13

Solution

$Given S_{1} + S_{2} + \dots + S_{10} = t_{1} + t_{2} + \dots + t_{15}$
$Let first sequence a_{1}, a_{1} + d_{1}, a_{1} + 2 d_{1}, \dots and second sequence is a_{1}, a_{1} + d_{2}, a_{2} + 2 d_{2}, \dots {(sinc s}_{1} {=t}_{1})$
$\begin{array}{l} Given S_{2} = 2 t_{2} \\ a_{1} + d_{1} = 2 (a_{1} + d_{2}) \\ a_{1} = d_{1} - 2 d_{2} - - - 1 \\ We have to find \\ \Rightarrow \frac{S_{2} - S_{1}}{t_{2} - t_{1}} = \frac{d_{1}}{d_{2}} \\ Now \frac{10}{2} [2 a_{1} + 9 d_{1}] = \frac{15}{2} [2 {at}_{1} + 14 d_{2}] \\ \Rightarrow a_{1} = 9 d_{1} - 21 d_{2} - \dots - 2 \\ For 1 and 2 \Rightarrow d_{1} - 2 d_{2} = 9 d_{1} - 21 d_{2} \\ 8 d_{1} = 19 d_{2} \end{array}$
$\frac{d_{1}}{d_{2}} = \frac{19}{8}$

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