$Let S_{1}, S_{2}, S_{3} \dots \dots \dots \dots . and t_{1}, t_{2}, t_{3} \dots \dots \dots \dots . are two arithmetic sequences such that$ $S_{1} = t_{1} \neq 0; S_{2} = 2 t_{2} and \sum_{i = 1}^{10} S_{i} = \sum_{i = 1}^{15} t_{i} then, the value of \frac{S_{2} - S_{1}}{t_{2} - t_{1}} is:$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{8}{3}$

$\frac{3}{2}$

$\frac{19}{8}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Let S_{1,} S_{2}, S_{3} \dots \dots \dots \dots . are in A.P. with common difference d and t_{1}, t_{2}, t_{3} are$

$in A.P with common difference d_{2} .$

$\begin{array}{l} \sum_{i = 1}^{10} S_{i} = \frac{10}{2} [2 S_{1} + (10 - 1) d_{1}] \\ \sum_{i = 1}^{15} t_{i} = \frac{15}{2} [2 t_{1} + (15 - 1) d_{2}] \\ 10 (2 S_{1} + 9 d_{1}) = 15 (2 t_{1} + 14 d_{2}) \\ 2 (2 S_{1} + 9 d_{1}) = 3 (2 t_{1} + 14 d_{2}) \\ 2 S_{1} + 9 d_{1} = 3 (t_{1} + 7 d_{2}) \\ 2 S_{1} + 9 d_{1} = 3 t_{1} + 21 d_{2} \\ As, S_{1} = t_{1} \\ 2 t_{1} + 9 d_{1} = 3 t_{1} + 21 d_{2} \\ 9 d_{1} = t_{1} + 21 d_{2} \\ d_{1} = S_{2} - S_{1} = 2 t_{2} - t_{1} \\ d_{2} = t_{2} - t_{1} \end{array}$

$\begin{array}{l} 9 (2 t_{2} - t_{1}) = t_{1} + 21 t_{2} - 21 t_{1} \\ 18 t_{2} - 9 t_{1} = 21 t_{2} - 20 t_{1} \\ 11 t_{1} = 3 t_{2} \\ Thus, \frac{S_{2} - S_{1}}{t_{2} - t_{1}} = \frac{2 t_{2} - t_{1}}{t_{2} - t_{1}} = \frac{2 \cdot \frac{11}{3} - 1}{\frac{11}{3} - 1} = \frac{22 - 3}{8} = \frac{19}{8} \end{array}$