Let S1,S2,S3…………. and t1,t2,t3…………. are two arithmetic sequences such that S1=t1≠0;S2=2t2 and ∑i=110 Si=∑i=115 ti then, the value of S2−S1t2−t1 is:

A
$\frac{8}{3}$
B
$\frac{3}{2}$
C
$\frac{19}{8}$
D
2

Solution:

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

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

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

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

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

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us