Maths / Arithmetic Progression / Sum of First n Terms

QUESTION

Find the sum of the given series

$\dpi{120} \fn_jvn \large \frac{a-b}{a+b}, \frac{3a-2b}{a+b}, \frac{5a-3b}{a+b}, ...... 11\; terms$

 OPTIONS A. $\dpi{120} \fn_jvn \large 11(\frac{11a-5b }{a+b})$ B. $\dpi{120} \fn_jvn \large 11(\frac{11a+6b }{a+b})$ C. $\dpi{120} \fn_jvn \large 11(\frac{11a-6b }{a+b})$ D. $\dpi{120} \fn_jvn \large 11(\frac{a-6b }{a+b})$
Right Option : C

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large \frac{a-b}{a+b}, \frac{3a-2b}{a+b}, \frac{5a-3b}{a+b}, ...... 11\; terms$

$\dpi{120} \fn_jvn \large a= \frac{a-b}{a+b}$

$\dpi{120} \fn_jvn \large d_1= \frac{3a-2b}{a+b}- \frac{a-b}{a+b}= \frac{3a-2b-a+b}{a+b}= \frac{2a-b}{a+b}$

$\dpi{120} \fn_jvn \large d_2= \frac{5a-3b}{a+b}-\frac{3a-2b}{a+b}= \frac{5a-3b-3a+2b}{a+b}= \frac{2a-b}{a+b}$

$\dpi{120} \fn_jvn \large d= \frac{2a-b}{a+b}$

$\dpi{120} \fn_jvn \large s_n= \frac{n}{2}(2a+(n-1)d)$

$\dpi{120} \fn_jvn \large s_{11}= \frac{11}{2}(2\frac{a-b}{a+b}+(11-1)\frac{2a-b}{a+b})$

$\dpi{120} \fn_jvn \large s_{11}= \frac{11}{2}(\frac{2a-2b}{a+b}+\frac{20a-10b}{a+b})$

$\dpi{120} \fn_jvn \large s_{11}= \frac{11}{2}(\frac{2a-2b +20a-10b}{a+b})$

$\dpi{120} \fn_jvn \large s_{11}= \frac{11}{2}(\frac{22a-12b }{a+b})$

$\dpi{120} \fn_jvn \large s_{11}=11(\frac{11a-6b }{a+b})$

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