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1) IF A SERIES OF TERMS IN ARTHMETIC PROGRESSION BE COLLECTED INTO GROUPS OF N TERMS AND THE TERMS IN EACH GROUP BE ADDED TOGETHER, THEN THE RESULT FORMS AN ARTHEMETIC PROGRESSION, THEN THE RATIO OF THE COMMON DIFFERENCE OF THAT ARTHEMETIC PROGRESSION TO THE ORGINAL COMMON DIFFERENCE. (A)n 2 :1 (B)n 2 +1:1 (C)n 2 -1:1 (D) n-1:n+1
Dear dilip let first term is a and common difference is d so first term of new series = n/2[2a +(n-1)d] and second term of new series =2n/2[2a +(2n-1)d] -n/2[2a +(n-1)d] so common difference of new series d' =2n/2[2a +(2n-1)d] -n/2[2a +(n-1)d] -n/2[2a +(n-1)d] =2n/2[2a +(2n-1)d -n[2a +(n-1)d] or d'/d =n2/1 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin
Dear dilip
let first term is a and common difference is d
so first term of new series = n/2[2a +(n-1)d]
and second term of new series =2n/2[2a +(2n-1)d] -n/2[2a +(n-1)d]
so common difference of new series d' =2n/2[2a +(2n-1)d] -n/2[2a +(n-1)d] -n/2[2a +(n-1)d]
=2n/2[2a +(2n-1)d -n[2a +(n-1)d]
or d'/d =n2/1
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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