Class 11 sequence and series.
Answers
Answer:
11.
Step-by-step explanation:
A.P. : 5, A1, A2, A3, ..... , Ap, 41.
Here,
First term = 5
Last term = 41
No. of terms inserted = p
So,
Total no. of terms (including 5 and 41) = p + 2.
Common difference = (Last term-First term)/(No. of terms - 1 )
= (41-5)/ [(p+2)-1] = 36/(p+1).
A3 = First term + 3 × (common difference)
= 5 + 3 × [36/(p+1)] = 5 + [108/(p+1)] = [5(p+1) + 108] / (p+1) = [5p + 5 + 108] / (p+1) = [5p + 113] / (p+1) .
Ap-1 = First term + (p-1) × (common difference)
= 5 + (p-1) × [36/(p+1)] = 5 + [(36p-36)/(p+1)] = [5(p+1) + 36p - 36] / (p+1) = [5p + 5 + 36p - 36] / (p+1) = [41p - 31] / (p+1).
Then,
A3/Ap-1 == { [5p + 113] / (p+1) } ÷ { [41p - 31] / (p+1) }
== [5p + 113] / [41p - 31] .
Since, A3/Ap-1 == 2/5,
2/5 = [5p + 113] / [41p - 31]
Cross multiplying,
25p + 565 = 82p - 62
627 = 57p
So,
p = 627/57 = 11.
•°• p = 11.
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You're indeed right... This class 11 is really sooo harddd... I wonder, how you people are able to manage these JEE coachings as well as Class 11 syllabus at the same time !!
And, how were you able to understand that it was written in Russian !!!!! Copy paste option wasn't available too !!! How were you able to figure out what was written !!!! I was like OMG ... when I saw your answer !
And you were able to answer that coin question too !!! O my...!! I'm really so impressed !!
Bro, you're really a Pro !!!!!!
:-D
Step-by-step explanation:
answer is 11 , why u judge hmm..