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Grade: 10
        
There are n arithmetiv means between 11 and 53 such that each of them is an integer. How many distinct arithmetic progressions are possible from the above data ?
one month ago

Answers : (4)

Aditya Gupta
1072 Points
							
there must be 7 such aps corresponding to common difference d= 1, 2, 3, 6, 7, 14, 21, which all happen to be factors of 42, as 42= 53 – 11= (n+1)*d
one month ago
SAMBHAV MISHRA
34 Points
							
Answer is 14 bro. .. Try again !
Explain again ? as soon as possible
one month ago
Aditya Gupta
1072 Points
							
my answer is correct. ab tumme hi buddhi kam h to bhai main kya karu. 
11, a1, a2, …, an, 53
let common diff be d.
then 11+d= a1
an+d= 53
and an= a1+ (n – 1)d
or an + d= a1+ nd
or 53 = a1+ nd
or 53= 11+d+ nd
42= d(n+1)
d= 42/(n+1)
ab maharaj d ek integer h or 42 ko divide karta h, to d sirf 1, 2, 3, 6, 7, 14, 21, 42 hi ho sakta h kuki iske alawa 42 k koi factors hi nahi h. but obviously d cant be 42 as n cant be zero. hence there will be ONLY 7 POSSIBLE VALUES.
kuch bhi comment karne or disapprove karne se pehle thodi buddhi ka bhi prayog kar liya karo agar ho tumhare pas. or agar ab bhi tumhe mera answer gakat lagta ho to reason k sath prove krdo or main yahan p tutoring karna chhor dunga.
one month ago
SAMBHAV MISHRA
34 Points
							
first of all mind your language.... and second answer is 14 only.. here is the explanation
n arithmetic means are in between a and b; d = b-a/n+1
evaluate for how many values of (n+1), d is an integer. 
@aditya gupta tumhari budhi ghass khane gayi hai !
one month ago
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