Please solve this question
and Also give a concept how to solve this type of questions
Answers
Step-by-step explanation:
Solution:-
See the above attachment for understanding the solution clearly
(1 3/5)^2+(2 1/5)^2 + (3 1/5)^2 +4^2+(4 4/5)^2 +...
It is in the mixed fraction we have to convert it into improper fraction
=>(8/5)^2 +(12/5)^2+(16/5)^2+4^2+(24/5)^2+...
=>(8/5)^2 +(12/5)^2+(16/5)^2+(20/5)^2+(24/5)^2+...
=>(8^2+12^2+16^2+20^2+24^2+...)/5^2
=>(8^2+12^2+16^2+20^2+24^2+...)/25
Take Numerator
(8^2+12^2+16^2+20^2+24^2+...)
The series
8+12+16+20+24+...
First term 8
Common difference = 12-8=4
General term in AP = a+(n-1)d
=>8+(n-1)4
=>8+4n-4
=>4n+4
=>4(n+1)
Tn =[ {(4n+1)}^2]/25
Now it is given that
the sum of first ten terms = 16 m/5
=> Sum of [{(4n+1)}^2 ]/25 upto 10 terms
=>Sum of 16(n+1)^2 /25 upto 10 terms
=>(16/25) Sum of (n^2+2n+1) upto 10 terms
=>(16/25)[{(n(n+1)(2n+1)/6}+{2n(n+1)/2}+10] upto 10 terms
=>(16/25)[(10×11×21/6)+(2×10×11/2)+10]
=>(16/25)[(385+110+10)]
=>(16/25)(505)
=>(16/5)(101)
It is given that
(16/5)(101) = 16 m/5
On comparing both sides then
m=101
Answer:-
The value of m for the given problem is 101
Used Concept:-
- General term in AP = a+(n-1)d
- Sum of first n natural numbers = n(n+1)/2
- Sum of squares of the first n natural numbers = n(n+1)(2n+1)/6
1.In this type of series ,The terms in the question are in the form of AP then find the general term of the AP and solve it with the given Condition then.
Solution:-
See the above attachment for understanding the solution clearly
(1 3/5)^2+(2 1/5)^2 + (3 1/5)^2 +4^2+(4 4/5)^2 +...
It is in the mixed fraction we have to convert it into improper fraction
=>(8/5)^2 +(12/5)^2+(16/5)^2+4^2+(24/5)^2+...
=>(8/5)^2 +(12/5)^2+(16/5)^2+(20/5)^2+(24/5)^2+...
=>(8^2+12^2+16^2+20^2+24^2+...)/5^2
=>(8^2+12^2+16^2+20^2+24^2+...)/25
Take Numerator
(8^2+12^2+16^2+20^2+24^2+...)
The series
8+12+16+20+24+...
First term 8
Common difference = 12-8=4
General term in AP = a+(n-1)d
=>8+(n-1)4
=>8+4n-4
=>4n+4
=>4(n+1)
Tn =[ {(4n+1)}^2]/25
Now it is given that
the sum of first ten terms = 16 m/5
=> Sum of [{(4n+1)}^2 ]/25 upto 10 terms
=>Sum of 16(n+1)^2 /25 upto 10 terms
=>(16/25) Sum of (n^2+2n+1) upto 10 terms
=>(16/25)[{(n(n+1)(2n+1)/6}+{2n(n+1)/2}+10] upto 10 terms
=>(16/25)[(10×11×21/6)+(2×10×11/2)+10]
=>(16/25)[(385+110+10)]
=>(16/25)(505)
=>(16/5)(101)
It is given that
(16/5)(101) = 16 m/5
On comparing both sides then
m=101
Answer:-
The value of m for the given problem is 101
Used Concept:-
General term in AP = a+(n-1)d
Sum of first n natural numbers = n(n+1)/2
Sum of squares of the first n natural numbers = n(n+1)(2n+1)/6
1.In this type of series ,The terms in the question are in the form of AP then find the general term of the AP and solve it with the given Condition then.
Step-by-step explanation: