If 7 + 10 + 13 + …… + x = 710, then value of x is
Answers
Step-by-step explanation:
Given :-
7+10+13+...+X = 710
To find:-
Find the value of X ?
Solution:-
Given that
7+10+13+...+X = 710
First term (a) = 7
Common difference (d) = 10-7 = 3
=>13-10 = 3
Since the common difference is same throughout the series,
7,10,13,...,X are in the Arithmetic Progression.
We have ,
a = 7 ,d = 3 and last term= X
Let an = X
We know that
The general term = an = a+(n-1)d
=> X = 7+(n-1)(3)
=> X = 7+3n-3
=> X = 3n+4 -----------(1)
We know that
The sum of first n terms = Sn = (n/2)(a+an)
=> Sn = (n/2)(7+3n+4)
=> Sn = (n/2)(3n+11)
According to the given problem
The Sum of the given terms = 710
=> (n/2)(3n+11) = 710
=> (n)(3n+11) = 710×2
=> 3n²+11n = 1420
=>3n²+11n-1420 = 0
=> 3n²-60n+71n -1420 = 0
=>3n(n-20)+71(n-20) = 0
=> (n-20)(3n+71) = 0
=> n-20 = 0 or 3n+71 = 0
=> n = 20 or 3n = -71
=> n = 20 or n = -71/3
n can't be a negative number.
So, n = 20
On Substituting the value of n in (1) then
X = 3(20)+4
=> X = 60+4
=> X = 64
Therefore, X = 64
Answer:-
The value of X for the given problem is 64
Check:-
If X = 64 then the AP: 7+10+13+...+64
Sum of the terms = (20/2)(7+64)
=> 10×71
=> 710
Verified the given relations in the given problem.
Used formulae:-
- The general term = an = a+(n-1)d
- The sum of first n terms = Sn = (n/2)(a+an)
- a = First term
- d = Common difference
- n = number of terms
- an = general or last or nth term
Step-by-step explanation:
the value of x is 64 pls mark it brainly