If 7+10+13+....+x=710, then the value of x is?
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Answers
Step-by-step explanation:
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