The sum of 'n' terms of an AP is given by Sn=(3n^2+5n). Which of its terms is 164?
Answers
Answer :
27th
Note :
★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.
★ If a1 , a2 , a3 , . . . , an are in AP , then
a2 - a1 = a3 - a2 = a4 - a3 = . . .
★ The common difference of an AP is given by ; d = a(n) - a(n-1) .
★ The nth term of an AP is given by ;
a(n) = a + (n - 1)d .
★ If a , b , c are in AP , then 2b = a + c .
★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .
or S(n) = (n/2)×(a + l) , l is the last term .
★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .
★ A linear polynomial in variable n always represents the nth term of an AP .
★ A quadratic polynomial in variable n always represents the sum of n terms of an AP .
Solution :
- Given : S(n) = 3n² + 5n
- To find : If a(n) = 164 , then n = ?
We have ,
S(n) = 3n² + 5n
Also ,
We know that , a(n) = S(n) - S(n - 1)
Thus ,
=> a(n) = (3n² + 5n) - [ 3(n - 1)² + 5(n - 1) ]
=> a(n) = 3n² + 5n - 3(n - 1)² - 5(n - 1)
=> a(n) = 3n² + 5n - 3(n² - 2n + 1) - 5(n - 1)
=> a(n) = 3n² + 5n - 3n² + 6n - 3 - 5n + 5
=> a(n) = 6n + 2
If a(n) = 164 , then
=> 6n + 2 = 164
=> 6n = 164 - 2
=> 6n = 162
=> n = 162/6
=> n = 27