the common difference of an arithmetic sequence is 6 and its one term 25 .Can the sum of any 15 terms in this sequence be 2028.Why?
Answers
Step-by-step explanation:
Sum of n terms in this A.P. is
S=n/2(2a+(n-1)d)=2018
15/2(2a+(15–1)6)=2018
2a=((2018×2)/15)–84
a=2776/30
Result will be in decimal which is not possible acc. to given data so it is not possible to have 2018 as the sum of any 15 terms for this sequence
Step-by-step explanation:
Given :-
The common difference of an arithmetic sequence is 6 and its one term 25.
To find :-
Can the sum of any 15 terms in this sequence be 2028.Why?
Solution :-
Given that
The common difference of an arithmetic sequence = (d) = 6
It's one term 25.
Sum of 15 terms = 2028
We know that
The sum of first n terms of an AP is
Sn=(n/2)[2a+(n-1)d]
Sum of 15 terms = S15
=> (15/2)[2a+(15-1)(6)] = 2028
=> (15/2)[2a+14(6)] = 2028
=> (15/2)[2a+84] = 2028
=> 2a+84 = 2028×2/15
=> 2a+84 = 4056/15
=> 2a+84 = 1352/5
=> 2a+84 = 270.4
=> 2a = 270.4-84
=> 2a = 190.4
=> a = 190.4/2
=> a =95.2
Here the first term is 95.2
It is a decimal number
It is not possible for the give dada
So 2028 is not in the given AP.
Answer :-
The sum of 15 terms can not be equal to 2028
Used formulae:-
The sum of first n terms of an AP is
Sn=(n/2)[2a+(n-1)d]